PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS

Transcription

1 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS SHIGEKI AKIYAMA, HORST BRUNOTTE, ATTILA PETHŐ, AND WOLFGANG STEINER Abstract. We determine periodic and aperiodic points of certain piecewise affine maps in the Euclidean plane. Using these maps, we prove for λ { ±± 5, ±, ± 3} that all integer sequences (a k ) k Z satisfying 0 a k + λa k + a k+ < are periodic.. introduction In the past few decades, discontinuous piecewise affine maps have found considerable interest in the theory of dynamical systems. For an overview, we refer the reader to [, 7,, 3, 7, 8], for particular instances to [9, 6, 5] (polygonal dual billiards), [5] (polygonal exchange transformations), [0, 3,, 8] (digital filters) and [9,, ] (propagation of round-off errors in linear systems). The present paper deals with a conjecture on the periodicity of a certain kind of these maps: Conjecture.. [4, 7] For every real λ with λ <, all integer sequences (a k ) k Z satisfying (.) 0 a k + λa k + a k+ < for all k Z are periodic. This conjecture originated on the one hand from a discretization process in a rounding-off scheme occurring in computer simulation of dynamical systems (we refer the reader to [9, 7] and the literature quoted there), and on the other hand in the study of shift radix systems (see [4, ] for details). Extensive numerical evidence on the periodicity of integer sequences satisfying (.) was first observed in [6]. We summarize the situation of the Conjecture.. Since we have approximately ( ) ( ) ( ) ak 0 ak λ a k+ and the eigenvalues of the matrix are exp(±θπi) with θ [0, ], the sequence may be viewed as a discretized rotation on Z, and it is natural to parametrize λ = cos(θπ). There are five different classes of λ of apparently increasing difficulty: () θ is rational and λ is rational. () θ is rational and λ is quadratic. (3) θ is rational and λ is cubic or of higher degree. (4) θ is irrational and λ is rational. (5) None of the above. The first case consists of the three values λ =, 0,, where the conjecture is trivially true. Already in case () the problem is far from trivial. A computer assisted proof for λ = 5 was given by Lowenstein, Hatjispyros and Vivaldi [9]. A short proof (without use of computers) of the golden mean case λ = + 5 was given by the authors [3]. The main goal of this paper is to settle the conjecture for all the cases of (), i.e., the quadratic parameters a k λ = ± ± 5, ±, ± 3. Date: January 8, 008. Indeed, they showed that all trajectories of the map (x, y) ( ( λ)x y, x) on Z are periodic.

2 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER The proofs are sensitive to the choice of λ, and we have to work tirelessly in computation and drawings, especially in the last case ± 3. However, an important feature of our proof is that it can basically be checked by hand. The (easiest) case + 5 in Section gives a prototype of our discussion and should help the reader to understand the idea for the remaining values. For case (3), it is possible that Conjecture. can be proved using the same method, which involves a map on [0, ) d, where d denotes the degree of λ. However, it seems to be difficult in case d 3 to find self inducing structures, which are essential for this method. In [], a similar embedding into a higher dimensional torus is used for efficient orbit computations. Goetz [, 3, 4] found a piecewise π/7 rotation on an isosceles triangle in a cubic case having a self inducing structure, but we do not see a direct connection to our problem. The problem currently seems hopeless for cases (4) and (5). However, a nice observation on rational values of λ with prime-power denominator p n is exhibited in [9]. The authors show that the dynamical system given by (.) can be embedded into a p-adic rotation dynamics, by multiplying a p-adic unit. These investigations were extended in [30]. Furthermore, in [7] the case λ = q/p with p prime was related to the concept of minimal modules, the lattices of minimal complexity which support periodic orbits. Now we come back to the content of the present paper. The proof in [9] is based on a discontinuous non-ergodic piecewise affine map on the unit square, which dates back to Adler, Kitchens and Tresser []. Let λ = bλ + c with b, c Z. Set x = {λa k } and y = {λa k }, where {z} = z z denotes the fractional part of z. Then we have a k+ = a k λa k + y and {λa k+ } = { λa k λ a k + λy} = { x + (λ b)y} = { x + cy/λ} = { x λ y}, where λ is the algebraic conjugate of λ. Therefore we are interested in the map T : [0, ) [0, ) given by T (x, y) = (y, { x λ y}). Obviously, it suffices to study the periodicity of (T k (z)) k Z for points z = (x, y) (Z[λ] [0, )) in order to prove the conjecture. Using this map, Kouptsov, Lowenstein and Vivaldi [8] showed for all quadratic λ corresponding to rational rotations λ = ±± 5, ±, ± 3 that the trajectories of almost all points are periodic, by heavy use of computers. Of course, such metric results do not settle Conjecture., which deals with countably many points in [0, ), which may be exceptional. The main goal of this article is to show that no point with aperiodic trajectory has coordinates in Z[λ], which proves Conjecture. for these eight values of λ. This number theoretical problem is solved by introducing a map S, which is the composition of the first hitting map to the image of a suitably chosen self inducing domain under a (contracting) scaling map and the inverse of the scaling map. A crucial fact is that the inverse of the scaling constant is a Pisot unit in the quadratic number field Q(λ). This number theoretical argument greatly reduces the classification problem of periodic orbits, see e.g. Theorem.. All possible period lengths can be determined explicitly and one can even construct concrete aperiodic points in (Q(λ) [0, )). We can associate to each aperiodic orbit a kind of β-expansion with respect to the scaling constant. Note that the set of aperiodic points can be constructed similarly to a Cantor set, and that it is an open question of Mahler [3] whether there exist algebraic points in the triadic Cantor set. The paper is organized as follows. In Section, we reprove the conjecture for the simplest nontrivial case, i.e., where λ equals the golden mean. An exposition of our domain exchange method is given in Section 3, where the ideas of Section are extended to a general setting. In the subsequent seven sections we prove the conjecture for the cases λ = γ, ±/γ, ±, ± 3. Some parts of the proofs for λ = ± 3 are put into the Appendix. We conclude this paper by an observation relating the famous Thue-Morse sequence to the trajectory of points for λ = ±γ, ±/γ, 3.. The case λ = γ = + 5 = cos 4π 5 We consider first the golden mean λ = γ = + 5, λ = λ +. Note that T is given by ( ) 0 (.) T (x, y) = (x, y)a + (0, x y/γ ) with A =. /γ

3 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 3 D 0 R T(R) T T(D ) D T(D 0 ) Figure.. The piecewise affine map T and the set R, λ = γ = + 5. Therefore, we have T (x, y) = (x, y)a if y γx and T (z) = za + (0, ) for the other points z [0, ), see Figure.. A particular role is played by the set R = {(x, y) [0, ) : y < γx, x + y >, x < yγ} {(0, 0)}. If z R, z (0, 0), then we have T k+ (z) = T k (z)a + (0, ) for all k {0,,, 3, 4}, hence T 5 (z) = za 5 + (0, )(A 4 + A 3 + A + A + A 0 ) = z + (0, )(A 5 A 0 )(A A 0 ) = z since A 5 = A 0. It can be easily verified that the minimal period length is 5 for all z R except ( γ γ +, γ γ + ) and (0, 0), which are fixed points of T. Therefore, it is sufficient to consider the domain D = [0, ) \ R in the following. According to the action of T, we partition D into two sets D 0 and D, with D 0 = {(x, y) [0, ) : y γx} \ {(0, 0)}, In Figure., we scale D 0 and D by the factor /γ and follow their T -trajectory until the return to D/γ. Let P be the set of points in D which are not eventually mapped to D/γ, i.e., P = D α T (D α ) D β T (D β ) T (D β ), where D α is the closed pentagon {(x, y) D 0 : y /γ, x + y, y ( + x)/γ} and D β is the open pentagon R/γ \ {(0, 0)}. (In Figure., D α is split up into {T k (D α ) : k {0,, 4, 6, 8}}, and D β is split up into {T k (D β) : k {0, 3, 6, 9, }}.) All points in P are periodic (with minimal period lengths, 3, 0 or 5). Figures. and. show that the action of the first return map on D/γ is similar to the action of T on D, more precisely, { T (z) T (z/γ ) if z D 0, (.) γ = T 6 (z/γ ) if z D. For z D \ P, let s(z) = min{m 0 : T m (z) D/γ }. (Figure. shows s(z) 5.) By the map S : D \ P D, z γ T s(z) (z), we can completely characterize the periodic points. For z [0, ), denote by π(z) the minimal period length if (T k (z)) k Z is periodic and set π(z) = else. Theorem.. (T k (z)) k Z is periodic if and only if z R or S n (z) P for some n 0. We postpone the proof to Section 3, where the more general Proposition 3.3 and Theorem 3.4 are proved (with U(z) = z/γ, R(z) = z, ˆT (z) = T (z), ˆπ(z) = π(z), and z D or T (z) D for all z D, σ n (), see below). (.) and Figure. suggest to define a substitution (or morphism) σ on the alphabet A = {0, }, i.e., a map σ : A A (where A denotes the set of words with letters in A), by σ : in order to code the trajectory of the scaled domains until their return to D/γ : We have T k (D l /γ ) D σ(l)[k] and T σ(l) (z/γ ) = T (z)/γ for all z D l, where w[k] denotes the k-th

4 4 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER β 7 β β 3 T( D γ ) β 0 β 4 T 6 ( D γ ) T 4 ( D γ ) T( D0 γ ) T (D α ) T 8 (D α ) T 6 (D α ) T 4 (D α ) D α T 3 ( D γ ) T ( D γ ) β β D 0 γ β 6 D β β 9 β 3 T 7 (D α ) T 3 (D α ) β β 5 β 4 β 8 T(D α ) T 9 (D α ) D γ T 5 (D α ) T 5 ( D γ ) Figure.. The trajectory of the scaled domains and the (gray) set P, λ = γ. ( β k stands for T k (D β).) letter of the word w and w denotes its length. Furthermore, we have T k (D l /γ ) D/γ = for k < σ(l). Extend the definition of σ naturally to words in A by setting σ(vw) = σ(v)σ(w), where vw denotes the concatenation of v and w. Then we get the following lemma, which resembles Proposition by Poggiaspalla [4]. Lemma.. For every integer n 0 and every l {0, }, we have T σn (l) (z/γ n ) = T (z)/γ n for all z D l, T k (D l /γ n ) D σ n (l)[k] for all k, k σ n (l) T k (D l /γ n ) D/γ n = for all k, k < σ n (l). The proof is again postponed to Section 3, Lemma 3.. This lemma allows to determine the minimal period lengths: If z D α, then T σn (00000) (z/γ n ) = T σn (0000) (T (z)/γ n ) = = T 0 (z)/γ n = z/γ n for all n 0. The only points of the form T k (z/γ n ), k 5 σ n (0), which lie in D/γ n are the points T m (z)/γ n, m 9, which are all different from z/γ n if π(z) = 0. Therefore, we obtain π(z/γ n ) = 5 σ n (0) in this case. A point z lies in the trajectory of z/γ n if and only if S n ( z) = T m (z) for some m Z, see Lemma 3.. This implies π( z) = 5 σ n (0) for these z as well. The period lengths of all points are given by the following theorem.

5 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 5 Theorem.3. If λ = γ, then the minimal period lengths π(z) of (T k (z)) k Z are if z = (0, 0) or z = ( γ γ +, γ γ + ) 5 if z R \ {(0, 0), ( γ γ +, γ γ + )} (5 4 n + )/3 if S n (z) = T m ( /γ γ +, γ + ) for some n 0, m {0, } 5(5 4 n + )/3 if S n (z) T m( D α \ {( /γ γ +, γ + )}) for some n 0, m {0, } (0 4 n )/3 if S n (z) = T m ( γ +, γ + ) for some n 0, m {0,, } 5(0 4 n )/3 if S n (z) T m( D β \ {( γ +, γ + )}) for some n 0, m {0,, } if S n (z) D \ P for all n 0. The minimal period length of (a k ) k Z is π({γa k }, {γa k }) (which does not depend on k). Proof. By Theorem., Proposition 3.3 and the remarks preceding the theorem, it suffices to calculate σ n (0) and σ n (). Clearly, we have σ n (0) = for all n 0 and thus σ n () = σ n (00) = 4 σ n () + = 4(5 4 n )/3 + = (5 4 n )/3. If S n (z) T m (D α ), then π(z) = σ n (0) and π(z) = 5 σ n (0) respectively. If S n (z) T m (D β ), then π(z) = σ n (0) and π(z) = 5 σ n (0) respectively. Now consider aperiodic points z [0, ), i.e., S n (z) D \ P for all n 0. We can write S(z) = γ T s(z) (z) = γ ( za s(z) + t(z) ) for some t(z) by using (.). Note that T (z) = za for z D 0 and T (z) = za + (0, ) for z D. For z D/γ, we have s(z) = 0 and t(z) = 0. For z T k (D /γ ), k 5, we have s(z) = 6 k, (0, ) if s(z) {, }, t(z) = (0, )A + (0, ) = (/γ, /γ ) if s(z) = 3, (0, )A 3 + (0, )A + (0, ) = (0, /γ) if s(z) {4, 5}. We obtain inductively n S n (z) = γ n za s(z)+s(s(z))+...+s(sn (z)) + γ (n k) t(s k (z))a s(sk+ (z))+ +s(s n (z)). k=0 If z Q(γ), then we have ( ) za s(z)+s(s(z))+ +s(sn (z)) (S n (z)) = γ n + n k=0 (S n (z)) max h Z (za h ) n γ n + k=0 ( ) t(s k (z))a s(sk+ (z))+ +s(s n (z)) γ (n k) max h Z,w D\P (t(w)a h ) γ n k, where z = (x, y ) if z = (x, y) and x, y are the algebraic conjugates of x, y. Since t(z)a h { (0, 0), (0, ), (, /γ), (/γ, /γ), ( /γ, ), (, 0), (/γ, /γ ), (/γ, /γ ), ( /γ, /γ), ( /γ, 0), (0, /γ), (0, /γ), ( /γ, /γ ), ( /γ, /γ ), (/γ, /γ), (/γ, 0) } and za h takes only the values z, za, za, za 3 and za 4, we obtain (S n (z)) max h Z (za h ) n γ n + k=0 γ C(z) < γ(n k) γ n + γ for some constant C(z). If z ( Q Z[γ]) for some integer Q, then S n (z) ( Q Z[γ]). Since there exist only finitely many points w ( Q Z[γ] [0, )) with w < C(z) + γ, we must have (S n (z)) γ for some n 0, which proves the following proposition. Proposition.4. Let z ( Q Z[γ] [0, )) be an aperiodic point. Then there exists an aperiodic point z ( Q Z[γ]) D with z γ.

6 6 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER Figure.3. Aperiodic points, λ = γ. Figure.4. Aperiodic points, λ = /γ. For every denominator Q, it is therefore sufficient to check the periodicity of the (finite set of) points z ( Q Z[γ]) D with z γ in order to determine if all points in ( QZ[γ] [0, )) are periodic. For Q =, we have to consider z = (x, y) D with x, y Z[γ] [0, ) and x, y γ, hence (x, y) {0, /γ}. Since (0, 0) and (/γ, /γ) are in R, it only remains to check the periodicity of (0, /γ) and (/γ, 0). These two points lie in P, thus Conjecture. is proved for λ = γ. For Q =, a short inspection shows that all points z ( Z[γ] [0, )) are periodic as well. The situation is completely different for Q = 3, and we have S(0, /3) = (0, γ /3), S(0, γ /3) = γ ( (0, γ /3)A 5 + (0, /γ) ) = (0, /3), S(0, /3) = γ ( (0, /3)A 5 + (0, /γ) ) = ( 0, /(3γ ) ), S 4 (0, /3) = S ( 0, /(3γ ) ) = (0, /3). This implies S n (0, /3) D \ P for all n 0 and π(0, /3) = by Theorem.3. Theorem.5. π(z) is finite for all points z (Z[γ] [0, )), but (T k (0, /3)) k Z is aperiodic. 3. General description of the method In this section, we generalize the method presented in Section in order to make it applicable for λ = γ, ±/γ, ±, ± 3. For the moment, we only need that T : X X is a bijective map on a set X. Fix D X, let R = {z X : T m (z) D for all m 0} set r(z) = min{m 0 : T m (z) D} for z X \ R, and R : X \ R D, R(z) = T r(z) (z). Let ˆT be the first return map (of the iterates by T ) on D, i.e., ˆT : D D, ˆT (z) = RT (z) = T r(t (z))+ (z), in particular ˆT (z) = T (z) if T (z) D. Let A be a finite set, {D l : l A} a partition of D and define a coding map ι : D A Z by ι(z) = (ι k (z)) k Z such that ˆT k (z) D ιk (z) for all k Z. Let U : D D, ε {, } and σ a substitution on A such that, for every l A and z D l, U ˆT (z) = ˆT ε σ(l) U(z),

7 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 7 ˆT εk U(z) U(D) for all k, k < σ(l), and { ι0 (U(z)) ι σ(l) = (U(z)) ι σ(l) (U(z)) if ε =, ι σ(l) (U(z)) ι (U(z)) ι (U(z)) if ε =. Then the following lemma holds. Lemma 3.. For every integer n 0, every l A and z D l, we have U n ˆT (z) = ˆT ε n σ n (l) U n (z), ˆT εnk U n (z) U n (D) for all k, k < σ n (l), and ι 0 (U n (z)) ι (U n (z)) ι σn (l) (U n (z)) = σ n (l) if ε =, ι 0 (U n (z)) ι (U n (z)) ι σn (l) (U n (z)) = (σ σ) n/ (l) if ε =, ε n =, ι σn (l) (U n (z)) ι (U n (z)) ι (U n (z)) = (σ σ) (n )/ σ(l) if ε =, ε n =, where σ(l) = l m l l if σ(l) = l l l m. Proof. The lemma is trivially true for n = 0, and for n = by the assumptions on σ. If we suppose inductively that it is true for n, then let σ(l) = l l l m if ε =, σ(l) = l m l l if ε =, and we obtain (by another induction) for all j, j m, (3.) ˆT ε n σ n (l l j l j ) U n (z) = ˆT εn σ n (l j ) U n ˆT ε(j ) U(z) = U n ˆT εj U(z). If ε =, then this follows immediately from the induction hypothesis; if ε =, then this follows by setting k = σ n (l j ) in (3.) ˆT ( ) nk U n ˆT ( ˆT j U(z) ) = ˆT ( )n (k σ n (l j) ) U n ˆT j U(z). Therefore, we have ˆT εn σ n (l) U n (z) = ˆT εn σ n (l l m l m ) U n (z) = U n ˆT εm U(z) = U n ˆT ε σ(l) U(z) = U n ˆT (z). If ε =, then (3.) implies that ι 0 (U n (z)) ι σn (l) (U n (z)) = ( ι 0 (U n U(z)) ι σ n (l ) (U n U(z)) ) ( ι0 (U n ˆT m U(z)) ι σ n (l m) (U n ˆT m U(z)) ) = σ n (l ) σ n (l m ) = σ n (l); if ε = and ε n =, then (3.) and (3.) provide ι 0 (U n (z)) ι σn (l) (U n (z)) = ( ι σ n (l ) (U n T U(z)) ι (U n T U(z)) ) (ι σ n (l m) (U n ˆT m U(z)) ι (U n ˆT m U(z)) ) = (σ σ) (n )/ σ(l ) (σ σ) (n )/ σ(l m ) = (σ σ) n/ (l); if ε = and ε n =, then ι σn (l) (U n (z)) ι (U n (z)) = ( ι 0 (U n T m U(z)) ι σ n (l m ) (U n T m U(z)) ) (ι 0 (U n ˆT U(z)) ι σ n (l ) (U n ˆT U(z)) ) = (σ σ) (n )/ (l m ) (σ σ) (n )/ (l ) = (σ σ) (n )/ σ(l). By (3.), (3.) and the induction hypothesis, the only points in ( ˆT εnk U n (z)) k< σn (l) lying in U n (D) are U n ˆT εj (z), j < σ(l). Since ˆT εj (z) U(D) for these j, the lemma is proved. Remark. If z = ˆT (z) D l, then U n ˆT ( z) = ˆT ε n σ n (l) U n ( z), thus U n ˆT (z) = T εn σ n (l) U n (z). As in Section, a key role will be played by the map S. Assume that U is injective, let P = {z D : ˆT m (z) U(D) for all m Z}, fix ŝ(z) = min{m 0 : ˆT m (z) U(D)} or ŝ(z) = max{m 0 : ˆT m (z) U(D)} for every z D \ P, let s(z) Z be such that ˆT ŝ(z) (z) = T s(z) (z), and define S : D \ P D, z U ˆT ŝ(z) (z) = U T s(z) (z). Remark. Allowing s(z) and ŝ(z) to be negative decreases the δ in Proposition 3.5 in some cases.

8 8 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER Lemma 3.. If S n R(z) exists, then we have some m 0 such that U n S n R(z) = T m (z), and z = T m (z) for some m Z if and only if S n R( z) = ˆT k S n R(z) for some k Z. Proof. Suppose that S n R(z) exists. Then we have U n S n R(z) = U n ˆT ŝ(s n R(z)) S n R(z) = ˆT m U n S n R(z) = = ˆT m + +m n R(z) = T m (z) for some m,..., m n, m 0. If S n R( z) = ˆT k S n R(z) for some k Z, then let m, m 0 be such that U n S n R(z) = T m (z), U n S n R( z) = T m ( z), and we have T m ( z) = U n S n R( z) = U n ˆT k S n R(z) = ˆT k U n S n R(z) = T k+m (z) for some k, k Z, hence z = T m (z) with m = k + m m. If z = T m (z) for some m Z and n = 0, then we have S n R( z) = ˆT k n S n R(z) for some k n Z. If we suppose inductively that this is true for n, then S n R( z) = S ˆT kn S n R(z) = S ˆT kn ŝ(sn R(z)) US n R(z) = SU ˆT kn S n R(z) = ˆT kn S n R(z) for some k n, k n Z, and the statement is proved. If rt is constant on every D l, l A, then we can define τ : A N by τ(l) = r(t (z)) + for z D l (cf. the definition of ˆT ) and extend τ naturally to words w A by τ(w) = l A w lτ(l). Let π(z), ˆπ(z) be the minimal period lengths of (T k (z)) k Z and ( ˆT k (z)) k Z respectively, with π(z) =, ˆπ(z) = if the sequences are aperiodic. Then the following proposition holds. Proposition 3.3. If ˆπ(S n R(z)) = p and l l p = ι 0 (S n R(z)) ι p (S n R(z)), then we have ˆπ(R(z)) = σ n (l l l p ) and π(z) = τ(σ n (l l l p )) (if τ is well defined). Proof. Since U n S n R(z) = T m (z) = ˆT ˆm R(z) for some m, ˆm Z, and T τ(σn (l l l p )) U n S n R(z) = ˆT σn (l l l p ) U n S n R(z) = U n ˆT p S n R(z) = U n S n R(z), we have ˆπ(R(z)) σ n (l l p ) and π(z) τ(σ n (l l p )) (if τ exists). Since p is minimal, we can show similarly to the proof of Lemma 3. that these period lengths are minimal. We obtain the following characterization of periodic points z R. Note that all points in P R are periodic in our cases, hence the characterization is complete. Theorem 3.4. Let R, S, T, D, P, R, σ be as in the preceding paragraphs of this section. Assume that ˆπ(z) is finite for all z P, and that for every z D \ P there exist m Z, l A, such that ˆT m (z) D l and σ n (l) for n. Then we have for z R: (T k (z)) k Z is periodic if and only if S n R(z) P for some n 0. Proof. If S n R(z) P, then we have ˆπ(R(z)) = ˆπ(S n R(z)) <, which implies π(z) <. Suppose now that S n R(z) D \ P for all n 0. Then we have m n Z and l n A such that ˆT m n S n R(z) D ln and σ n (l n ) for n (because A is finite). We have U n m ˆT n S n R(z) = ˆT m n U n S n R(z) U n (D ln ) for some m n Z, hence ˆT m n+k U n S n R(z) U n (D) for all k, k < σ n (l n ), which implies π(z) ˆπ(R(z)) = ˆπ(U n S n R(z)) σ n (l n ) for all n 0, thus π(z) =. Assume now λ {±, ±± 5, ± 3}, let λ be its algebraic conjugate, T : [0, ) [0, ), ( ) T (x, y) = (x, y)a + (0, x + λ 0 (3.3) y ) with A = λ, U(z) = V (κv (z)) with 0 < κ <, κ Z[λ], κκ =, and V (z) = ±κ n (z v) some v Z[λ], n Z. Let t(z) = V ( T s(z) (z) ) V (z)a s(z)

9 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 9 for z D \ P. Since U (z) = V (V (z)/κ), we have Note that A h = A 0 for some h {5, 8, 0, }, S(z) = U T s(z) (z) = V ( V (z)a s(z) + t(z) κ T (x, y) = (x, y)a + ( λ x + y, 0) with A = ). ( ) λ, 0 and T (x, y) = ( x, ỹ) with (ỹ, x) = T (y, x). Since ŝ(z) < max l A σ(l), there exists only a finite number of values for t(z), and we obtain the following proposition. Proposition 3.5. Let T, V, κ be as above and the assumptions of Theorem 3.4 be satisfied. Suppose that π(z) = for some z ( Q Z[λ] [0, )) \ R, where Q is a positive integer. Then there exists an aperiodic point z ( Q Z[λ]) D with V ( z) δ, where δ = max{ (t(z)ah ) : z D \ P, π(z) =, h Z} κ Proof. First note that δ exists since t(z) and A h take only finitely many values. If π(z) = for some z ( Q Z[λ] [0, )) \ R, then S n R(z) D \ P for all n 0 by Theorem 3.4. In particular, S n R(z) is aperiodic as well. We use the abbreviations s n = s(s n R(z)) and t n = t(s n R(z)). Then we obtain inductively, for n, V S n R(z) = V Sn R(z)A sn + t n κ = V R(z)As0+s+ +sn n κ n + If we look at the algebraic conjugates, then note that κ >, and we obtain ( ) V R(z)A s0+s+ +sn (V S n R(z)) < κ n + δ, k=0 t k A s k++ +s n κ n k. thus (V S n R(z)) δ for some n 0 (as in Section ), and we can choose z = S n R(z). Remarks. The last proof shows that, for every z (Q(λ) [0, )) \ R with π(z) =, there are only finitely many possibilities for V S n R(z), hence (S n R(z)) n 0 is eventually periodic. For every z D with π(z) =, we have V (z) = ( n V S n (z)κ n t k A s k++ +s n κ k) A s0 sn k=0 = k=0. t k A P k j=0 s(sj (z)) κ k, which is a κ-expansion (κ < ) of V (z) with (two-dimensional) digits t k A s 0 s s k. As a consequence of Lemma 3. and the definition of U, for every aperiodic point z [0, ) \ R and every c > 0, there exists some m Z such that T m (z) v < c. In all our cases, we have ε = κκ. 4. The case λ = /γ = 5 = cos π 5 Now we apply the method of Section 3 for λ = /γ, i.e., λ = γ. To this end, set D = {(x, y) [0, ) : x + y 3 γ} = D 0 D with D 0 = {(x, y) D : x + γy > }, D = {(x, y) D : x + γy }. Figure 4. shows that ˆT is given by ˆT (z) = T τ(l) (z) if z D l, l A = {0, }, with τ(0) = and τ() = 4. The set which is left out by the iterates of D 0 and D is R = {(0, 0)} D A D B, with D A = {z [0, ) : T k+ (z) = T k (z)a + (0, ) for all k 0}, D B = {z [0, ) : T k+ (z) = T k (z)a + (0, ) for all k 0}.

10 0 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER T 3 (D ) D 0 D ˆT T 4 (D ) T(D 0 ) D B D A T (D ) T(D ) Figure 4.. The map ˆT, ˆT (D 0 ) = T (D 0 ), ˆT (D ) = T 4 (D ), and the (gray) set R, λ = /γ D β D α 3 0 Figure 4.. The trajectory of the scaled domains and P, λ = /γ. (l k stands for ˆT k U(D l ).) As in Section, we have T 5 (z) = z for all z R. If we set U(z) = z ( γ + γ, ) = (, ) γ V (z) = (, ) z, κ = /γ, ε =, and σ : (, ) z γ, then Figure 4. shows that σ satisfies the conditions in Section 3, and P = D α D β with D α = U(D A ), D β = U(D B ). All points in P are periodic and σ n (l) as n for all l A. Therefore, all conditions of Proposition 3.3 and Theorem 3.4 are satisfied, and we obtain the following theorem. Theorem 4.. If λ = /γ, then the period lengths π(z) are if z {(0, 0), ( γ +, γ + ), ( γ +, γ + )} 5 for the other points of the pentagons D A and D B (5 4 n + )/3 if S n R(z) = ( γ γ +, γ γ + ) for some n 0 0(5 4 n + )/3 for the other points with S n R(z) D α for some n 0 (5 4 n )/3 if S n 3 R(z) = ( γ +, 3 γ + ) for some n 0 5(5 4 n )/3 for the other points with S n R(z) D β for some n 0 if S n R(z) D \ P for all n 0.

11 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS Proof. We easily calculate ( ) ( ) σ n (0) 0 σ n = 4 n /3 (0) /3 ( ) /3 +, /3 ( ) σ n () 0 σ n () ( ) = 4 n /3 + /3 ( ) /3, /3 hence τ(σ n (0)) = 5 3 4n 3, τ(σn ()) = 0 3 4n + 3. If Sn R(z) D α, then π(z) = τ(σ n ()) and π(z) = τ(σ n ()) respectively; if S n R(z) D β, then π(z) = τ(σ n (0)) and π(z) = 5τ(σ n (0)) respectively. For z U(D), we have ŝ(z) = s(z) = 0 and t(z) = (0, 0). For the other z D \ P, we choose ŝ(z) as follows and obtain the following s(z), t(z): z ˆT U(D 0 ) ˆT U(D ) : ŝ(z) =, s(z) = 5, t(z) = V ( ˆT (z)) V (z) = ( /γ, 0) z ˆT U(D ) : ŝ(z) =, s(z) =, t(z) = V ( ˆT (z)) V (z)a = (/γ, 0) z ˆT 4 U(D ) : ŝ(z) =, s(z) =, t(z) = V ( ˆT (z)) V (z)a = (0, /γ) z ˆT U(D 0 ) ˆT 3 U(D ) : ŝ(z) =, s(z) = 5, t(z) = V ( ˆT (z)) V (z) = (0, /γ ) Observe the symmetry between positive and negative ŝ(z) which is due to the symmetry of T (x, y) and T (y, x) and the symmetry of D. With {(/γ, 0)A h : h Z} = {(/γ, 0), (0, /γ), ( /γ, ), (, ), (, /γ)}, we obtain δ max{ (t(z)a h ) : z D \ P, h Z}/γ = (/γ ) /γ = γ, as in Section. The following theorem shows that aperiodic points with t(z) = ( /γ, 0) exist, hence δ = γ. Theorem 4.. π(z) is finite for all z (Z[γ] [0, )), but π ( /(3γ), /(3γ) ) =. Proof. By Proposition 3.5, we have to show that all z Z[γ] D with V (z) γ are periodic. Since V (D) = {(x, y) : x > 0, y > 0, x + y /γ}, we have to consider x, y Z[γ] (0, /γ) with x, y γ. No such x, y exist, hence the conjecture is proved for λ = /γ. Note that π(z) is finite for all z ( Z[γ] [0, )) as well. If V (z) = ( /(3γ), /(3γ) ), then we have V S(z) = γ ( V (z)a 5 + (0, /γ ) ) = ( γ/3, /(3γ 3 ) ) V S (z) = γ ( V S(z)A 5 + ( /γ, 0) ) = ( /(3γ), /(3γ) ) V S 3 (z) = γ ( V S (z)a 5 + (0, /γ ) ) = ( /(3γ 3 ), γ/3) ) V S 4 (z) = γ ( V S 3 (z)a 5 + (0, /γ ) ) = ( /(3γ), /(3γ) ) = V (z), hence S n (z) D \ P for all n 0 and π(z) = by Theorem 4.. Let λ = (λ = ) and set 5. The case λ = = cos 3π 4 D = {(x, y) [0, ) : < x y < 0, 0 < x y < } = l A={0,,,3} D l, D 0 = {(x, y) D : x < }, D = {(x, y) D : x >, y }, D = {(x, y) D : x >, y > }, D 3 = {(x, y) D : x = }. Figure 5. shows that ˆT (z) = T τ(l) (z) if z D l, with τ(0) = 5, τ() = 9, τ() = 3, τ(3) =, and R = {(0, 0)} 3 k=0 T k (D A ) 5 k=0 T k (D B ) with D A = {(0, y) : / < y < / }, D B = {(0, / )}. If we set U(z) = ( )z, V (z) = z, κ =, ε =, and σ : , then Figure 5. shows that σ satisfies the conditions in Section 3, and P = {(x, y) D : x, y } = D α D β ˆT (D β ) D ζ with D α = D, D β = {(x, ) : < x < } and D ζ = {(, )}. All points in P are periodic and σ n (l) as n for all l A. Therefore, all conditions of Proposition 3.3 and Theorem 3.4 are satisfied, and we obtain the following theorem.

12 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER B 0 A A 0 B B A ˆT B 5 A 3 B 3 3 Figure 5.. The map ˆT and the set R, λ =. (l k stands for T k (D l ).) D α ˆT(D β ) D ζ D β Figure 5.. The trajectory of the scaled domains and P, λ =. (l k stands for ˆT k U(D l ).)

13 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 3 Theorem 5.. If λ =, then the minimal period length π(z) is if z = (0, 0) 4 if z = T m (0, /), 0 m 3 8 for the other points of T m (D A ), 0 m 3 6 if z = T m (0, / ), 0 m 5 3 n + ( ) n if S n R(z) = (/, / ), n 0 8( 3 n + ( ) n ) for the other points with S n R(z) D α 4(3 n+ + + ( ) n ) if S n R(z) {(/, ), (, /)}, n 0 8(3 n+ + + ( ) n ) for the other points with S n R(z) D β ˆT (D β ) 3 n ( ) n if S n R(z) = (, ), n 0 if S n R(z) D \ P for all n 0. Proof. We easily calculate ( ) ( ) σ n (0) 0 σ n = 3 n 3/4 (0) /4 ( ) + ( ) n /4, /4 ( ) ( ) ( ) σ n () 0 σ n = 3 n 3/4 + ( ) n 3/4 () /4 3/4 and obtain τ(σ n (0)) = 3 n+ ( ) n, τ(σ n (3)) = τ(σ n (030)) = 3 n ( ) n. If S n R(z) D α and n, then π(z) = τ(σ n ()) = τ(σ n (0)) and π(z) = 8τ(σ n (0)) respectively; if S n R(z) D β, then π(z) = τ(σ n (3)) = τ(σ n (000030)) and π(z) = τ(σ n (000030)) respectively; if S n R(z) = (, ), then π(z) = τ(σ n (3)). The given π(z) hold for n = 0 as well. For z D \ (U(D) P), we choose ŝ(z) as follows and obtain the following s(z), t(z): z ˆT U(D 0 D D 3 ) : ŝ(z) =, s(z) = 5, t(z) = ˆT (z) za 5 = (, ) z ˆT U(D 0 D D 3 ) : ŝ(z) =, s(z) = 5, t(z) = ˆT (z) za 5 = (, ) This gives δ = ( + )/ = + since {t(z)a h : z D \P, h Z} = ±{(0, 0), (, ), (, 0), (0, ), (, )}. Theorem 5.. π(z) is finite for all z (Z[ ] [0, )), but (T k ( 3 4, 4 )) k Z is aperiodic. Proof. We have to consider z Z[ ] D with z δ = +. The only such point is (, ) = D ζ, hence Conjecture. holds for λ =. It can be shown that all points in ( Z[ ] [0, )) and ( 3 Z[ ] [0, )) are periodic as well. For z = ( 3 4, 4 ), we have S(z) = ( za 5 + (, ) ) /κ = ( ( 9 6 ) ( ) =, 3 ),, S (z) = ( S(z)A 5 + (, ) ) /κ = ( ( ), 5 ) ( = , 3 4 S 3 (z) = ( S (z)a 5 + (, ) ) /κ = ( 3 4, ) and S 4 (z) = ( 3 4, 4 ) = z. ), Let λ = (λ = ) and set 6. The case λ = = cos π 4 D = {(x, y) [0, ) : x + y > or x + y > } = l A={0,,} D l, with D 0 = {(x, y) D : x + y > } and D = {(x, y) D : x + y < }. Figure 6. shows that ˆT (z) = T τ(l) (z) if z D l, with τ(0) =, τ() =, τ() = 3, and R = {(0, 0)} D A D B 3 T k (D Γ ) 9 T k (D ), k=0 k=0 D A = {(x, y) : 0 x, y 3 } \ {(0, 0), (3, 3 )}, D B = {z [0, ) : T k+ (z) = T k (z)a + (0, ) for all k Z}, D Γ = {z [0, ) : T k+ (z) = T k (z)a + (0, ) for all k Z},

14 4 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER Figure 5.3. Aperiodic points, λ =. Figure 5.4. Aperiodic points, λ =. D = {(/, 0)}. Set κ =, V (z) = ((, ) z)/κ = ( + )((, ) z), i.e., U(z) = (, ) ( ) ( (, ) z ) = ( )z + (, ). Then Figure 6. shows that the conditions in Section 3 are satisfied by σ : with ε = and P = D α 5 k=0 ˆT k (D β ) k=0 ˆT k (D ζ ) with D α = {z [0, ) : T k+ (z) = T k (z)a + (0, 3) for all k Z}, D β = {(x, x) : 5 3 < x < } and D ζ = {(8 5, 8 5 )}. All points in P are periodic and σ n (l) as n for all l A. Therefore, all conditions of Proposition 3.3 and Theorem 3.4 are satisfied, and we obtain the following theorem. Theorem 6.. If λ =, then the minimal period length π(z) is if z {(0, 0), (/, / ), (, )} 4 if z = T m (3/, 3/ ) for some m {0,,, 3} 0 if z = T m (/, 0) for some m {0,,..., 9} 8 for the other points in R 3 n+ 5( ) n if S n R(z) = (3 3/, 3 3/ ) for some n 0 8( 3 n+ 5( ) n ) for the other points with S n R(z) D α 4(3 n ( ) n ) if S n R(z) = ˆT m( (9 5 )/, 5 3 ) for some m {0,..., 5}, n 0 8(3 n ( ) n ) for the other points with S n R(z) ˆT m (D β ) 3 n ( ) n if S n R(z) = ˆT m (8 5, 8 5 ) for some m {0,, }, n 0 if S n R(z) D \ P for all n 0. Proof. As for λ =, we have ( ) ( ) ( ) σ n (0) 0 σ n = 3 n 3/4 + ( ) n /4, (0) /4 /4 ( ) σ n () 0 σ n () ( ) ( ) = 3 n 3/4 + ( ) n 3/4, /4 3/4 hence τ(σ n (0)) = 3 n+ 5( ) n and τ(σ n ()) = τ(σ n (00)) = 3 n ( ) n. For S n R(z) D α, we have π(z) = τ(σ n (0)) and π(z) = 8τ(σ n (0)) respectively; if S n R(z) T m (D β ), then π(z) = τ(σ n (00000)) and π(z) = τ(σ n (00000)) respectively; if S n R(z) = ˆT m (D ζ ), then π(z) = τ(σ n (00)).

15 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS T(D Γ ) T (D Γ ) ˆT D B D A D Γ T 3 (D Γ ) Figure 6.. The map ˆT and the set R, λ =. (l k stands for T k (D l ).) 0 ˆT (D ζ ) ˆT 3 (D β ) ˆT(D β ) D β D α 0 0 D ζ 0 ˆT 4 (D β ) ˆT 5 (D β ) ˆT (D β ) ˆT(D ζ ) 0 Figure 6.. The trajectory of the scaled domains and P, λ =. (l k stands for ˆT k U(D l ).) For z D \ (U(D) P), we choose ŝ(z) as follows and obtain the following s(z), t(z): z ˆT U(D 0 D D ) : ŝ(z) =, s(z) =, t(z) = V ( ˆT (z)) V (z)a = (, 0) z ˆT U(D 0 D D ) : ŝ(z) =, s(z) =, t(z) = V ( ˆT (z)) V (z)a = (0, )

16 6 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER ˆT Figure 7.. The map ˆT, λ = /γ. (l k stands for T k (D l ).) This gives δ = / = since {t(z)a h : z D \ P, h Z} = ±{(0, 0), (, 0), (0, ), (, ), (, )}. Theorem 6.. π(z) is finite for all z (Z[ ] [0, )), but (T k ( 3 4, 5 4 )) k Z is aperiodic. Proof. Since V (D) = {(x, y) : x > 0, y > 0, x + y < or x + y < }, there exists no z Z[ ] D with (V (z)). Therefore Conjecture. holds for λ =. It can be shown that all points in ( Z[ ] [0, )) and ( 3 Z[ ] [0, )) are periodic as well. For z = ( 3 4, 5 4 ), we have V (z) = ( + 4, 4 ), V S(z) = ( + )(V (z)a + (0, )) = ( ( + ) 4, 3 ) ( + ) =,, V S (z) = ( 4, + 4 ), V S 3 (z) = ( 4, + 4 ) and V S 4 (z) = ( + 4, 4 ) = V (z). Let λ = /γ (λ = γ) and set 7. The case λ = /γ = cos 3π 5 D = {(x, y) [0, ) : γx < y < x/γ} = l A={0,,,3} D l, with D 0, D, D, D 3 satisfying the (in)equalities D 0 D D D 3 y > x /γ 0 < y < x /γ y = x /γ y = 0, /γ < x < /γ Figure 7. shows that ˆT (z) = T τ(l) (z) if z D l, with τ(0) = 6, τ() = 4, τ() = 7, τ(3) = 5, and R = {(0, 0)}. If we set U(z) = z/γ, V (z) = z, κ = /γ, ε =, and σ : then Figure 7. shows that σ satisfies the conditions in Section 3, and P = D α D β 3 k=0 ˆT k (D ζ ) D ϑ k=0 ˆT k (D η ) D µ with D α = {z D : ˆT k (z) D 0 for all k Z}, D β = {z D : ˆT k (z) D for all k Z}, D ζ = {(x, 0) : /γ 3 < x < /γ }, D η = D 3, D ϑ = {(/γ 3, 0)} and D µ = {(/γ, 0)}. All points in P are periodic and σ n (l) as n for all l A. Therefore, all conditions of Proposition 3.3 and Theorem 3.4 are satisfied, and we obtain the following theorem.

17 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS ˆT(D ζ ) ˆT(D ϑ ) D β D α ˆT 0 3 (D ζ ) ˆT (D ζ ) D ϑ D ζ D µ D η Figure 7.. The trajectory of the scaled domains and P, λ = /γ. (l k stands for ˆT k U(D l ).) Theorem 7.. If λ = /γ, then the minimal period length π(z) is if z = (0, 0) (5 4 n + 4)/3 if S n R(z) = ( γ γ +, /γ γ +) for some n 0 0(5 4 n + 4)/3 for the other points with S n R(z) D α 4(5 4 n )/3 if S n R(z) = ( γ γ +, γ +) for some n 0 0(5 4 n )/3 for the other points with S n R(z) D β 5(4 n+ )/3 if S n R(z) = (0, /) for some n 0 0(4 n+ )/3 for the other points with S n R(z) D ϑ 5( 4 n+ + 7)/3 if S n R(z) = ˆT m (/(γ), 0) for some m {0,,, 3} and n 0 0( 4 n+ + 7)/3 for the other points with S n R(z) ˆT m (D ζ ) (0 4 n + )/3 if S n R(z) = (/γ, 0) for some n 0 (5 4 n+ + 9)/3 if S n R(z) = ˆT m (/γ 3, 0) for some m {0, } and n 0 if S n R(z) D \ P for all n 0. Proof. As for λ = /γ, we have ( ) ( ) σ n (0) 0 σ n = 4 n /3 (0) /3 ( ) /3 +, /3 ( ) σ n () 0 σ n () ( ) = 4 n /3 + /3 ( ) /3, /3 hence τ(σ n (0)) = 0 3 4n + 8 3, τ(σn ()) = 0 3 4n 8 3, τ(σn ()) = 0 3 4n + 3, τ(σn (3)) = 0 3 4n 5 3. For S n R(z) D α, we have π(z) = τ(σ n (0)) and π(z) = 5τ(σ n (0)) respectively; if S n R(z) D β, then π(z) = τ(σ n ()) and 5τ(σ n ()) respectively; if S n R(z) D η, then π(z) = τ(σ n (3)) and τ(σ n (3)) respectively; if S n R(z) D ζ, then π(z) = τ(σ n (000)) and τ(σ n (000)) respectively; if S n R(z) = ˆT m (/γ 3, 0), then π(z) = τ(σ n (0)); if S n R(z) = (/γ, 0), then π(z) = τ(σ n ()). Note that ˆT m U(D 3 ) plays no role in the calculation of δ since U(D 3 ) U(P) and thus π(z) < for all z ˆT m U(D 3 ). For the other z D \ (P U(D)), we choose ŝ(z) as follows: z ˆT U(D 0 D D ) : ŝ(z) =, s(z) = 0, t(z) = ˆT (z) z = ( /γ, /γ ) z ˆT U(D D ) : ŝ(z) =, s(z) = 6, t(z) = ˆT (z) + za = (, /γ) z ˆT 4 U(D ) : ŝ(z) =, s(z) = 6, t(z) = ˆT (z) + za = (/γ, 0) z ˆT U(D 0 ) ˆT 3 U(D ) : ŝ(z) =, s(z) = 0, t(z) = ˆT (z) z = ( /γ, 0)

18 8 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER Figure 7.3. Aperiodic points, λ = /γ. Figure 7.4. Aperiodic points, λ = γ. This gives again δ = γ /γ = γ since {(/γ, 0)A h : h Z} = ±{(/γ, 0), (0, /γ), (/γ, ), (, ), (, /γ)}. Theorem 7.. π(z) is finite for all z (Z[γ] [0, )), but ( T k( /4, /(4γ 3 ) )) k Z is aperiodic. Proof. Conjecture. holds for λ = /γ since no z Z[γ] D satisfies z γ. It can be shown that all points in ( Z[γ] [0, )) and ( 3 Z[γ] [0, )) are periodic as well. If z = ( /4, /(4γ 3 ) ), then we have S(z) = ( γ /4, /(4γ) ), S (z) = γ ( S(z) (/γ, 0) ) = ( (3γ )/4, γ/4 ) and S 3 (z) = γ ( S (z) (/γ, /γ ) ) = ( /4, /(4γ 3 ) ) = z. Let λ = γ (λ = /γ) and set 8. The case λ = γ = cos π 5 D = {(x, y) [0, ) : x < y, γx + y 4 γ} = D 0 D with D 0 = {(x, y) D : x > /γ 5 }, D = {(x, y) D : x /γ 5 }. Figure 8. shows that ˆT (z) = T τ(l) (z) if z D l, with τ(0) = 4, τ() = 8, and R = {(0, 0)} D A D B 4 k=0 T k (D Γ ) k=0 T k (D ) 4 k=0 T k (D E ) 0 k=0 T k (D Z ) with D A = {z : T k+ (z) = T k (z)a + (0, ) for all k Z}, D B = {z : T k+ (z) = T k (z)a + (0, )}, D = {z [0, ) : T k+ (z) = T k (z)a + (0, ), T k (z) = T k (z)a + (0, ) for all k Z}, D Γ = {(x, y) : 0 x, y /γ 4 } \ {(0, 0), (/γ 4, /γ 4 )}, D E = {(x, x) : /γ 5 < x < }, D Z = {( /γ 5, /γ 5 )}. Set κ = /γ, V (z) = γ 4( (, ) z), i.e. U(z) = (, ) ( (, ) z ) /γ = z/γ + (/γ, /γ). Then Figure 8. shows that the conditions in Section 3 are satisfied by ε = and σ : All points in P = D α D β are periodic, with D α = {z D : ˆT k (z) D 0 for all k Z}, D β = {z D : ˆT k (z) D for all k Z}. Since σ n (l) as n for all l A, all conditions of Proposition 3.3 and Theorem 3.4 are satisfied, and we obtain the following theorem.

19 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 9 Z 8 E 8 9 T(D Γ ) 4 E T(D ) 5 Z 5 E E 0 E E 4 E Z 0 T 3 (D Γ ) Z E 0 E 9 3 E 0 9 E D B 0 Z 0 6 E E Z 9 E E 3 E 5 Z E D A T (D Γ ) E Z E 4 E Z E 0 D 0 D Γ E Z E E E 7 Z 3 T 4 (D Γ ) ˆT 8 Figure 8.. The map ˆT and the set R, λ = γ. (l k stands for T k (D l ).) 0 3 D α D β 4 0 Figure 8.. The trajectory of the scaled domains and P, λ = γ. (l k stands for ˆT k U(D l ).)

20 0 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER Theorem 8.. If λ = γ, then the minimal period length π(z) is if z {(0, 0), (/γ, /γ ), (/γ, /γ )} if z {( 5 γ γ +, /γ γ + ), ( /γ γ +, 5 γ γ + )} 5 if z = T m (/(γ 4 ), /(γ 4 )) for some m {0,,, 3, 4} 0 for the other points of D A, D B, T m (D Γ ), T m (D ) if z = T m ( /γ 5, /γ 5 ) for some m {0,,..., 0} 5 if z = T m ( /(γ 5 ), /(γ 5 )) for some m {0,,..., 4} 50 for the other points of T m (D E ) (35 4 n + 8)/3 if S n R(z) is the center of D α 0(35 4 n + 8)/3 for the other points of D α 4(35 4 n 4)/3 if S n R(z) is the center of D β 0(35 4 n 4)/3 for the other points of D β if S n R(z) D \ P for all n 0 Proof. As for λ = /γ and λ = /γ, we have ( ) ( ) ( ) σ n (0) 0 σ n = 4 n /3 /3 +, (0) /3 /3 ( ) ( ) σ n () 0 σ n = 4 n /3 + () /3 ( ) /3, /3 hence τ(σ n (0)) = (70 4 n + 56)/3, τ(σ n ()) = (40 4 n 56)/3. For S n R(z) D α, we have π(z) = τ(σ n (0)) and 5τ(σ n (0)) respectively; if S n R(z) D β, then π(z) = τ(σ n ()) and 5τ(σ n ()) respectively. We choose ŝ(z) as follows and obtain the following s(z), t(z): z ˆT U(D 0 D ) : ŝ(z) =, s(z) = 70, t(z) = V ( ˆT (z)) V (z) = ( /γ, /γ ) z ˆT U(D ) : ŝ(z) =, s(z) = 4, t(z) = V ( ˆT (z)) V (z)a = (/γ, /γ) z ˆT 4 U(D ) : ŝ(z) =, s(z) = 4, t(z) = V ( ˆT (z)) V (z)a = (, 0) z ˆT U(D 0 ) ˆT 3 U(D ) : ŝ(z) =, s(z) = 70, t(z) = V ( ˆT (z)) V (z) = ( /γ, 0) This gives again δ = γ /γ = γ since {(, 0)A h : h Z} = ±{(, 0), (0, ), (, /γ), (/γ, /γ), (/γ, )}. Theorem 8.. π(z) is finite for all z (Z[γ] [0, )), but π ( /(3γ ), /(3γ 5 ) ) =. Proof. Since V (D) = {(x, y) : x > y > 0, γx + y γ}, we have no point z Z[γ] D with V (z) γ, and Conjecture. holds for λ = γ. If V (z) = ( γ /3, /(3γ) ), then we have V S(z) = γ ( ( )) ( V (z) γ, 0 = 3, γ ), V S (z) = γ ( ( V S(z) 3 γ, )) ( γ + γ = 3γ, ), 3γ V S 3 (z) = γ ( V S (z) ( γ, γ )) = ( 3γ 3, 3γ 3 ) and V S 4 (z) = γ ( V S 3 (z) ( γ, 0)) = V (z). 9. The case λ = 3 = cos 5π 6 The case λ = 3 is much more involved than the previous cases. Therefore we show only that all points in (Z[ 3] [0, )) are periodic and refrain from calculating the period lengths. Furthermore we postpone the determination of ˆT and R to Appendix A. Let D = {(x, y) : x 3y < 3, y 3x < 3, y 3x < , x 3y < } and D = D \ D, where D is defined by the inequalities x 3y > , y 3x > , y 3x > , x 3y > The sets D and D have to be treated separately because their trajectories are disjoint, and both sets contain aperiodic points. The trajectories of aperiodic points in D come arbitrarily

21 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS (, ) {60,733} 5 0 ˆT {375,3307} {375,3307} (7 4 3, 7 4 3) (73/ 4 3/, 73/ 4 3/) 0 {375,3307} ˆT {937,9459} (7 4 3, 7 4 3) Figure 9.. The first return map on D and D respectively, λ = 3. (l k stands for T k (D l ).) close to (, ), whereas (7 4 3, 7 4 3) is a limit point in D. (Note that = ( 3 + )( 3) ) The scaling maps are U (z) = ( 3)z + ( 3, 3 ) = V (κv (z)) for z D, U (z) = ( 3)z + (3 3 95, ) = V (κv (z)) for z D, with κ = 3, V (z) = ( (, ) z ) /κ 4, V (z) = ( z (7 4 3, 7 4 3) ) /κ 5. Then we have V (D) = {(x, y) : x > 3y, y > 3x, x > 3y, y > 3x }, V (D ) = {(x, y) : x > 3y, y > 3x, x > 3y 3, y > 3x 3}.

22 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER The first return map ˆT induces a partition of D into sets D 0,..., D 9 and a partition of D into sets D 0,..., D 4, as in Figure 9.. These sets are defined by the following (in)equalities: V (D 0 ) V (D ) V (D ) V (D 3 ) V (D 4 ) x > 3y x > 3y x > 3y + 3 y > 3x + 3 y > 3x + 3 x < x > x >, y < 3 y > 3 V (D 5 ) V (D 6 ) V (D 7 ) V (D 8 ) V (D 9 ) x = 3y x = y = 3 x = 3y + 3 y = 3x + 3 x < 3 / 3 x > x > V (D 0 ) V (D ) V (D ) V (D 3 ) V (D 4 ) y > 3x y < 3x, x < 3 + x > 3 + y = 3x x = 3 + The return times of z D l to D are given by the following tables. D 0 D D D 3 D 4 D 5 D 6 D 7 D 8 D 9 60, , D 0 D D D 3 D , Note that the return times are not constant on all D l. E.g., the return time for z D 0 is 60 if V (z) = (, y) and 733 else, see Appendix A for details. Since we do not calculate the period lengths, it is not necessary to distinguish between the parts of D l with different period lengths. 9.. The scaling domain D. Figure 9. shows the trajectory of the open scaled sets in D. Here, V (D ) is split up into the three stripes x < 3, 3 < x < and x >, and D denotes the set given by V (D ) = {(x, y) V (D) : x > 3y, x < }. We see that σ : codes the trajectory of U (D l ), l {0,,, 3, 4}, with ˆT σ(l) U (z) = U ˆT (z) for z Dl. All points in D α, D β and D γ are periodic. Figure 9.4 shows that D ε, D ζ, D η and the grey part of U (D ) split up further, but all their points are periodic as well. The trajectory of the scaled lines is depicted in Figure 9.3, where again V (D ) is split up into the stripes x < 3, 3 x < and x. Here, D denotes boundary lines of D, and D 6 is given by V (D 6 ) = {(, y) V (D)}. We see that σ codes the trajectory of U (D l ), l {5, 6, 7, 8, 9}, as well and satisfies the conditions in Section 3 (with respect to D ). All points in D ι, D κ, D λ, D µ, D ν, D ξ, D o, D π, D ρ (and their orbits) are periodic. The finitely many remaining points in P = {z D : ˆT m (z) U (D ) for all m Z} are clearly periodic as well. Since σ n (l) for all l {0,..., 9}, we can use Proposition 3.5 to show the following proposition. Proposition 9.. π(z) is finite for all z Z[ 3] D, but π(v ( 3 + /4, 7/4)) =. Proof. First we show that only D 0 and D contain aperiodic points: D 3, D 4, D 7, D 8, D 9 lie in P. The only part of D which is not in P or ˆT m U (P ), lies in ˆT U (D ). By iterating this argument on ˆT U (D ), the possible set of aperiodic points in D becomes smaller and smaller, and converges to V (, 3) D. A similar reasoning shows that all points in D 5 and D 6 are periodic. Therefore it is sufficient to determine t(z) for points in the trajectories of U (D 0 D ). z ˆT U (D 0 ) ˆT 3 U (D ) : ŝ(z) =, s(z) 0 mod, t(z) = ( 3)( 3, ) z ˆT 4 U (D ) : ŝ(z) =, s(z) 5 mod, t(z) = V ( ˆT (z)) V (z)a 5 = ( 3, ) z ˆT U (D ) : ŝ(z) =, s(z) 5 mod, t(z) = (, 3) z ˆT U (D 0 ) ˆT U (D ) : ŝ(z) =, s(z) 0 mod, t(z) = ( 3)(, 3)

23 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS η 9 η 4 β β β ζ 5 ζ 0 η 8 4 β ζ β β 9 β 7 β η 5 η 3 β η 3 β 5 ζ ζ 3 8 D α 4 8 β η 6 β 7 β 3 ζ 7 ζ 0 D γ 0 β 9 ˆT(D γ ) β 8 ζ 8 3 β 3 ζ 3 ζ 9 β 4 β η 3 0 β 3 β 3 ζ 3 η 7 ζ β β ε ζ ε 0 ζ 0 η 0 β 0 ε ζ Figure 9.. The trajectory of the open scaled sets in D and the set P, λ = 3. (l k stands for ˆT k U (D l ) if l {0,,, 3, 4}, for ˆT k (D l ) else.)

24 4 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER ι ι 0 ι ι ν 9 ν λ 0 λ ν ι ι 6 ν ν 5 ι 9 ι 8 6 ι ν ν 5 ν µ 0 µ ν 6 ν ν ξ ι 7 ν ξ 0 ν 6 ν 7 µ ξ 7 λ ξ 4 ξ ξ ν λ κ ν 3 ν 5 ρ 5 ρ ν 9 κ 0 ξ ν 7 µ 3 ν ν ρ 3 ν 7 ν ν ν 9 ν 0 ν 4 ρ 0 ν 6 ν 8 ρ ν 0 ρ 4 ξ 3 ρ 6 Figure 9.3. The trajectory of the scaled lines and the set P, λ = 3. (l k stands for ˆT k U (D l ) if l {, 5, 6, 7, 8, 9}, for ˆT k (D l ) else.)

25 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 5 η 0 ρ 0 ρ 6 ζ 4 ε 0 0 ε 5 ζ η η 8 π 8 π 8 ζ 8 ζ 8 ρ 0 ζ ρ 4 η ρ π 4 π 4 ρ 6 ζ 4 ζ 57 o 5 η 3 ζ 67 π 57 π 4 o 35 ζ 4 o 0 ζ ρ 4 D ε D δ o 0 ζ 5 ζ 6 ρ 8 ρ 8 π 6 ζ 6 η 6 π 6 ρ ζ 0 ζ 47 o 5 π 47 π 0 η 0 ζ 0 o 0 o 30 ζ o 5 6 ζ 37 η 7 π 37 π 5 Figure 9.4. Small parts of P, λ = 3. (l k stands for ˆT k (D l ) for l {, 6}.) We have δ = ( 3 + )/( 3 + ) = since {( 3, )A h : h Z} = ±{( 3, ), (, 3), ( 3, ), (, 0), (0, ), (, 3)}. The only point z V (Z[ 3] D ) with z is (, ) V (D α ). If V (z) = ( 3 + /4, 7/4), then we have V S(z) = ( + 3) ( V (z) + ( 3)(, 3) ) = (3/ + 3/4, 3 3/4 + /), V S (z) = ( + 3) ( V S(z) + ( 3)(, 3) ) = (7/4, 3 + /4), V S 3 (z) = (+ 3) ( V S (z)+( 3)( 3, ) ) = (3 3/4+/, 3/+ 3/4), V S 4 (z) = V (z). Remark. The primitive part of σ is again 0 00, The scaling domain D. Figure 9.5 shows the trajectory of the the scaled domains in D. Here, V (D ) is split up into x 3 + and x > 3 +. With ε = and σ :

26 6 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER ω 7 ω ω 0 ψ 0 6 ω 5 ψ 0 D χ ω D ϕ ω ψ ψ ω ω Figure 9.5. The trajectory of the scaled domains in D and the set P, λ = 3. (l k stands for ˆT k (D l ) if l {ψ, ω}, for ˆT k U (D l ) else.) the conditions in Section 3 are satisfied. The set P = {z D : ˆT m (z) U (D ) for all m Z} consists of the orbits of D ϕ, D χ, D ψ, D ω and several isolated (periodic) points. Since σ n (l) for all l {0,,, 3, 4}, we can use Proposition 3.5 to show the following proposition. Proposition 9.. π(z) is finite for all z Z[ 3] D, but π(v (5/7, 3 3/7)) =.

27 PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS Figure 9.6. Aperiodic points, λ = 3. Figure 9.7. Aperiodic points in D D, λ = 3. Proof. Similarly to D, we see that all points in D 3 and D 4 are periodic. Choose ŝ(z) as follows: z ˆT 0 U (D 0 ) ˆT 5 U (D ) : ŝ(z) =, s(z) 7 mod, t(z) = (, 3) z ˆT 9 U (D 0 ) ˆT 4 U (D ) : ŝ(z) =, s(z) 3 mod, t(z) = ( 3, 3 ) z ˆT 8 U (D 0 ) ˆT 3 U (D ) : ŝ(z) = 3, s(z) 0 mod, t(z) = ( 3, 3) z ˆT 7 U (D 0 ) : ŝ(z) = 4, s(z) 5 mod, t(z) = 3( 3, ) z ˆT 6 U (D 0 ) : ŝ(z) = 5, s(z) 0 mod, t(z) = ( 3, ) For the remaining z ˆT m U (D 0 D ), ŝ(z), s(z) and t(z) are obtained by symmetry. The sets {( 3, 3 )A h : h Z} and {( 3, 3)A h : h Z} are ±{( 3, 3 ), ( 3, ), (, 3 + ), ( 3 +, 3 + ), ( 3 +, ), (, 3 )}, ±{( 3, 3), (3, 3 + ), ( 3 +, 3 + 3), (3 + 3, + 3), ( + 3, 3), ( 3, 3)}, hence δ = 4/( 3 + ) = ( 3 ). The only x Z[ 3] with 0 < x < 5 and x ( 3 ) are, + 3, + 3 and Therefore the only z V (Z[ 3] D ) with z ( 3 ) are (, ), the center of V U (D χ ), ( + 3, + 3), the center of D 4, ( + 3, + 3), the center of D χ, and (3 + 3, 3 + 3), a fixed point of ˆT 3. If V (z) = (5/7, 3 3/7), then we have V S(z) = ( + 3)V (z) and V S (z) = ( + 3) ( V S(z)A 3 + ( 3, 3 ) ) = (5/7, 3 3/7) = V (z). By combining Propositions 9. and 9. and the fact that all points in R are periodic (see Appendix A), we obtain the following theorem. Theorem 9.3. Conjecture. holds for λ = 3. Remark. The eigenvalues corresponding to the primitive part of σ (l {0,, }) are 5, and. 0. The case λ = 3 = cos π 6 Let D = {(x, y) [0, ) : x + 3y > or y + 3x > 5 3 6}, U as in Section 9 and U(z) = U (z) = ( 3) z + (4 3 6, 4 3 6) = V (κv (z)), κ = ( 3), V (z) = ( (, ) z ) /κ. Then we have V (D) = {(x, y) : x > 0, y > 0, x + 3y < or y + 3x < }.

28 8 S. AKIYAMA, H. BRUNOTTE, A. PETHŐ, AND W. STEINER ˆT Figure 0.. The map ˆT on D, λ = 3. (l k stands for T k (D l ).) Figure 0. shows the first return map ˆT on D, which is determined in Appendix B. The sets D 0,..., D 6 satisfy the (in)equalities V (D 0 ) V (D ) V (D ) 3x + y < 3x + y >, x < 3 x > 3, x + 3y 3 V (D 3 ) V (D 4 ) V (D 5 ) V (D 6 ) 3x + y =, x < / 3x + y =, x > / x = 3 x + 3y = 3 The remaining point z = V (/, 3/) has return time 83 and satisfies ˆT 0 (z) = z. Figure 0. shows that the first return map on U (D) differs from U ˆT U on several lines. Therefore we add the lines D 7, D 8, D 9 satisfying the following (in)equalities V (D 7 ) V (D 8 ) V (D 9 ) 3x + y = 3x + y =, x > 3 3x + y =, x < 3 and define D 0 = D 0 \ V ({(x, y) : 3x + y = }), D = D D 6. For z D l, l {0, } and z D l, l =, we have ˆT σ (l) U (z) = U ˆT (z) with σ : Figure 0.3 shows that the substitution σ given by σ(l) = σ σ (l) with σ : satisfies the conditions in Section 3 (with ε = ). The coding of the return path of the remaining point is σ ( ). Theorem 0.. π(z) is finite for all z (Z[ 3] [0, )), but π(v (/7, 3/7 + /7) =. Proof. First we show that all points on the lines U n (D l ), l {3,..., 9}, n 0, are periodic. The only possibly aperiodic part of D 5 is ˆT U (D 7 ), and the only possibly aperiodic part of U (D 7 ) is ˆT 3 U (D 5 ). Inductively, the set of aperiodic points in D 5 converges to V ( 3, / 3) D 5 and is therefore empty. Therefore, all points in U n (D 5 ) and U n U (D 7 ) are periodic. Similar arguments show that all points in U n (D 3 ) in U n U (D 9 ) are periodic, then the same holds for U n (D 4 ) and U n U (D 5 ), for U n (D 6 ) and U n U (D 8 ), and finally for U n (D 8 ) and U n U (D 6 ). Then it is clear that all points in U n U (D 3 D 4 ) and U n (D 7 D 9 ) are periodic as well.