PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS

PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS

arXiv:0704.3674v3 [math.DS] 16 Sep 2008

˝ AND WOLFGANG STEINER SHIGEKI AKIYAMA, HORST BRUNOTTE, ATTILA PETHO, Abstract. We determine periodic and aperiodic points of certain piecewise affine maps in the √ √ √ 5 , ± 2, ± 3} that all integer Euclidean plane. Using these maps, we prove for λ ∈ { ±1± 2 sequences (ak )k∈Z satisfying 0 ≤ ak−1 + λak + ak+1 < 1 are periodic.

1. 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 [1, 7, 12, 13, 17, 18], for particular instances to [29, 16, 25] (polygonal dual billiards), [15] (polygonal exchange transformations), [10, 31, 11, 8] (digital filters) and [19, 21, 22] (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 1.1. [4, 27] For every real λ with |λ| < 2, all integer sequences (ak )k∈Z satisfying (1.1)

0 ≤ ak−1 + λak + ak+1 < 1

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 [19, 27] and the literature quoted there), and on the other hand in the study of shift radix systems (see [4, 2] for details). Extensive numerical evidence on the periodicity of integer sequences satisfying (1.1) was first observed in [26]. We summarize the situation of the Conjecture 1.1. Since we have approximately 0 1 ak−1 ak ≈ ak −1 −λ ak+1 and the eigenvalues of the matrix are exp(±θπi) with θ ∈ [0, 1], the sequence may be viewed as a discretized rotation on Z2 , and it is natural to parametrize −λ = 2 cos(θπ). There are five different classes of λ of apparently increasing difficulty: (1) θ is rational and λ is rational. (2) 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 = −1, 0, 1, where the conjecture is trivially true. √ Already in case (2) the problem is far from trivial. A computer assisted proof for −λ = 5−1 was 2 1 given by Lowenstein, Hatjispyros and Vivaldi [19]. A short proof (without use of computers) of √ 1+ 5 the golden mean case λ = 2 was given by the authors [3]. The main goal of this paper is to settle the conjecture for all the cases of (2), i.e., the quadratic parameters √ √ √ ±1 ± 5 , ± 2, ± 3. λ= 2 Date: September 16, 2008. 1Indeed, they showed that all trajectories of the map (x, y) 7→ (⌊(−λ)x⌋ − y, x) on Z2 are periodic. 1

˝ AND W. STEINER S. AKIYAMA, H. BRUNOTTE, A. PETHO,

2

The proofs are sensitive to the choice √ of λ, and we have to work tirelessly in computation and important feature of our proof is that it drawings, especially in the last case ± 3. However, an √ 1+ 5 can basically be checked by hand. The (easiest) case 2 in Section 2 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 1.1 can be proved using the same method, which involves a map on [0, 1)2d−2 , 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 [22], a similar embedding into a higher dimensional torus is used for efficient orbit computations. Goetz [12, 13, 14] 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 pn is exhibited in [9]. The authors show that the dynamical system given by (1.1) can be embedded into a p-adic rotation dynamics, by multiplying a p-adic unit. These investigations were extended in [30]. Furthermore, in [27] 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 [19] is based on a discontinuous non-ergodic piecewise affine map on the unit square, which dates back to Adler, Kitchens and Tresser [1]. Let λ2 = bλ + c with b, c ∈ Z. Set x = {λak−1 } and y = {λak }, where {z} = z − ⌊z⌋ denotes the fractional part of z. Then we have ak+1 = −ak−1 − λak + y and {λak+1 } = {−λak−1 − λ2 ak + 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, 1)2 → [0, 1)2 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, 1))2 in order to prove the conjecture. Using this map, Kouptsov, Lowenstein and Vivaldi [18] showed for all quadratic λ corresponding to rational √ √ √ 5 2, ± 3 that the trajectories of almost all points are periodic, by heavy rotations λ = ±1± , ± 2 use of computers. Of course, such metric results do not settle Conjecture 1.1, which deals with countably many points in [0, 1)2 , 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 1.1 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 2.1. All possible period lengths can be determined explicitly and one can even construct concrete aperiodic points in (Q(λ) ∩ [0, 1))2 . 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 [23] whether there exist algebraic points in the triadic Cantor set. The paper is organized as follows. In Section 2, 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 2 are extended to a general √ setting. √ In the subsequent seven sections we√prove the conjecture for the cases λ = −γ, ±1/γ, ± 2, ± 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, ±1/γ, 3. 2. The case λ = γ =

√ 1+ 5 2 ,

= −2 cos 4π 5

λ2 = λ + 1. Note that T is given by 0 −1 T (x, y) = (x, y)A + (0, ⌈x − y/γ⌉) with A = . 1 1/γ

We consider first the golden mean λ = γ = (2.1)

√ 1+ 5 2


T (R)

R

D0

3

T →

T (D1 )

D1

T (D0 )

Figure 2.1. The piecewise affine map T and the set R, λ = γ =

√ 1+ 5 2 .

Therefore, we have T (x, y) = (x, y)A if y ≥ γx and T (z) = zA + (0, 1) for the other points z ∈ [0, 1)2 , see Figure 2.1. A particular role is played by the set R = {(x, y) ∈ [0, 1)2 : y < γx, x + y > 1, x < yγ} ∪ {(0, 0)}.

If z ∈ R, z 6= (0, 0), then we have T k+1 (z) = T k (z)A + (0, 1) for all k ∈ {0, 1, 2, 3, 4}, hence T 5 (z) = zA5 + (0, 1)(A4 + A3 + A2 + A1 + A0 ) = z + (0, 1)(A5 − A0 )(A − A0 )−1 = z

since A5 = A0 . It can be easily verified that the minimal period length is 5 for all z ∈ R except 2 2 ( γ 2γ+1 , γ 2γ+1 ) and (0, 0), which are fixed points of T . Therefore, it is sufficient to consider the domain D = [0, 1)2 \ R in the following. According to the action of T , we partition D into two sets D0 and D1 , with D0 = {(x, y) ∈ [0, 1)2 : y ≥ γx} \ {(0, 0)}, In Figure 2.2, we scale D0 and D1 by the factor 1/γ 2 and follow their T -trajectory until the return to D/γ 2 . Let P be the set of points in D which are not eventually mapped to D/γ 2 , i.e., P = Dα ∪ T (Dα ) ∪ Dβ ∪ T (Dβ ) ∪ T 2 (Dβ ),

where Dα is the closed pentagon {(x, y) ∈ D0 : y ≥ 1/γ 2, x + y ≤ 1, y ≤ (1 + x)/γ} and Dβ is the open pentagon R/γ 2 \ {(0, 0)}. (In Figure 2.2, Dα is split up into {T k (Dα˜ ) : k ∈ {0, 2, 4, 6, 8}}, and Dβ is split up into {T k (Dβ˜ ) : k ∈ {0, 3, 6, 9, 12}}.) All points in P are periodic (with minimal period lengths 2, 3, 10 or 15). Figures 2.1 and 2.2 show that the action of the first return map on D/γ 2 is similar to the action of T on D, more precisely, ( T (z/γ 2) if z ∈ D0 , T (z) (2.2) = γ2 T 6 (z/γ 2 ) if z ∈ D1 . For z ∈ D \ P, let s(z) = min{m ≥ 0 : T m (z) ∈ D/γ 2 }. (Figure 2.2 shows s(z) ≤ 5.) By the map S : D \ P → D,

z 7→ γ 2 T s(z) (z),

we can completely characterize the periodic points. For z ∈ [0, 1)2 , denote by π(z) the minimal period length if (T k (z))k∈Z is periodic and set π(z) = ∞ else. Theorem 2.1. (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 2 , R(z) = z, Tˆ(z) = T (z), π ˆ (z) = π(z), and z ∈ D1 or T (z) ∈ D1 for all z ∈ D, |σ n (1)| → ∞, see below). (2.2) and Figure 2.2 suggest to define a substitution (or morphism) σ on the alphabet A = {0, 1}, i.e., a map σ : A → A∗ (where A∗ denotes the set of words with letters in A), by σ:

0 7→ 0

1 7→ 101101

in order to code the trajectory of the scaled domains until their return to D/γ 2 : We have T k−1 (Dℓ /γ 2 ) ⊆ Dσ(ℓ)[k] and T |σ(ℓ)|(z/γ 2 ) = T (z)/γ 2 for all z ∈ Dℓ , where w[k] denotes the k-th


4

β˜7 β˜1 1 T(D γ2 )

β˜10

β˜13 1 T 6( D γ2 )

β˜4 1 T 4( D γ2 )

0 T(D γ2 )

T 2 (Dα˜ )

T 6 (Dα˜ )

T 8 (Dα˜ )

T 4 (Dα˜ ) 1 T 3( D γ2 )

Dα˜

1 T 2( D γ2 )

β˜2

β˜12 β˜11

β˜3

β˜6 D0 γ2

T 7 (Dα˜ ) Dβ˜

β˜5

β˜9

β˜14

T 9 (Dα˜ )

T (Dα˜ ) D1 γ2

β˜8

T 3 (Dα˜ )

5

T (Dα˜ )

1 T 5( D γ2 )

Figure 2.2. The trajectory of the scaled domains and the (gray) set P, λ = γ. (β˜k stands for T k (Dβ˜ ).) letter of the word w and |w| denotes its length. Furthermore, we have T k (Dℓ /γ 2 ) ∩ D/γ 2 = ∅ for 1 ≤ 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 1 by Poggiaspalla [24]. Lemma 2.2. For every integer n ≥ 0 and every ℓ ∈ {0, 1}, we have n • T |σ (ℓ)| (z/γ 2n ) = T (z)/γ 2n for all z ∈ Dℓ , k−1 • T (Dℓ /γ 2n ) ⊆ Dσn (ℓ)[k] for all k, 1 ≤ k ≤ |σ n (ℓ)| • T k (Dℓ /γ 2n ) ∩ D/γ 2n = ∅ for all k, 1 ≤ k < |σ n (ℓ)|.

The proof is again postponed to Section 3, Lemma 3.1. This lemma allows to determine the minimal period lengths: If z ∈ Dα , then T |σ

n

(0101010101)|

(z/γ 2n) = T |σ

n

(101010101)|

(T (z)/γ 2n ) = · · · = T 10 (z)/γ 2n = z/γ 2n

for all n ≥ 0. The only points of the form T k (z/γ 2n ), 1 ≤ k ≤ 5|σ n (01)|, which lie in D/γ 2n are the points T m (z)/γ 2n , 1 ≤ m ≤ 9, which are all different from z/γ 2n if π(z) = 10. Therefore, we obtain π(z/γ 2n ) = 5|σ n (01)| in this case. A point z˜ lies in the trajectory of z/γ 2n if and only if S n (˜ z ) = T m (z) for some m ∈ Z, see Lemma 3.2. This implies π(˜ z ) = 5|σ n (01)| for these z˜ as well. The period lengths of all points are given by the following theorem.


5

Theorem 2.3. If λ = γ, then the minimal period lengths π(z) of (T k (z))k∈Z are 2 2 1 if z = (0, 0) or z = ( γ 2γ+1 , γ 2γ+1 ) 2

2

5 if z ∈ R \ {(0, 0), ( γ 2γ+1 , γ 2γ+1 )} 2 (5 · 4n + 1)/3 if S n (z) = T m ( γ1/γ 2 +1 , γ 2 +1 ) for some n ≥ 0, m ∈ {0, 1} 2 n n m for some n ≥ 0, m ∈ {0, 1} 5(5 · 4 + 1)/3 if S (z) ∈ T Dα \ {( γ1/γ 2 +1 , γ 2 +1 )} 1 1 n n m (10 · 4 − 1)/3 if S (z) = T ( γ 2 +1 , γ 2 +1 ) for some n ≥ 0, m ∈ {0, 1, 2} 5(10 · 4n − 1)/3 if S n (z) ∈ T m Dβ \ {( γ 21+1 , γ 21+1 )} for some n ≥ 0, m ∈ {0, 1, 2} ∞ if S n (z) ∈ D \ P for all n ≥ 0. The minimal period length of (ak )k∈Z is π({γak−1 }, {γak }) (which does not depend on k). Proof. By Theorem 2.1, Proposition 3.3 and the remarks preceding the theorem, it suffices to calculate |σ n (0)| and |σ n (1)|. Clearly, we have |σ n (0)| = 1 for all n ≥ 0 and thus |σ n (1)| = |σ n−1 (101101)| = 4|σ n−1 (1)| + 2 = 4(5 · 4n−1 − 2)/3 + 2 = (5 · 4n − 2)/3.

If S n (z) ∈ T m (Dα ), then π(z) = |σ n (01)| and π(z) = 5|σ n (01)| respectively. If S n (z) ∈ T m (Dβ ), then π(z) = |σ n (101)| and π(z) = 5|σ n (101)| respectively. Now consider aperiodic points z ∈ [0, 1)2 , i.e., S n (z) ∈ D \ P for all n ≥ 0. We can write S(z) = γ 2 T s(z) (z) = γ 2 zAs(z) + t(z)

for some t(z) by using (2.1). Note that T (z) = zA for z ∈ D0 and T (z) = zA + (0, 1) for z ∈ D1 . For z ∈ D/γ 2 , we have s(z) = 0 and t(z) = 0. For z ∈ T k (D1 /γ 2 ), 1 ≤ k ≤ 5, we have s(z) = 6−k,  if s(z) ∈ {1, 2},  (0, 1) (0, 1)A2 + (0, 1) = (1/γ, 1/γ 2) if s(z) = 3, t(z) =  (0, 1)A3 + (0, 1)A2 + (0, 1) = (0, −1/γ) if s(z) ∈ {4, 5}.

We obtain inductively

S n (z) = γ 2n zAs(z)+s(S(z))+...+s(S

n−1

(z))

+

n−1 X

γ 2(n−k) t(S k (z))As(S

k+1

(z))+···+s(S n−1 (z))

.

k=0

If z ∈ Q(γ)2 , then we have ′ ′ n−1 k+1 n−1 n−1 zAs(z)+s(S(z))+···+s(S (z)) X t(S k (z))As(S (z))+···+s(S (z)) + (S n (z))′ = γ 2n γ 2(n−k) k=0

k(S n (z))′ k∞ ≤ ′

′

h ′

maxh∈Z k(zA ) k∞ + γ 2n

′

′

′

n−1 X k=0

maxh∈Z,w∈D\P k(t(w)Ah )′ k∞ , γ 2n−k

where z = (x , y ) if z = (x, y) and x , y are the algebraic conjugates of x, y. Since t(z)Ah ∈ (0, 0), (0, 1), (1, 1/γ), (1/γ, −1/γ), (−1/γ, −1), (−1, 0), (1/γ, 1/γ 2), (1/γ 2 , −1/γ 2), (−1/γ 2 , −1/γ), (−1/γ, 0), (0, 1/γ),

(0, −1/γ), (−1/γ, −1/γ 2), (−1/γ 2 , 1/γ 2 ), (1/γ 2 , 1/γ), (1/γ, 0)

and zAh takes only the values z, zA, zA2 , zA3 and zA4 , we obtain n−1

k(S n (z))′ k∞ ≤

maxh∈Z k(zAh )′ k∞ X γ 2 C(z) + < 2n + γ 2n 2(n−k) γ γ γ k=0

1 1 Z[γ])2 for some integer Q ≥ 1, then S n (z) ∈ ( Q Z[γ])2 . Since for some constant C(z). If z ∈ ( Q 1 2 ′ there exist only finitely many points w ∈ ( Q Z[γ] ∩ [0, 1)) with kw k∞ < C(z) + γ, we must have k(S n (z))′ k∞ ≤ γ for some n ≥ 0, which proves the following proposition. 1 Proposition 2.4. Let z ∈ ( Q Z[γ] ∩ [0, 1))2 be an aperiodic point. Then there exists an aperiodic 1 Z[γ])2 ∩ D with k˜ z ′ k∞ ≤ γ. point z˜ ∈ ( Q


6 1

1

0.9

0.9

0.8

0.8

0.7

0.7

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1 0

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Figure 2.3. Aperiodic points, λ = γ.

1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 2.4. Aperiodic points, λ = −1/γ.

For every denominator Q ≥ 1, it is therefore sufficient to check the periodicity of the (finite set 1 1 Z[γ])2 ∩ D with kz ′ k∞ ≤ γ in order to determine if all points in ( Q Z[γ] ∩ [0, 1))2 of) points z ∈ ( Q are periodic. For Q = 1, we have to consider z = (x, y) ∈ D with x, y ∈ Z[γ] ∩ [0, 1) and |x′ |, |y ′ | ≤ γ, hence (x, y) ∈ {0, 1/γ}2. Since (0, 0) and (1/γ, 1/γ) are in R, it only remains to check the periodicity of (0, 1/γ) and (1/γ, 0). These two points lie in P, thus Conjecture 1.1 is proved for λ = γ. For Q = 2, a short inspection shows that all points z ∈ ( 12 Z[γ] ∩ [0, 1))2 are periodic as well. The situation is completely different for Q = 3, and we have S(0, 1/3) = (0, γ 2 /3), S(0, γ 2 /3) = γ 2 (0, γ 2 /3)A5 + (0, −1/γ) = (0, 2/3), S(0, 2/3) = γ 2 (0, 2/3)A5 + (0, −1/γ) = 0, 1/(3γ 2) , S 4 (0, 1/3) = S 0, 1/(3γ 2) = (0, 1/3).

This implies S n (0, 1/3) ∈ D \ P for all n ≥ 0 and π(0, 1/3) = ∞ by Theorem 2.3.

Theorem 2.5. π(z) is finite for all points z ∈ (Z[γ] ∩ [0, 1))2 , but (T k (0, 1/3))k∈Z is aperiodic. 3. General description of the method In this section, we√generalize the method presented in Section 2 in order to make it applicable √ for λ = −γ, ±1/γ, ± 2, ± 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) 6∈ 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 Tˆ be the first return map (of the iterates by T ) on D, i.e., Tˆ : D → D,

Tˆ(z) = RT (z) = T r(T (z))+1 (z),

in particular Tˆ(z) = T (z) if T (z) ∈ D. Let A be a finite set, {Dℓ : ℓ ∈ A} a partition of D and define a coding map ι : D → AZ by ι(z) = (ιk (z))k∈Z such that Tˆk (z) ∈ Dιk (z) for all k ∈ Z. Let U : D → D, ε ∈ {−1, 1} and σ a substitution on A such that, for every ℓ ∈ A and z ∈ Dℓ , U Tˆ(z) = Tˆ ε|σ(ℓ)| U (z),


7

Tˆ εk U (z) 6∈ U (D) for all k, 1 ≤ k < |σ(ℓ)|, and ι0 (U (z)) ι1 (U (z)) · · · ι|σ(ℓ)|−1 (U (z)) if ε = 1, σ(ℓ) = ι−|σ(ℓ)| (U (z)) · · · ι−2 (U (z)) ι−1 (U (z)) if ε = −1. Then the following lemma holds.

Lemma 3.1. For every integer n ≥ 0, every ℓ ∈ A and z ∈ Dℓ , we have n n U n Tˆ(z) = Tˆ ε |σ (ℓ)| U n (z), n Tˆ ε k U n (z) 6∈ U n (D) for all k, 1 ≤ k < |σ n (ℓ)|, and

ι0 (U n (z)) ι1 (U n (z)) · · · ι|σn (ℓ)|−1 (U n (z)) = σ n (ℓ) ι0 (U n (z)) ι1 (U n (z)) · · · ι|σn (ℓ)|−1 (U n (z)) = (σ¯ σ )n/2 (ℓ) n n n ι−|σn (ℓ)| (U (z)) · · · ι−2 (U (z)) ι−1 (U (z)) = (σ¯ σ )(n−1)/2 σ(ℓ)

where σ ¯ (ℓ) = ℓm · · · ℓ2 ℓ1 if σ(ℓ) = ℓ1 ℓ2 · · · ℓm .

if ε = 1, if ε = −1, εn = 1, if ε = −1, εn = −1,

Proof. The lemma is trivially true for n = 0, and for n = 1 by the assumptions on σ. If we suppose inductively that it is true for n − 1, then let σ(ℓ) = ℓ1 ℓ2 · · · ℓm if ε = 1, σ(ℓ) = ℓm · · · ℓ2 ℓ1 if ε = −1, and we obtain (by another induction) for all j, 1 ≤ j ≤ m, (3.1)

n n−1 n n−1 Tˆε |σ (ℓ1 ···ℓj−1 ℓj )| U n (z) = Tˆ ε |σ (ℓj )| U n−1 Tˆ ε(j−1) U (z) = U n−1 Tˆεj U (z).

If ε = 1, then this follows immediately from the induction hypothesis; if ε = −1, then this follows by setting k = |σ n−1 (ℓj )| in n n n−1 (3.2) Tˆ(−1) k U n−1 Tˆ Tˆ −j U (z) = Tˆ (−1) (k−|σ (ℓj )|) U n−1 Tˆ −j U (z). Therefore, we have

n n n n−1 Tˆ ε |σ (ℓ)| U n (z) = Tˆ ε |σ (ℓ1 ···ℓm−1 ℓm )| U n (z) = U n−1 Tˆ εm U (z) = U n−1 Tˆ ε|σ(ℓ)| U (z) = U n Tˆ(z).

If ε = 1, then (3.1) implies that ι0 (U n (z)) · · · ι|σn (ℓ)|−1 (U n (z)) = ι0 (U n−1 U (z)) · · · ι|σn−1 (ℓ1 )|−1 (U n−1 U (z)) · · · ι0 (U n−1 Tˆm−1 U (z)) · · · ι|σn−1 (ℓm )|−1 (U n−1 Tˆm−1 U (z)) = σ n−1 (ℓ1 ) · · · σ n−1 (ℓm ) = σ n (ℓ);

if ε = −1 and εn = 1, then (3.1) and (3.2) provide

ι0 (U n (z)) · · · ι|σn (ℓ)|−1 (U n (z)) = ι−|σn−1 (ℓ1 )| (U n−1 T −1 U (z)) · · · ι−1 (U n−1 T −1 U (z)) · · · ι−|σn−1 (ℓ )| (U n−1 Tˆ−m U (z)) · · · ι−1 (U n−1 Tˆ−m U (z)) m

(n−2)/2

= (σ¯ σ)

if ε = −1 and εn = −1, then

σ(ℓ1 ) · · · (σ¯ σ )(n−2)/2 σ(ℓm ) = (σ¯ σ )n/2 (ℓ);

ι−|σn (ℓ)| (U n (z)) · · · ι−1 (U n (z)) = ι0 (U n−1 T −m U (z)) · · · ι|σn−1 (ℓm )|−1 (U n−1 T −m U (z)) · · · ι0 (U n−1 Tˆ −1 U (z)) · · · ι|σn−1 (ℓ1 )| (U n−1 Tˆ −1 U (z))

= (σ¯ σ )(n−1)/2 (ℓm ) · · · (σ¯ σ )(n−1)/2 (ℓ1 ) = (σ¯ σ )(n−1)/2 σ(ℓ).

n By (3.1), (3.2) and the induction hypothesis, the only points in (Tˆε k U n (z))1≤k 1, and we obtain

′

V R(z)As0 +s1 +···+sn−1 n ′ ∞ + δ, k(V S R(z)) k∞ < |κ′ |n

thus k(V S n R(z))′ k∞ ≤ δ for some n ≥ 0 (as in Section 2), and we can choose z˜ = S n R(z).

Remarks. • The last proof shows that, for every z ∈ (Q(λ) ∩ [0, 1))2 \ 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 ∞ n−1 X Pk X j tk A− j=0 s(S (z)) κk , V (z) = V S n (z)κn − tk Ask+1 +···+sn−1 κk A−s0 −···−sn−1 = − k=0

k=0

which is a κ-expansion (κ < 1) of V (z) with (two-dimensional) “digits” −tk A−s0 −s1 −···−sk . • As a consequence of Lemma 3.2 and the definition of U , for every aperiodic point z ∈ [0, 1)2 \ R and every c > 0, there exists some m ∈ Z such that kT m (z) − vk∞ < c. • In all our cases, we have ε = κκ′ . 4. The case λ = −1/γ =

√ 1− 5 2

= −2 cos 2π 5

Now we apply the method of Section 3 for λ = −1/γ, i.e., λ′ = γ. To this end, set D = {(x, y) ∈ [0, 1)2 : x + y ≥ 3 − γ} = D0 ∪ D1

with D0 = {(x, y) ∈ D : x + γy > 2}, D1 = {(x, y) ∈ D : x + γy ≤ 2}. Figure 4.1 shows that Tˆ is given by Tˆ(z) = T τ (ℓ)(z) if z ∈ Dℓ , ℓ ∈ A = {0, 1}, with τ (0) = 1 and τ (1) = 4. The set which is left out by the iterates of D0 and D1 is R = {(0, 0)} ∪ DA ∪ DB , with DA = {z ∈ [0, 1)2 : T k+1 (z) = T k (z)A + (0, 1) for all k ≥ 0},

DB = {z ∈ [0, 1)2 : T k+1 (z) = T k (z)A + (0, 2) for all k ≥ 0}.

10


T 4 (D1 )

D0

T (D0 ) Tˆ →

D1

T 3 (D1 ) DB

DA T 2 (D1 )

T (D1 )

Figure 4.1. The map Tˆ, Tˆ(D0 ) = T (D0 ), Tˆ(D1 ) = T 4 (D1 ), and the (gray) set R, λ = −1/γ. 02 1

2

00 1

4

1

Dβ

15

03

0

Dα 11 13

01

Figure 4.2. The trajectory of the scaled domains and P, λ = −1/γ. (ℓk stands for Tˆ k U (Dℓ ).) As in Section 2, we have T 5 (z) = z for all z ∈ R. If we set 1 1 z (1, 1) − z U (z) = 2 + = (1, 1) − , , γ γ γ γ2 V (z) = (1, 1) − z, κ = 1/γ 2 , ε = 1, and

σ : 0 7→ 010

1 7→ 01110

then Figure 4.2 shows that σ satisfies the conditions in Section 3, and P = Dα ∪ Dβ with Dα = U (DA ), Dβ = U (DB ). 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 4.1. If λ = −1/γ, 1 5 2(5 · 4n + 1)/3 10(5 · 4n + 1)/3 (5 · 4n − 2)/3 5(5 · 4n − 2)/3 ∞

then the period lengths π(z) are if z ∈ {(0, 0), ( γ 21+1 , γ 21+1 ), ( γ 22+1 , γ 22+1 )} for the other points of the pentagons DA and DB 2 2 if S n R(z) = ( γ 2γ+1 , γ 2γ+1 ) for some n ≥ 0 for the other points with S n R(z) ∈ Dα for some n ≥ 0 if S n R(z) = ( γ 23+1 , γ 23+1 ) for some n ≥ 0 for the other points with S n R(z) ∈ Dβ for some n ≥ 0 if S n R(z) ∈ D \ P for all n ≥ 0.


Proof. We easily calculate n 2/3 |σ (0)|0 n 1/3 + , =4 1/3 −1/3 |σ n (0)|1

|σ n (1)|0 |σ n (1)|1

11

2/3 −2/3 + , =4 2/3 1/3 n

2 n n n hence τ (σ n (0)) = 35 4n − 23 , τ (σ n (1)) = 10 3 4 + 3 . If S R(z) ∈ Dα , then π(z) = τ (σ (1)) and n n n π(z) = τ (σ (11111)) respectively; if S R(z) ∈ Dβ , then π(z) = τ (σ (0)) and π(z) = 5τ (σ n (0)) respectively.

For z ∈ U (D), we have sˆ(z) = s(z) = 0 and t(z) = (0, 0). For the other z ∈ D \ P, we choose sˆ(z) as follows and obtain the following s(z), t(z): z ∈ Tˆ 2 U (D0 ) ∪ Tˆ2 U (D1 ) : sˆ(z) = −2, s(z) = −5, t(z) = V (Tˆ −2 (z)) − V (z) = (−1/γ 2 , 0) z ∈ TÛ (D1 ) : sˆ(z) = −1, s(z) = −1, t(z) = V (Tˆ −1 (z)) − V (z)A−1 = (1/γ, 0) z ∈ Tˆ4 U (D1 ) : sˆ(z) = 1, s(z) = 1, t(z) = V (Tˆ(z)) − V (z)A = (0, 1/γ)

z ∈ TÛ (D0 ) ∪ Tˆ3 U (D1 ) : sˆ(z) = 2, s(z) = 5, t(z) = V (Tˆ2 (z)) − V (z) = (0, −1/γ 2)

Observe the symmetry between positive and negative sˆ(z) which is due to the symmetry of T (x, y) and T −1 (y, x) and the symmetry of D. With {(1/γ, 0)Ah : h ∈ Z} = {(1/γ, 0), (0, −1/γ), (−1/γ, 1), (1, −1), (−1, 1/γ)},

we obtain δ ≤ max{k(t(z)Ah )′ k∞ : z ∈ D \ P, h ∈ Z}/γ = (1/γ 2 )′ /γ = γ, as in Section 2. The following theorem shows that aperiodic points with t(z) = (−1/γ 2 , 0) exist, hence δ = γ. Theorem 4.2. π(z) is finite for all z ∈ (Z[γ] ∩ [0, 1))2 , but π 1 − 1/(3γ), 1 − 2/(3γ) = ∞.

Proof. By Proposition 3.5, we have to show that all z ∈ Z[γ]2 ∩ D with kV (z)′ k∞ ≤ γ are periodic. Since V (D) = {(x, y) : x > 0, y > 0, x + y ≤ 1/γ}, we have to consider x, y ∈ Z[γ] ∩ (0, 1/γ) with |x′ |, |y ′ | ≤ γ. No such x, y exist, hence the conjecture is proved for λ = −1/γ. Note that π(z) is finite for all z ∈ ( 12 Z[γ] ∩ [0, 1))2 as well. If V (z) = 1/(3γ), 2/(3γ) , then we have V S(z) = γ 2 V (z)A5 + (0, −1/γ 2 ) = γ/3, 1/(3γ 3) V S 2 (z) = γ 2 V S(z)A−5 + (−1/γ 2 , 0) = 2/(3γ), 1/(3γ) V S 3 (z) = γ 2 V S 2 (z)A−5 + (0, −1/γ 2) = 1/(3γ 3), γ/3) V S 4 (z) = γ 2 V S 3 (z)A5 + (0, −1/γ 2 ) = 1/(3γ), 2/(3γ) = V (z), hence S n (z) ∈ D \ P for all n ≥ 0 and π(z) = ∞ by Theorem 4.1. √ 5. The case λ = 2 = −2 cos 3π 4 √ √ ′ Let λ = 2 (λ = − 2) and set [ √ √ √ √ D = {(x, y) ∈ [0, 1)2 : 2 − 2 < x − 2y < 0, 0 < 2x − y < 2 − 2} = Dℓ , ℓ∈A={0,1,2,3} √ √ √ D0 = {(x, y) ∈ D : x < 2 − 1}, D1 = {(x, y) ∈ D : x > 2 − 1, y ≤ 2 − 1}, √ √ √ D2 = {(x, y) ∈ D : x > 2 − 1, y > 2 − 1}, D3 = {(x, y) ∈ D : x = 2 − 1}.

Figure 5.1 shows that Tˆ(z) = T τ (ℓ)(z) if z ∈ Dℓ , with τ (0) = 5, τ (1) = 9, τ (2) √ = 3, τ (3) =√11, S5 S3 {(0, y) : 1 − 1/ 2 < y < 1/ 2}, and R = {(0,√0)} ∪ k=0 T k (DA ) ∪ √k=0 T k (DB ) with DA = √ DB = {(0, 1/ 2)}. If we set U (z) = ( 2 − 1)z, V (z) = z, κ = 2 − 1, ε = −1, and σ : 0 7→ 010

1 7→ 000

2 7→ 0

3 7→ 030,

then Figure 5.2 shows that σ satisfies the conditions in Section 3, and √ P = {(x, y) ∈ D : x, y ≥ 2 − 1} = Dα ∪ Dβ ∪ Tˆ(Dβ ) ∪ Dζ √ √ √ √ √ with Dα = D2 , Dβ = {(x, 2 − 1) : 2 − 1 < x < 2 − 2} and Dζ = {( 2 − 1, 2 − 1)}. 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

15 13

02

38

32 22

B 0 12

A1

23

16 14

20 A2

A0 3

18 B2

10

04 3

30

B3 10 4 00 B

5

34

3 36

37

Tˆ →

19 311

21

05

39

1

01

03 11

B5

17 A3

33

B1

Figure 5.1. The map Tˆ and the set R, λ =

√

2. (ℓk stands for T k (Dℓ ).)

Dα

01 11 31 20 30 00

10

32

Tˆ(Dβ ) Dζ Dβ 02 21

12

03

13 33

Figure 5.2. The trajectory of the scaled domains and P, λ =

√

2. (ℓk stands for Tˆ −k U (Dℓ ).)


13

√ Theorem 5.1. If λ = 2, then the minimal period length π(z) is 1 if z = (0, 0) 4 if z = T m (0, 1/2), 0 ≤ m ≤ 3 m 8 for the other points √ of T (DA ), 0 ≤ m ≤ 3 m ≤m≤5 6 if z = T (0, 1/ √2), 0 √ 2 · 3n + (−1)n if S n R(z) = (1/ 2, 1/ 2), n ≥ 0 8(2 · 3n + (−1)n ) for the other points√with S n R(z) √ ∈ Dα n+1 n 4(3 + 1 + (−1) ) if S n R(z) ∈ {(1/2, 2 − 1), ( 2 − 1, 1/2)}, n ≥ 0 n ˆ 8(3n+1 + 1 + (−1)n ) for the other points with √ S R(z) ∈ Dβ ∪ T (Dβ ) √ n+1 n n 2·3 + 4 + (−1) if S R(z) = ( 2 − 1, 2 − 1), n ≥ 0 ∞ if S n R(z) ∈ D \ P for all n ≥ 0. Proof. We easily calculate n 1/4 |σ (0)|0 n n 3/4 , + (−1) =3 −1/4 1/4 |σ n (0)|1

n |σ (1)|0 n −3/4 n 3/4 + (−1) =3 3/4 1/4 |σ n (1)|1

and obtain τ (σ n (0)) = 2 · 3n+1 − (−1)n , τ (σ n (3)) = τ (σ n−1 (030)) = 2 · 3n+1 + 4 + (−1)n . If S n R(z) ∈ Dα and n ≥ 1, then π(z) = τ (σ n (2)) = τ (σ n−1 (0)) and π(z) = 8τ (σ n−1 (0)) respectively; if S n R(z) ∈ Dβ , then π(z) = τ (σ n (13)) = τ (σ n−1 (000030)) and π(z) = 2τ (σ n−1 (000030)) √ √ respectively; if S n R(z) = ( 2 − 1, 2 − 1), 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 sˆ(z) as follows and obtain the following s(z), t(z): √ √ z ∈ Tˆ −2 U (D0 ∪ D1 ∪ D3 ) : sˆ(z) = −1, s(z) = −5, t(z) = Tˆ −1 (z) − zA−5 = ( 2 − 1, 2 − 2) √ √ z ∈ Tˆ −1 U (D0 ∪ D1 ∪ D3 ) : sˆ(z) = 1, s(z) = 5, t(z) = Tˆ(z) − zA5 = (2 − 2, 2 − 1) √ √ √ This gives δ = (2 + 2)/ 2 = 2 + 1 since √ √ √ √ √ √ {t(z)Ah : z ∈ D \ P, h ∈ Z} = ±{(0, 0), (2 − 2, 2 − 1), ( 2 − 1, 0), (0, 1 − 2), (1 − 2, 2 − 2)}. √ √ √ Theorem 5.2. π(z) is finite for all z ∈ (Z[ 2] ∩ [0, 1))2 , but (T k ( 3−4 2 , 2 42−1 ))k∈Z is aperiodic. √ √ 2 ′ Proof. We √ have to consider z ∈ Z[ 2] ∩ D with kz k∞ ≤√δ = 2 + 1. The only such point is √ that all points in ( 2 − 1, 2 − 1) = Dζ , hence Conjecture 1.1 holds for λ = 2. It can be shown √ √ √ √ 1 1 3− 2 2 2−1 2 2 ( 2 Z[ 2] ∩ [0, 1)) and ( 3 Z[ 2] ∩ [0, 1)) are periodic as well. For z = ( 4 , 4 ), we have √ √ 9 − 6√2 √ √ √ √ 5 3 2 − 3 3 − 2 5 = , S(z) = zA + (2 − 2, 2 − 1) /κ = ( 2 + 1) , 2− , 4 √ 4 4 4 √ √ 5 − 3 2 √ √ √ √ 5 2 2 − 1 3 − 2 S 2 (z) = S(z)A5 + (2 − 2, 2 − 1) /κ = ( 2 + 1) = , , 2− , 4 4 4 4 √ √ √ √ √ √ S 3 (z) = S 2 (z)A−5 + ( 2 − 1, 2 − 2) /κ = ( 3−4 2 , 3 42−3 ) and S 4 (z) = ( 3−4 2 , 2 42−1 ) = z. √ 6. The case λ = − 2 = −2 cos π4

√ √ Let λ = − 2 (λ′ = 2) and set [ √ √ D = {(x, y) ∈ [0, 1)2 : 2x + y > 2 or x + 2y > 2} = Dℓ , ℓ∈A={0,1,2} √ √ with D0 = {(x, y) ∈ D : x + 2y > 2} and D1 = {(x, y) ∈ D : x + 2y < 2}. Figure 6.1 shows that Tˆ(z) = T τ (ℓ) (z) if z ∈ Dℓ , with τ (0) = 1, τ (1) = 21, τ (2) = 31, and [9 [3 T k (D∆ ), T k (DΓ ) ∪ R = {(0, 0)} ∪ DA ∪ DB ∪ k=0 k=0 √ √ √ DA = {(x, y) : 0 ≤ x, y ≤ 3 − 2 2} \ {(0, 0), (3 − 2 2, 3 − 2 2)}, DB = {z ∈ [0, 1)2 : T k+1 (z) = T k (z)A + (0, 1) for all k ∈ Z}, DΓ = {z ∈ [0, 1)2 : T k+1 (z) = T k (z)A + (0, 2) for all k ∈ Z},


14 1

1

0.9

0.9

0.8

0.8

0.7

0.7

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

0

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Figure 5.3. Aperiodic points, λ =

0.9

√ 2.

1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

√ Figure 5.4. Aperiodic points, λ = − 2.

√ √ √ D∆ = {(1/ 2, 0)}. Set κ = 2 − 1, V (z) = ((1, 1) − z)/κ = ( 2 + 1)((1, 1) − z), i.e., √ √ √ √ U (z) = (1, 1) − ( 2 − 1) (1, 1) − z = ( 2 − 1)z + (2 − 2, 2 − 2).

Then Figure 6.2 shows that the conditions in Section 3 are satisfied by σ : 0 7→ 010 1 7→ 000 2 7→ 020 S5 ˆ k S2 ˆ k with ε = −1 and P = Dα ∪ k=0 T (Dβ ) ∪ k=0 T (Dζ ) with

Dα = {z ∈ [0, 1)2 : T k+1 (z) = T k (z)A + (0, 3) for all k ∈ Z}, √ √ √ √ √ Dβ = {(x, 2 − 2x) : 5 − 3 2 < x < 2 2 − 2} and Dζ = {(8 − 5 2, 8 − 5 2)}. 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. √ period length is Theorem 6.1. If λ = − 2, then the minimal √ √ √ π(z) √ 1 if z ∈ {(0, 0), (1/ √2, 1/ 2),√ (2 − 2, 2 − 2)} 4 if z = T m (3/2 − 2, 3/2 − 2) for some m ∈ {0, 1, 2, 3} √ 10 if z = T m (1/ 2, 0) for some m ∈ {0, 1, . . . , 9} 8 for the other points √ in R √ 2 · 3n+1 − 5(−1)n if S n R(z) = (3 − 3/ 2, 3 − 3/ 2) for some n ≥ 0 8(2 · 3n+1 − 5(−1)n ) for the other points with√S n R(z) ∈ D √α 4(3n+2 + 5 − 5(−1)n ) if S n R(z) = Tˆ m (9 − 5 2)/2, 5 − 3 2 for some m ∈ {0, . . . , 5}, n ≥ 0 n ˆm 8(3n+2 + 5 − 5(−1)n ) for the other points with √∈ T (Dβ ) √ S R(z) n+2 n n m ˆ 2·3 + 20 − 5(−1) if S R(z) = T (8 − 5 2, 8 − 5 2) for some m ∈ {0, 1, 2}, n ≥ 0 ∞ if S n R(z) ∈ D \ P for all n ≥ 0. √ Proof. As for λ = 2, we have n n 1/4 |σ (1)|0 |σ (0)|0 n −3/4 n 3/4 n n 3/4 , + (−1) = 3 , + (−1) = 3 3/4 1/4 |σ n (1)|1 −1/4 1/4 |σ n (0)|1 hence τ (σ n (0)) = 2 · 3n+1 − 5(−1)n and τ (σ n (2)) = τ (σ n−1 (020)) = 2 · 3n+1 + 20 + 5(−1)n . For S n R(z) ∈ Dα , we have π(z) = τ (σ n (0)) and π(z) = 8τ (σ n (0)) respectively; if S n R(z) ∈ T m (Dβ ), then π(z) = τ (σ n (002000)) and π(z) = 2τ (σ n (002000)) respectively; if S n R(z) = Tˆ m (Dζ ), then π(z) = τ (σ n (020)).


16 2

230 ∆9 210

16

116

T (DΓ) 120 220 10 1

26

2

∆5 2

121 0

14

26

24

13 213

0

231 01

T 2 (DΓ)

20

Tˆ →

10

23

227 2

∆ 13

DB 2

117 ∆6

23

217 17 27

28

222 22

114 14 2 ∆3

1

∆1

24

15

112 212 2 1

DA

DΓ

219 119 19

18

∆7 21818 1

229 ∆8 29

228 25 ∆4 225

211 T 3 (DΓ) 111 11

15 15 115 2

221 ∆0 21

√ Figure 6.1. The map Tˆ and the set R, λ = − 2. (ℓk stands for T k (Dℓ ).) 11 Tˆ 3 (D ) 21 β ˆ 00 T (Dβ) 01 Dζ Tˆ 2 (Dζ) 20 Dβ 0 Dα ˆ 4 T (Dβ)1

13 23 03

Tˆ 5 (Dβ) Tˆ 2 (Dβ) ˆ T (Dζ) 12 2 02 2

√ Figure 6.2. The trajectory of the scaled domains and P, λ = − 2. (ℓk stands for Tˆ −k U (Dℓ ).) For z ∈ D \ (U (D) ∪ P), we choose sˆ(z) as follows and obtain the following s(z), t(z): z ∈ Tˆ −2 U (D0 ∪ D1 ∪ D2 ) : sˆ(z) = −1, s(z) = −1, t(z) = V (Tˆ−1 (z)) − V (z)A−1 = (1, 0)

z ∈ Tˆ −1 U (D0 ∪ D1 ∪ D2 ) : sˆ(z) = 1, s(z) = 1, t(z) = V (Tˆ(z)) − V (z)A = (0, 1)


16

01

12

21

3

2

33 3 1

11 3

04

1

0

34

3

05 2

4

23

25

20

0 26 0

10

14

Tˆ →

02

06 30

22

27

35

Figure 7.1. The map Tˆ, λ = 1/γ. (ℓk stands for T k (Dℓ ).) This gives δ =

√ √ 2/ 2 = 1 since

√ √ {t(z)Ah : z ∈ D \ P, h ∈ Z} = ±{(0, 0), (1, 0), (0, 1), (1, − 2), (− 2, 1)}. √ √ Theorem 6.2. π(z) is finite for all z ∈ (Z[ 2] ∩ [0, 1))2 , but (T k ( 43 , 5−4 2 ))k∈Z is aperiodic. √ √ Proof. √ Since V (D) = {(x, y) : x > 0, y > 0, x + 2y < 1 or 2x + y < 1}, there √ exists no z ∈ Z[ 2]2 ∩ D with k(V (z))′ k∞√≤ 1. Therefore Conjecture 1.1 holds for λ = − 2. It can √ be shown that all points in ( 12 Z[ 2] ∩ [0, 1))2 and ( 13 Z[ 2] ∩ [0, 1))2 are periodic as well. For √ √ 1 z = ( 34 , 5−4 2 ), we have V (z) = ( 2+1 4 , 4 ), 1 3 − 2√2 √2 + 1 √2 − 1 √ √ V S(z) = ( 2 + 1)(V (z)A + (0, 1)) = ( 2 + 1) , = , , 4 4 4 4 V S 2 (z) = ( 14 ,

√ 2+1 4 ),

V S 3 (z) = (

√ √ 2−1 2+1 , 4 4 )

and V S 4 (z) = (

√ 2+1 1 4 , 4)

= V (z).

7. The case λ = 1/γ = −2 cos 3π 5

Let λ = 1/γ (λ′ = −γ) and set

D = {(x, y) ∈ [0, 1)2 : γx − 1 < y < x/γ} =

with D0 , D1 , D2 , D3 satisfying the (in)equalities D0 y > x − 1/γ 2

D1 0 < y < x − 1/γ 2

D2 y = x − 1/γ 2

[

ℓ∈A={0,1,2,3}

Dℓ ,

D3 y = 0, 1/γ 2 < x < 1/γ

Figure 7.1 shows that Tˆ(z) = T τ (ℓ)(z) if z ∈ Dℓ , with τ (0) = 6, τ (1) = 4, τ (2) = 7, τ (3) = 5, and R = {(0, 0)}. If we set U (z) = z/γ 2, V (z) = z, κ = 1/γ 2 , ε = 1, and σ : 0 7→ 010

1 7→ 01110

2 7→ 012

3 7→ 01112

then Figure 7.2 shows that σ satisfies the conditions in Section 3, and [1 [3 Tˆ k (Dη ) ∪ Dµ Tˆ k (Dζ ) ∪ Dϑ ∪ P = Dα ∪ Dβ ∪ k=0

k=0

with Dα = {z ∈ D : Tˆ k (z) ∈ D0 for all k ∈ Z}, Dβ = {z ∈ D : Tˆ k (z) ∈ D1 for all k ∈ Z}, Dζ = {(x, 0) : 1/γ 3 < x < 1/γ 2 }, Dη = D3 , Dϑ = {(1/γ 3 , 0)} and Dµ = {(1/γ 2, 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.


15 0

2

17

22 12

03 23

35 Tˆ(Dζ) Tˆ(Dϑ)

31 3

Dα 0

0

20 10

Tˆ 3 (Dζ) 11 2 ˆ T (Dζ) 21 01

3 0 Dϑ Dζ Dµ

14

32 34D

β

3

13

Dη

Figure 7.2. The trajectory of the scaled domains and P, λ = 1/γ. (ℓk stands for Tˆ k U (Dℓ ).) Theorem 7.1. If λ = 1/γ, then the minimal period length π(z) is 1 if z = (0, 0) n 2(5 · 4 + 4)/3 if S n R(z) = γ 2γ+1 , γ1/γ for some n ≥ 0 2 +1 10(5 · 4n + 4)/3 for the other points with S n R(z) ∈ Dα 2 4(5 · 4n − 2)/3 if S n R(z) = γ 2γ+1 , γ 21+1 for some n ≥ 0 20(5 · 4n − 2)/3 for the other points with S n R(z) ∈ Dβ 5(4n+1 − 1)/3 if S n R(z) = (0, 1/2) for some n ≥ 0 n+1 10(4 − 1)/3 for the other points with S n R(z) ∈ Dϑ n+1 5(2 · 4 + 7)/3 if S n R(z) = Tˆ m (1/(2γ), 0) for some m ∈ {0, 1, 2, 3} and n ≥ 0 n+1 10(2 · 4 + 7)/3 for the other points with S n R(z) ∈ Tˆm (Dζ ) n (10 · 4 + 11)/3 if S n R(z) = (1/γ 2 , 0) for some n ≥ 0 (5 · 4n+1 + 19)/3 if S n R(z) = Tˆ m (1/γ 3 , 0) for some m ∈ {0, 1} and n ≥ 0 ∞ if S n R(z) ∈ D \ P for all n ≥ 0. Proof. As for λ = −1/γ, we have n 2/3 |σ (0)|0 n 1/3 + = 4 , 1/3 −1/3 |σ n (0)|1

|σ n (1)|0 |σ n (1)|1

= 4n

2/3 −2/3 + , 2/3 1/3

8 20 n 8 10 n 11 20 n 5 n n n n hence τ (σ n (0)) = 10 3 4 + 3 , τ (σ (1)) = 3 4 − 3 , τ (σ (2)) = 3 4 + 3 , τ (σ (3)) = 3 4 − 3 . n n n n For S R(z) ∈ Dα , we have π(z) = τ (σ (0)) and π(z) = 5τ (σ (0)) respectively; if S R(z) ∈ Dβ , then π(z) = τ (σ n (1)) and 5τ (σ n (1)) respectively; if S n R(z) ∈ Dη , then π(z) = τ (σ n (3)) and 2τ (σ n (3)) respectively; if S n R(z) ∈ Dζ , then π(z) = τ (σ n (0002)) and 2τ (σ n (0002)) respectively; if S n R(z) = Tˆ m (1/γ 3 , 0), then π(z) = τ (σ n (02)); if S n R(z) = (1/γ 2 , 0), then π(z) = τ (σ n (2)).

Note that Tˆ m U (D3 ) plays no role in the calculation of δ since U (D3 ) ⊂ U (P) and thus π(z) < ∞ for all z ∈ Tˆ m U (D3 ). For the other z ∈ D \ (P ∪ U (D)), we choose sˆ(z) as follows: z ∈ Tˆ 2 U (D0 ∪ D1 ∪ D2 ) : sˆ(z) = −2, s(z) = −10, t(z) = Tˆ −2 (z) − z = (−1/γ, −1/γ 2) z ∈ TÛ (D1 ∪ D2 ) : sˆ(z) = −1, s(z) = −6, t(z) = Tˆ −1 (z) + zA−1 = (1, 1/γ) z ∈ Tˆ4 U (D1 ) : sˆ(z) = 1, s(z) = 6, t(z) = Tˆ(z) + zA = (1/γ, 0)

z ∈ TÛ (D0 ) ∪ Tˆ3 U (D1 ) : sˆ(z) = 2, s(z) = 10, t(z) = Tˆ 2 (z) − z = (−1/γ 2, 0)


18 1

1

0.9

0.9

0.8

0.8

0.7

0.7

0.6

0.6

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

0

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

Figure 7.3. Aperiodic points, λ = 1/γ.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 7.4. Aperiodic points, λ = −γ.

This gives again δ = γ 2 /γ = γ since {(1/γ, 0)Ah : h ∈ Z} = ±{(1/γ, 0), (0, 1/γ), (1/γ, 1), (1, 1), (1, 1/γ)}. Theorem 7.2. π(z) is finite for all z ∈ (Z[γ] ∩ [0, 1))2 , but T k 1/4, 1/(4γ 3) k∈Z is aperiodic.

Proof. Conjecture 1.1 holds for λ = 1/γ since no z ∈ Z[γ]2 ∩D satisfies kz ′ k∞ ≤ γ. It can be shown 3 that all points in ( 12 Z[γ] ∩ [0, 1))2 and ( 13 Z[γ] ∩ [0, 1))2 are periodic aswell. If z = 1/4, 1/(4γ ) , then we have S(z) = γ 2 /4, 1/(4γ) , S 2 (z) = γ 2 S(z) − (1/γ 2 , 0) = (3γ − 2)/4, γ/4 and S 3 (z) = γ 2 S 2 (z) − (1/γ, 1/γ 2) = 1/4, 1/(4γ 3) = z. 8. The case λ = −γ = −2 cos π5

Let λ = −γ (λ′ = 1/γ) and set

D = {(x, y) ∈ [0, 1)2 : x < y, γx + y ≥ 4 − γ} = D0 ∪ D1 with D0 = {(x, y) ∈ D : x > 1 − 1/γ 5}, D1 = {(x, y) ∈ D : x ≤ 1 − 1/γ 5}. Figure 8.1 shows that Tˆ(z) = T τ (ℓ)(z) if z ∈ Dℓ , with τ (0) = 42, τ (1) = 28, and [10 [24 [1 [4 T k (DZ ) T k (DE ) ∪ T k (D∆ ) ∪ T k (DΓ ) ∪ R = {(0, 0)} ∪ DA ∪ DB ∪ k=0

k+1

k=0

k

k=0

k=0

k+1

k

with DA = {z : T (z) = T (z)A + (0, 1) for all k ∈ Z}, DB = {z : T (z) = T (z)A + (0, 2)}, D∆ = {z ∈ [0, 1)2 : T 2k+1 (z) = T 2k (z)A + (0, 2), T 2k (z) = T 2k−1 (z)A + (0, 1) for all k ∈ Z}, DΓ = {(x, y) : 0 ≤ x, y ≤ 1/γ 4} \ {(0, 0), (1/γ 4, 1/γ 4 )}, DE = {(x, x) : 1 − 1/γ 5 < x < 1}, DZ = {(1 − 1/γ 5 , 1 − 1/γ 5 )}. Set κ = 1/γ 2, V (z) = γ 4 (1, 1) − z), i.e. U (z) = (1, 1) − (1, 1) − z /γ 2 = z/γ 2 + (1/γ, 1/γ).

Then Figure 8.2 shows that the conditions in Section 3 are satisfied by ε = 1 and σ : 0 7→ 010

1 7→ 01110.

All points in P = Dα ∪ Dβ are periodic, with Dα = {z ∈ D : Tˆ k (z) ∈ D0 for all k ∈ Z}, Dβ = {z ∈ D : Tˆ k (z) ∈ D1 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.


29

27 T (D∆ )

T (DΓ )

Z8 E 8 08

214 22 E 022 8 1

024 110 010 E 10 E 24 216 Z 10 T 3 (DΓ )

5 25 Z 211 19 5 E 5 5 19E 0 1 0

014 10 00

E

Z0 E 11

016

E 12 012

15 023 223 E

E 15 15 Z1 0 11

E6 06 26 14

16 020

04 E

24 4

17 Z

017 E 17

07 E7

10

210

Z6 E 20 212

03 E 3 13 23

21

7

011

T 2 (DΓ )

E 18 018

E 21 0 213

111

21 1 01 E

DA

DΓ

025

E 16

E9 09 Z9 9 1

Z4

217

DB

Z2 12

E0 20

14

013 E 13 22 E2 02

19

D∆

T 4 (DΓ )

28

Z3

042

00

Tˆ →

128

Figure 8.1. The map Tˆ and the set R, λ = −γ. (ℓk stands for T k (Dℓ ).) 01 1

3

1

10

1

Dβ 12

Dα

00

03 1

5

14 02

Figure 8.2. The trajectory of the scaled domains and P, λ = −γ. (ℓk stands for Tˆk U (Dℓ ).)

20


Theorem 8.1. If λ = −γ, then the minimal period length π(z) is 1 if z ∈ {(0, 0), (1/γ 2, 1/γ 2 ), (2/γ 2 , 2/γ 2 )} 2/γ 2 2/γ 2 5−γ 2 if z ∈ {( γ5−γ 2 +1 , γ 2 +1 ), ( γ 2 +1 , γ 2 +1 )} 5 if z = T m (1/(2γ 4 ), 1/(2γ 4 )) for some m ∈ {0, 1, 2, 3, 4} 10 for the other points of DA , DB , T m (DΓ ), T m (D∆ ) 11 if z = T m (1 − 1/γ 5 , 1 − 1/γ 5 ) for some m ∈ {0, 1, . . . , 10} 25 if z = T m (1 − 1/(2γ 5), 1 − 1/(2γ 5)) for some m ∈ {0, 1, . . . , 24} 50 for the other points of T m (DE ) n 2(35 · 4 + 28)/3 if S n R(z) is the center of Dα 10(35 · 4n + 28)/3 for the other points of Dα 4(35 · 4n − 14)/3 if S n R(z) is the center of Dβ 20(35 · 4n − 14)/3 for the other points of Dβ ∞ if S n R(z) ∈ D \ P for all n ≥ 0 Proof. As for λ = −1/γ and λ = 1/γ, we have n 2/3 |σ (0)|0 n 1/3 + , = 4 1/3 −1/3 |σ n (0)|1

|σ n (1)|0 |σ n (1)|1

= 4n

2/3 −2/3 + , 2/3 1/3

hence τ (σ n (0)) = (70 · 4n + 56)/3, τ (σ n (1)) = (140 · 4n − 56)/3. For S n R(z) ∈ Dα , we have π(z) = τ (σ n (0)) and 5τ (σ n (0)) respectively; if S n R(z) ∈ Dβ , then π(z) = τ (σ n (1)) and 5τ (σ n (1)) respectively. We choose sˆ(z) as follows and obtain the following s(z), t(z): z ∈ Tˆ 2 U (D0 ∪ D1 ) : sˆ(z) = −2, s(z) = −70, t(z) = V (Tˆ−2 (z)) − V (z) = (−1/γ 2, −1/γ 2 ) z ∈ TÛ (D1 ) : sˆ(z) = −1, s(z) = −42, t(z) = V (Tˆ−1 (z)) − V (z)A−2 = (1/γ, 1/γ)

z ∈ Tˆ4 U (D1 ) : sˆ(z) = 1, s(z) = 42, t(z) = V (Tˆ(z)) − V (z)A2 = (1, 0)

z ∈ TÛ (D0 ) ∪ Tˆ3 U (D1 ) : sˆ(z) = 2, s(z) = 70, t(z) = V (Tˆ 2 (z)) − V (z) = (−1/γ, 0) This gives again δ = γ 2 /γ = γ since {(1, 0)Ah : h ∈ Z} = ±{(1, 0), (0, 1), (1, −1/γ), (1/γ, 1/γ), (1/γ, −1)}. Theorem 8.2. π(z) is finite for all z ∈ (Z[γ] ∩ [0, 1))2 , but π 1 − 1/(3γ 2), 1 − 1/(3γ 5) = ∞.

Proof. Since V (D) = {(x, y) : x > y > 0, γx + y ≤ γ}, we have no point z ∈ Z[γ]2 ∩ D with kV (z)′ k∞ ≤ γ, and Conjecture 1.1 holds for λ = −γ. If V (z) = γ 2 /3, 1/(3γ) , then we have 1 2 γ 1 1 γ 2 + 1 2 V S(z) = γ 2 V (z) − , V S 2 (z) = γ 2 V S(z) − , = , ,0 = , , γ 3 3 γ2 γ2 3γ 3γ 1 1 4 2 3 V S 3 (z) = γ 2 V S 2 (z) − γ12 , γ12 = 3γ−2 = V (z). 3 , 3γ 3 and V S (z) = γ V S (z) − γ , 0 9. The case λ =

√ 3 = −2 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, 1))2 are periodic and refrain from calculating the period lengths. Furthermore we postpone the determination of Tˆ and R to Appendix A. Let √ √ √ √ √ √ √ √ D = {(x, y) : 2x− 3y < 2− 3, 2y− 3x < 2− 3, y− 3x < 195−113 3, x− 3y < 195−113 3} and D1 = D \ D2 , where D2 is defined by the inequalities √ √ √ √ √ √ √ √ 2x − 3y > 267 − 154 3, 2y − 3x > 267 − 154 3, y − 3x > 98 − 57 3, x − 3y > 98 − 57 3. The sets D1 and D2 have to be treated separately because their trajectories are disjoint, and both sets contain aperiodic points. The trajectories of aperiodic points in D1 come arbitrarily


21

(1, 1)

00

Tˆ →

50

60

10

1{3175,3307}

6 3

3

0

0

53021

1{3175,3307}

10

20 80 90 0 40 7

0{1601,1733}

79799 49771

63593 97907

63593

7406

23230 8

√ √ (72 − 41 3, 72 − 41 3)

11473

√ √ (73/2 − 41 3/2, 73/2 − 41 3/2)

20

2{3175,3307}

40 1 30 0

0

0

43593 Tˆ → 0

1

15524

318171 {19327,19459}

√ √ (72 − 41 3, 72 − 41 3)

Figure 9.1. The first return map on D1 and D2 respectively, λ =

√

3. (ℓk stands for T k (Dℓ ).)

√ √ √ close √to (1, 1), whereas √ 4 (72 − 41 3, 72 − 41 3) is a limit point in D2 . (Note that 72 − 41 3 = 1 − ( 3 + 1)(2 − 3) ≈ 0.9859.) The scaling maps are √ √ √ U1 (z) = (2 − 3)z + ( 3 − 1, 3 − 1) = V1−1 (κV1 (z)) for z ∈ D1 , √ √ √ −1 U2 (z) = (2 − 3)z + (113 3 − 95, 113 3 − 195) = V2 (κV2 (z)) for z ∈ D2 , √ √ 5 √ 4 with κ = 2 − 3, V1 (z) = (1, 1) − z /κ , V2 (z) = z − (72 − 41 3, 72 − 41 3) /κ . Then we have √ √ √ √ V1 (D) = {(x, y) : 2x > 3y, 2y > 3x, x > 3y − 2, y > 3x − 2}, √ √ √ √ √ √ V2 (D2 ) = {(x, y) : 2x > 3y, 2y > 3x, x > 3y − 2 − 3, y > 3x − 2 − 3}.


22

The first return map Tˆ induces a partition of D1 into sets D0 , . . . , D9 and a partition of D2 into sets D0 , . . . , D4 , as in Figure 9.1. These sets are defined by the following (in)equalities: V1√ (D1 ) V1√ (D0 ) x > 3y − 1 x > 3y − 1 x 3x − 1

y

V1 (D2 )√ √ √V1 (D3 )√ 3y + 3 − 1 2y > 3x + √3 − 1 x>2 x > 2, y < 2 3 − 1

V1 (D √ 7) y = 2 3 −√1 x < 3 − 1/ 3

√ V2 (D1 ) √ 3x − 1, x < 3 + 1

2x =

V1 (D8 )√ √ 3y + 3 − 1 x>2

V2√ (D2 ) x> 3+1

2y =

√V1 (D4 )√ 2y > 3x√+ 3 − 1 y >2 3−1

√V1 (D9 )√ 3x + 3 − 1 x>2

V2√(D3 ) V2√ (D4 ) y = 3x − 1 x = 3 + 1

The return times of z ∈ Dℓ to D are given by the following tables. D0 D1 D2 D3 D4 D5 D6 D7 D8 D9 1601, 1733 3175, 3307 3230 7406 9771 3021 3593 9799 11473 7907 D0 D1 D2 D3 D4 19459 15524 3175, 3307 18171 3593 Note that the return times are not constant on all Dℓ . E.g., the return time for z ∈ D0 is 1601 if V1 (z) = (1, y) and 1733 else, see Appendix A for details. Since we do not calculate the period lengths, it is not necessary to distinguish between the parts of Dℓ with different period lengths. 9.1. The scaling domain D1 . Figure 9.2 shows the of the open scaled sets in D1 . √ √ trajectory Here, V1 (D1 ) is split up into the three stripes x < 3 −√1, 3 − 1 < x < 2 and x > 2, and D˜1 denotes the set given by V1 (D˜1 ) = {(x, y) ∈ V1 (D) : x > 3y − 1, x < 2}. We see that 0 7→ 010 3 7→ 012100001210 5 7→ 7 01210000000001210 6 → 7 σ1 : 1 7→ 01110 4 → 2 7→ 01210

01510 7 01610 8 9

7→ 01210000500001210 7 → 01210012621001210 7→ 0121005001210

codes the trajectory of U1 (Dℓ ), ℓ ∈ {0, 1, 2, 3, 4}, with Tˆ|σ1 (ℓ)| U1 (z) = U1 Tˆ(z) for z ∈ Dℓ . All points in Dα , Dβ and Dγ are periodic. Figure 9.4 shows that Dε˜, Dζ˜, Dη˜ and the grey part of U1 (D˜1 ) 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 V1 (D1 ) is split up into √ the stripes x < 3 − 1, 3 − 1 ≤ x < 2 and x ≥ 2. Here, D¯1 denotes boundary lines of D1 , and D˜6 is given by V1 (D˜6 ) = {(2, y) ∈ V1 (D)}. We see that σ1 codes the trajectory of U1 (Dℓ ), ℓ ∈ {5, 6, 7, 8, 9}, as well and satisfies the conditions in Section 3 (with respect to D1 ). All points in Dι , Dκ , Dλ , Dµ , Dν , Dξ , Do , Dπ , Dρ (and their orbits) are periodic. The finitely many remaining points in P1 = {z ∈ D1 : Tˆ m (z) 6∈ U1 (D1 ) for all m ∈ Z} are clearly periodic as well. Since |σ1n (ℓ)| → ∞ for all ℓ ∈ {0, . . . , 9}, we can use Proposition 3.5 to show the following proposition. √ √ Proposition 9.1. π(z) is finite for all z ∈ Z[ 3]2 ∩ D1 , but π(V1−1 ( 3 + 1/4, 7/4)) = ∞. Proof. First we show that only D0 and D1 contain aperiodic points: D3 , D4 , D7 , D8 , D9 lie in P1 . The only part of D2 which is not in P1 or Tˆ m U1 (P1 ), lies in Tˆ2 U1 (D2 ). By iterating this argument on Tˆ2 U1 (D√ 2 ), the possible set of aperiodic points in D2 becomes smaller and smaller, and converges to V1−1 (2, 3) 6∈ D2 . A similar reasoning shows that all points in D5 and D6 are periodic. Therefore it is sufficient to determine t(z) for points in the trajectories of U1 (D0 ∪ D1 ). √ √ z ∈ TÛ1 (D0 ) ∪ Tˆ 3 U1 (D1 ) : sˆ(z) = 2, s(z) ≡ 0 mod 12, t(z) = (1 − 3)( 3, 2) √ z ∈ Tˆ 4 U1 (D1 ) : sˆ(z) = 1, s(z) ≡ 5 mod 12, t(z) = V1 (Tˆ(z)) − V1 (z)A5 = ( 3, 2) √ z ∈ TÛ1 (D1 ) : sˆ(z) = −1, s(z) ≡ −5 mod 12, t(z) = (2, 3) √ √ z ∈ Tˆ2 U1 (D0 ) ∪ Tˆ 2 U1 (D1 ) : sˆ(z) = −2, s(z) ≡ 0 mod 12, t(z) = (1 − 3)(2, 3)


23

00

03

37 12 4 47 410

→

˜ 15

4

25

η˜4 20 β0 β

24

40 β 5

44

β4 ζ˜4

3 ˜5 ζ

17

16

411

4 46 22 β 36 η˜ 17 1 311 β 6 4 β 1 1 2 η˜3 1 β 6 3 ˜ ζ ζ˜1 8

45 35 ˜ 14 02

˜ 11

˜ 12

ζ˜2

34

0

18

Dα

49

ζ˜0

η˜9 312

˜ 10

η˜6 22 β2

48

β 19 η˜5 1 η˜ β 15

32 ˜7 ζ

β 20

38

413

42 β 7

39 414

Dγ

01

β 8˜8 ζ

Tˆ(Dγ )

β9 13

β ˜13 ζ ˜9 ζ

˜ 13 η˜210 3 β 21 3 2 β 16 ζ˜3 15 4 β3 43 3 3

η˜0

ε˜1

7

η˜

β 14 ζ˜14

β 10 ζ˜11

ζ˜12

ε˜0 ˜10 ζ

→

ε˜2

β 11 β 12

Figure 9.2. The trajectory of the open scaled sets in D1 and the set P1 , λ = (ℓk stands for Tˆ k U1 (Dℓ ) if ℓ ∈ {0, ˜1, 2, 3, 4}, for Tˆk (Dℓ ) else.)

√

3.


24

77

ι0 ι3

ι5

710

50 ¯ 15

¯ 10

˜ 65 ν 24 913

812

817

ι7

79

60 ν0 ˜

ν 29

717

ι8

55

→

98

712

74 80

ν4 90

ν5

70 85 95

˜ 64

75 54

ν 28

816 912 716 811

94 84

62 ν2 ˜ ν 26

78

ι9 ι4 711 76 97 96

86 1 ν 23 7191 8 0 µ 6 µ1 ν ν 1 ˜ 61

¯ 14

ι6

ι2 ι1

814

713

99

910

82 92 2 7 ν7 87 µ2

714

813 ν 25

89

ξ0

ν 21 ξ6

λ0 ¯ 11 λ1

ξ7

51

52 ¯ 12

ν8

ξ4 λ3

15 κ1 ν 13 ν 9 1 5 ν ρ˜ ρ˜ κ0

53

¯ 13

ν 19

ν 20ν 14 ρ˜0 ν 16

˜ 63

µ3 22 ν 15 810 7911815 83

ν 27

ρ˜3 ν 17 ν 11 ν3

ρ˜2

ν 12

ρ˜4

88

ξ1

ξ2

λ2

ν 10

ξ5

ξ3 ρ˜6 →

ν 18

93 73

Figure 9.3. The trajectory of the scaled lines and the set P1 , λ = (ℓk stands for Tˆ k U1 (Dℓ ) if ℓ ∈ {¯1, 5, ˜6, 7, 8, 9}, for Tˆ k (Dℓ ) else.)

√ 3.


η˜0 ρ˜0

ε˜0

˜14 ρ˜6 ζ →

˜ 10

ε˜2

60 ζ˜0 ˜

˜15

˜65 η˜9

π 18 ζ 18 ρ10

η18 π 2828 ζ ρ0 ζ 14 ρ6 η14

ρ16 ζ

ζ 22 14 ρ

25

η32 π 57 π 42

o35 ζ 42 o10

π 14 π 24 ζ 57 o25

Dδ

Dε

ρ12 ζ 20

24

ζ 67

o20ζ 52

15

ζ 47 o

π 47 π0 0 η0 ζ 0 o

ρ18 ζ 26

ρ8

η

o5 ζ 37

ζ 16 π 26 16 π 16

η

π 37 27

o30 ζ 62 π 52

Figure 9.4. Small parts of P1 , λ =

√

3. (ℓk stands for Tˆ k (Dℓ ) for ℓ 6∈ {˜1, ˜6}.)

√ √ We have δ1 = ( 3 + 1)2/( 3 + 1) = 2 since √ √ √ √ √ {( 3, 2)Ah : h ∈ Z} = ±{( 3, 2), (2, 3), ( 3, 1), (1, 0), (0, 1), (1, 3)}. √ The only point√z ∈ V1 (Z[ 3]2 ∩ D1 ) with kz ′ k∞ ≤ 2 is (1, 1) ∈ V1 (Dα ). If V1 (z) = ( 3 + 1/4, 7/4), then we have √ √ √ √ √ V1 S(z) = (2 + 3) V1 (z) + (1 − 3)(2, 3) = (3/2 + 3/4, 3 3/4 + 1/2), √ √ √ √ V1 S 2 (z) = (2 + 3) V1 S(z) + (1 − 3)(2, 3) = (7/4, 3 + 1/4), √ √ √ √ √ V1 S 3 (z) = (2+ 3) V1 S 2 (z)+(1− 3)( 3, 2) = (3 3/4+1/2, 3/2+ 3/4), V1 S 4 (z) = V1 (z). Remark. The primitive part of σ1 is again 0 7→ 010, 1 7→ 01110.

9.2. The scaling domain D2 . Figure 9.5 shows √ the trajectory of the the scaled domains in D2 . √ Here, V2 (D2 ) is split up into x ≤ 3 + 1 and x > 3 + 1. With ε2 = 1 and 0 σ2 : 1 2

7→ 01222222210 3 7→ 012242210 7 → 012210 4 7→ 030 7→ 0


26

13 6 08 3 33 ω7 ω2

34 14 37 9 ω 0 0

03 ψ1

ψ0

06 ω5

Dχ

38 04 ω 3

15

31

20

35

32 12

21 43

40 10

16

→

39

30 00

01

41

ω6 07

ω1

010 ψ2 ψ3

11

42

02

Dϕ

ω 4 05

011

Figure 9.5. The trajectory of the scaled domains in D2 and the set P2 , λ = (ℓk stands for Tˆ k (Dℓ ) if ℓ ∈ {ψ, ω}, for Tˆ k U2 (Dℓ ) else.)

√

3.

the conditions in Section 3 are satisfied. The set P2 = {z ∈ D2 : Tˆ m (z) 6∈ U2 (D2 ) for all m ∈ Z} consists of the orbits of Dϕ , Dχ , Dψ , Dω and several isolated (periodic) points. Since |σ2n (ℓ)| → ∞ for all ℓ ∈ {0, 1, 2, 3, 4}, we can use Proposition 3.5 to show the following proposition. √ √ Proposition 9.2. π(z) is finite for all z ∈ Z[ 3]2 ∩ D2 , but π(V2−1 (5/7, 3 3/7)) = ∞.

PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 1

27

1

0.9 0.998 0.8 0.996

0.7 0.6

0.994

0.5 0.992 0.4 0.3

0.99

0.2 0.988 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Figure 9.6. Aperiodic points, λ =

0.9

√

3.

1

0.986 0.986

0.988

0.99

0.992

0.994

0.996

0.998

Figure 9.7. Aperiodic points in D1 ∪ D2 , λ =

1

√ 3.

Proof. Similarly to D1 , we see that all points in D3 and D4 are periodic. Choose sˆ(z) as follows: √ z ∈ Tˆ 10 U2 (D0 ) ∪ Tˆ 5 U2 (D1 ) : sˆ(z) = 1, s(z) ≡ 7 mod 12, t(z) = (2, 3) √ √ z ∈ Tˆ 9 U2 (D0 ) ∪ Tˆ 4 U2 (D1 ) : sˆ(z) = 2, s(z) ≡ 3 mod 12, t(z) = (1 − 3, 3 − 1) √ z ∈ Tˆ 8 U2 (D0 ) ∪ Tˆ 3 U2 (D1 ) : sˆ(z) = 3, s(z) ≡ 10 mod 12, t(z) = (1 − 3, −3) √ √ z ∈ Tˆ 7 U2 (D0 ) : sˆ(z) = 4, s(z) ≡ 5 mod 12, t(z) = 3( 3, 2) √ z ∈ Tˆ 6 U2 (D0 ) : sˆ(z) = 5, s(z) ≡ 0 mod 12, t(z) = −2( 3, 2)

For the z ∈ Tˆm U2 (D0 ∪ D√1 ), sˆ(z), s(z) and t(z) are obtained by symmetry. The sets √ remaining √ {(1 − 3, 3 − 1)Ah : h ∈ Z} and {( 3 − 1, 3)Ah : h ∈ Z} are √ √ √ √ √ √ √ √ ±{(1 − 3, 3 − 1), ( 3 − 1, 2), (2, 3 + 1), ( 3 + 1, 3 + 1), ( 3 + 1, 2), (2, 3 − 1)}, √ √ √ √ √ √ √ √ √ √ ±{( 3 − 1, 3), (3, 2 3 + 1), (2 3 + 1, 3 + 3), (3 + 3, 2 + 3), (2 + 3, 3), ( 3, 1 − 3)}, √ √ √ √ hence √ δ2 = 4/(√ 3 + 1) = 2(√ 3 − 1). The only x ∈ Z[ 3] with√0 < x < 5 and |x′ | ≤ 2( 3 − √ 1) are 2 ′ the only z ∈ V (Z[ 3] ∩ D ) with kz k ≤ 2( 3 − 1) 1, 1 + 3, 2 + 3 and 3 + 3. Therefore 2 2 ∞ √ √ √ √ are (1, 1), the center of V2 U2 (Dχ ), (1 + 3, 1 + 3), the center of D4 , (2 + 3, 2 + 3), the center √ √ of Dχ , and (3 + 3,√ 3 + 3), a fixed point of Tˆ 3 . √ 3/7), then we have V S(z) = (2 + 3)V√2 (z) and If V2 (z) = (5/7, 3 2 √ √ √ 2 3 V2 S (z) = (2 + 3) V2 S(z)A + (1 − 3, 3 − 1) = (5/7, 3 3/7) = V2 (z). By combining Propositions 9.1 and 9.2 and the fact that all points in R are periodic (see Appendix A), we obtain the following theorem. √ Theorem 9.3. Conjecture 1.1 holds for λ = 3.

Remark. The eigenvalues corresponding to the primitive part of σ2 (ℓ ∈ {0, 1, 2}) are 5, −2 and 1. √ 10. The case λ = − 3 = −2 cos π6 √ √ √ √ Let D = {(x, y) ∈ [0, 1)2 : x + 3y > 5 3 − 6 or y + 3x > 5 3 − 6}, U1 as in Section 9 and √ √ √ U (z) = U12 (z) = (2 − 3)2 z + (4 3 − 6, 4 3 − 6) = V −1 (κV (z)), √ κ = (2 − 3)2 , V (z) = (1, 1) − z /κ. Then we have √ √ V (D) = {(x, y) : x > 0, y > 0, x + 3y < 1 or y + 3x < 1}.


28

20

20 6

0

40

50

10 059

30 00

3183 Tˆ →

686 142 5353 486

285 285

√ Figure 10.1. The map Tˆ on D, λ = − 3. (ℓk stands for T k (Dℓ ).) Figure 10.1 shows the first return map Tˆ on D, which is determined in Appendix B. The sets D0 , . . . , D6 satisfy the (in)equalities √ V (D0 ) 3x + y < 1

V (D1 ) √ √ 3x + y > 1, x < 3 − 1

x>

V (D2 ) √ √ √ 3 − 1, 2x + 3y 6= 3

V (D4 ) V (D V√(D6 ) √ V (D3 ) √ 5) √ √ 3x + y = 1, x < 1/2 3x + y = 1, x > 1/2 x = 3 − 1 2x + 3y = 3 √ The remaining point z = V −1 (1/2, 1 − 3/2) has return time 183 and satisfies Tˆ 10 (z) = z. Figure 10.2 shows that the first return map on U1 (D) differs from U1 TÛ1−1 on several lines. Therefore we add the lines D7 , D8 , D9 satisfying the following (in)equalities V (D8 ) V (D9 ) √ √ √ √ 3x + 2y = 1, x > 2 − 3 3x + 2y = 1, x < 2 − 3 √ and define D˜0 = D0 \ V −1 ({(x, y) : 3x + 2y = 1}), D˜2 = D2 ∪ D6 . For z ∈ Dℓ˜, ℓ ∈ {0, 2} and z ∈ Dℓ , ℓ = 1, we have Tˆ|σ1 (ℓ)| U1 (z) = U1 Tˆ(z) with √ V (D7 ) 3x + y = 1

σ1 :

0 5

7 → 020 1 7→ 010440 7

7 → 0104 10 2 7→ 0109 10 7→ 050 8 → 7 0604 10 9

7→ 0609 309 40

Figure 10.3 shows that the substitution σ given by σ(ℓ) = σ1 σ2 (ℓ) with σ2 :

0 → 7 020 1 7→ 0104 10 2 7→ 5 5 3 → 7 050 90 80 4 → 7 0504 10 5 → 7

010910 0104704 10 6 7→ 0104 80

satisfies the conditions in Section 3 (with ε = 1). The coding of the return path of the remaining point is σ1 (0504 704 80). √ √ Theorem 10.1. π(z) is finite for all z ∈ (Z[ 3] ∩ [0, 1))2 , but π(V −1 (2/7, 3/7 + 1/7) = ∞. Proof. First we show that all points on the lines U1n (Dℓ ), ℓ ∈ {3, . . . , 9}, n ≥ 0, are periodic. The only possibly aperiodic part of D5 is TÛ1 (D7 ), and the only possibly aperiodic √1 (D7 ) is √ part of U Tˆ 23 U12 (D5 ). Inductively, the set of aperiodic points in D5 converges to V −1 ( 3−1, 1−1/ 3) 6∈ D5 and is therefore empty. Therefore, all points in U n (D5 ) and U n U1 (D7 ) are periodic. Similar arguments show that all points in U n (D3 ) in U n U1 (D9 ) are periodic, then the same holds for U n (D4 ) and U n U1 (D5 ), for U n (D6 ) and U n U1 (D8 ), and finally for U n (D8 ) and U n U1 (D6 ). Then it is clear that all points in U n U1 (D3 ∪ D4 ) and U n (D7 ∪ D9 ) are periodic as well.


˜1 0

91 ˜ 01 81

ζ

1

71 11 51

ζ

16 ˜26 916 ˜8 83 98 2 6 5 13 96 918 86 921 10 ζ ˜ 211 Tˆ(Dβ ) 911 4 ˜1

˜ 20 50

533

9

˜ 00

70

ζ6

90

˜ 23 9

˜ 00

Dβ

13

ζ0 910

˜ 24

ζ

80

10

ζ3

914

2

29

˜ 210 ζ 9

Dα

7

9

4

5 920 8

54 14 84

15

919 ζ8

˜ 29

95

55 ˜ 25

99

915 9

12 2 ˜ 22 9

917 97 5 ˜ 27 2 2 ζ5 1 2 ˆ 8 T (Dβ ) 2

˜ 03

923 ˜ 03 5

→

8

73 18 88 ˜ 213

˜ 212 8

7

ζ 11 17 ζ2

72 57

˜ 02 922 ˜ 02

√ Figure 10.2. Trajectory of U1 (D) and large parts of P, λ = − 3. (ℓk stands for Tˆk U1 (Dℓ ).)

Therefore we can limit our considerations to U1n (D˜0 ∪ D1 ∪ D2 ), and consider the scaling map U1 instead of U . If we define sˆ1 (z), s1 (z) and t1 (z) accordingly, we obtain: z ∈ Tˆ −1 U1 (D) : sˆ1 (z) = 1, s1 (z) ≡ 11 mod 12, t1 (z) = V (Tˆ(z)) − V (z)A−1 = (1, 0) √ z ∈ Tˆ 6 U1 (D1 ) ∪ Tˆ 11 U1 (D2 ) : sˆ1 (z) = 2, s1 (z) ≡ 5 mod 12, t1 (z) = (−1, 3 − 1) √ √ z ∈ Tˆ 5 U1 (D1 ) ∪ Tˆ 10 U1 (D2 ) : sˆ1 (z) = 3, s1 (z) ≡ 4 mod 12, t1 (z) = ( 3 − 1, 3 − 2) √ √ z ∈ Tˆ 4 U1 (D1 ) ∪ Tˆ 9 U1 (D2 ) : sˆ1 (z) = 4, s1 (z) ≡ 3 mod 12, t1 (z) = ( 3 − 1)(− 3, 2) √ √ z ∈ Tˆ 8 U1 (D2 ) : sˆ1 (z) = 5, s1 (z) ≡ 2 mod 12, t1 (z) = (2 − 3)( 3, −2) √ √ z ∈ Tˆ 7 U1 (D2 ) : sˆ1 (z) = 6, s1 (z) ≡ 1 mod 12, t1 (z) = (2 3 − 4, 3 3 − 4)


30

623 9 η 123 23 5 349 η40 364

ζ3

03

η24

539 33 43

ζ 102 13

Tˆ3 U1 (Dβ ) ζ 78 53

414 η46

423

20 200 0 6 5

530 114 15 314 η

ζ 75

ζ 49 ζ 23

63

355 ζ 14

ζ 64

ζ

U1 (Dβ ) η0

ζ0

ζ 55 U1 (Dα ˜) η52 ζ 20

ζ 58

358

00

614

η3 617 92

30

514 ζ 89 η31

323 η55

40 10

320

η34

ζ 61

ζ 17

517 117

η18 533 ζ 96

317 417

ζ 52

420

ζ 99 536

η21

120

ζ 26 ζ 11

η37 361

η6

611

η28

ζ 86 Tˆ11 U1 (Dβ )

019

→

431 547 634

134 550 249 434 249

620

η49 η59 352

375

520

326 311 η12 η43

511 111527

411

ζ 110

131 ζ 72 631 372

016

√ Figure 10.3. Trajectory of U (D) and small parts of P, λ = − 3. (ℓk stands for Tˆ k U (Dℓ ).)

For the remaining z, sˆ1 (z), s1 (z) and t1 (z) √ symmetrically. √ are given √ By looking at the following sets {t1 (z)Ah : h ∈ Z}, we obtain δ1 = (3 3 + 4)/( 3 + 1) = (5 + 3)/2: √ √ √ √ ±{(1, 0), (0, 1), (1, − 3), (− 3, 2), (2, − 3), (− 3, 1)}, √ √ √ √ √ √ √ √ ±{(1, 1 − 3), (1 − 3, 2 − 3), (2 − 3, 2 − 3), (2 − 3, 1 − 3), (1 − 3, 1), (1, −1)}, √ √ √ √ √ √ ±{(2 3 − 4, 3 3 − 4), (3 3 − 4, 2 3 − 5), (2 3 − 5, 2 3 − 2), √ √ √ √ (2 3 − 2, −1), (−1, 3 − 2), (2 − 3, 4 − 2 3)}. √ √ √ ′ The only √ 2x ∈ Z[ 3] with 0′ < x < 1 and |x | ≤ (5 + 3)/2 is 3 − 1.√ Therefore no point z ∈ V (Z[ 3] ∩ D) satisfies kz k∞ ≤ δ1 , and Conjecture 1.1 holds for λ = − 3.

PERIODICITY OF CERTAIN PIECEWISE AFFINE PLANAR MAPS 1

31

1

0.9 0.99 0.8 0.98

0.7 0.6

0.97

0.5 0.96 0.4 0.95

0.3 0.2

0.94 0.1 0.93

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

√ Figure 10.4. Aperiodic points, λ = − 3.

1

0.93

0.94

0.95

0.96

0.97

0.98

0.99

1

√ Figure 10.5. Aperiodic points in D, λ = − 3.

√ If V (z) = (2/7, 3/7 + 1/7), then we have √ √ √ √ √ V S1 (z) = (2 + 3) V (z)A3 + ( 3 − 1)(− 3, 2) = (3 3/7 − 5/7, 5 3/7 − 3/7), √ √ √ V S12 (z) = (2 + 3) V S1 (z)A11 + (1, 0) = ( 3/7 + 2/7, 3/7 − 1/7), √ √ √ √ V S13 (z) = (2 + 3) V S12 (z)A5 + (−1, 3 − 1) = ( 3/7 − 1/7, 3 3/7), √ √ V S14 (z) = (2 + 3) V S13 (z)A11 + (1, 0) = (2/7, 3/7 + 1/7) = V (z).

Remark. The eigenvalues corresponding to the primitive part of σ1 (ℓ ∈ {0, 1, 2}) are 5, −2 and 1. 11. The Thue-Morse sequence, the golden mean and

√

3

We conclude by exhibiting a relation between the Thue-Morse sequence and substitutions we used in golden mean cases (see [6] for a survey on links between fractal objects and automatic sequences). The Thue-Morse sequence is a fixed point of the substitution 0 7→ 01, 1 7→ 10: 0 1 10 1001 10010110 1001011001101001 10010110011010010110100110010110 · · · It can be written as 01 12 01 11 02 12 02 11 01 12 01 11 02 11 01 12 02 12 01 11 02 12 02 11 01 12 02 12 01 11 02 11 01 12 01 11 02 12 02 11 01 12 01 · · · By subtracting 1 from each term of the sequence of exponents (the run-lengths of 0’s and 1’s) we obtain the sequence 0 10 01110010 01001110011100111001001001110010 · · · which is easily shown to be the fixed point of the substitution 0 7→ 010, 1 7→ 01110 (see√[5]), which is equal to σ in the cases λ = −1/γ, λ = 1/γ, λ = −γ, and to σ1 in the case λ = 3. In case λ = γ, we have that σ ∞ (1) is the image of this word by the morphism 0 7→ 10, 1 7→ 110 since σ(10) = (10)(110)(10) and σ(110) = (10)(110)(110)(110)(10). Acknowledgments. We thank Professors Nikita Sidorov and Franco Vivaldi for valuable hints and for drawing our attention to several references. The second author wishes to express his heartfelt thanks to the members of the LIAFA for their hospitality in December 2006. The third author was supported partially by the Hungarian National Foundation for Scientific Research Grant No. T67580. The fourth author was supported by the grant ANR-06-JCJC-0073 of the French Agence Nationale de la Recherche.

32


References [1] R.L. Adler, B.P. Kitchens, C.P. Tresser, Dynamics of non-ergodic piecewise affine maps of the torus, Ergodic Theory Dyn. Syst. 21 (2001), 959–999. ˝ , J. M. Thuswaldner, On a generalization of the radix [2] S. Akiyama, T. Borb´ ely, H. Brunotte, A. Petho representation – a survey, in “High primes and misdemeanours: lectures in honour of the 60th birthday of Hugh Cowie Williams”, Fields Inst. Commun. 41 (2004), 19–27. ˝ , W. Steiner, Remarks on a conjecture on certain integer sequences, [3] S. Akiyama, H. Brunotte, A. Petho Period. Math. Hung. 52 (2006), 1–17. ˝ , J. Thuswaldner, Generalized radix representations and dynamical [4] S. Akiyama, H. Brunotte, A. Petho systems II, Acta Arith. 121 (2006), 21–61. [5] G. Allouche, J.-P. Allouche, J. Shallit, Kolam indiens, dessins sur le sable aux ˆıles Vanuatu, courbe de Sierpinski et morphismes de mono¨ıde, Ann. Inst. Fourier 56 (2006), 2115–2130. [6] J.-P. Allouche, G. Skordev, Von Koch and Thue-Morse revisited, Fractals 15 (2007), 405–409. [7] P. Ashwin, Elliptic behaviour in the sawtooth standard map, Phys. Lett., A 232 (1997), 409–416. [8] P. Ashwin, W. Chambers, G. Petrov, Lossless digital filters overflow oscillations: approximations of invariant fractals, Int. J. Bifurcation Chaos Appl. Sci. Eng. 7 (1997), 2603–2610. [9] D. Bosio, F. Vivaldi, Round-off errors and p−adic numbers, Nonlinearity 13 (2000), 309–322. [10] L. O. Chua, T. Lin, Chaos in digital filters, IEEE Trans. Circuits Syst. 35 (1988), 648–658. [11] A. C. Davies, Nonlinear oscillations and chaos from digital filters overflow, Phil. Trans. R. Soc. Lond., A 353 (1995), 85–99. [12] A. Goetz, Dynamics of piecewise isometries, Ill. J. Math. 44 (2000), 465–478. [13] A. Goetz, Stability of piecewise rotations and affine maps, Nonlinearity 14 (2001), 205–219. [14] A. Goetz, Piecewise Isometries — An Emerging Area of Dynamical Systems, Fractals in Graz 2001, ed. P. Grabner and W. Woess, Birkh¨ auser, Basel, 2003, 133–144. [15] E. Gutkin, N. Haydn, Topological entropy of polygon exchange transformations and polygonal billiards, Ergodic Theory Dyn. Syst. 17 (1997), 849–867. [16] E. Gutkin, N. Simanyi, Dual polygonal billiards and necklace dynamics, Commun. Math. Phys. 143 (1992), 431–449. [17] B. Khang, Dynamics of symplectic affine maps on tori, PhD Thesis, University of Illinois at UrbanaChampaign, 2000. [18] K.L. Kouptsov, J. H. Lowenstein, F. Vivaldi, Quadratic rational rotations of the torus and dual lattice maps, Nonlinearity 15 (2002), 1795–1842. [19] J.H. Lowenstein, S. Hatjispyros, F. Vivaldi, Quasi-periodicity, global stability and scaling in a model of Hamiltonian round-off, Chaos 7 (1997), 49–66. [20] J. H. Lowenstein, K. L. Kouptsov, F. Vivaldi, Recursive tiling and geometry of piecewise rotations by π/7, Nonlinearity 17 (2004), 371–395. [21] J. H. Lowenstein, F. Vivaldi, Anomalous transport in a model of hamiltonian round-off errors, Nonlinearity 11 (1998), 1321–1350. [22] J. H. Lowenstein, F. Vivaldi, Embedding dynamics for round-off errors near a periodic orbit, Chaos 10 (2000), 747–755. [23] K. Mahler, Some suggestions for further research, Bull. Aust. Math. Soc. 29 (1984), 101–108. [24] G. Poggiaspalla, Self-similarity in piecewise isometric systems, Dyn. Syst. 21 (2006), 147–189. [25] S. Tabachnikov, Dual billiards, Russ. Math. Surv. 48 (1993), 81–109. [26] F. Vivaldi, Periodicity and transport from round-off errors, Exp. Math. 3 (1994), 303–315. [27] F. Vivaldi, The arithmetic of discretized rotations, p-adic mathematical physics, AIP Conf. Proc. 826 (2006), Amer. Inst. Phys., Melville, NY, 162–173. [28] F. Vivaldi, J. H. Lowenstein, Arithmetical properties of a family of irrational piecewise rotations, Nonlinearity 19 (2006), 1069–1097. [29] F. Vivaldi, A. V. Shaidenko, Global stability of discontinuous dual billiards, Commun. Math. Phys. 110 (1987), 625–640. [30] F. Vivaldi, I. Vladimirow, Pseudo-randomness of round-off errors in discretized linear maps on the plane, Int. J. Bifurcation Chaos Appl. Sci. Eng. 13 (2003), 3373–3393. [31] C. W. Wu, L. O. Chua, Properties of admissible sequences in a second-order digital filter with overflow non-linearity, Int. J. Circuit Theory Appl. 21 (1993), 299–307. Dep. of Mathematics, Faculty of Science Niigata University, Ikarashi 2-8050, Niigata 950-2181, Japan [email protected] ¨ sseldorf, Germany Haus-Endt-Straße 88, D-40593 Du [email protected] Department of Computer Science, University of Debrecen, P.O. Box 12, H-4010 Debrecen, Hungary [email protected] LIAFA, CNRS, Universit´ e Paris Diderot – Paris 7, Case 7014, 75205 Paris Cedex 13, France [email protected]


Γ

i11 e27

18

a ˜25 b25

b

B3

e1818 d ˜10 43 a

c24e i

b349 26 d i

A

e42 c31

b3 c3 44 e

Γ

Γ

c6 c26 6 6 ˜ e37b a

i21 d14 b39 d4 e14 e4 b29 B 8

b21e24 23 b22 e 25 21 8 e b20 a ˜ i 22 a ˜ i7 a ˜20 i9 23 b Γ14 Γ15 a ˜23 26 e Γ16 i10 B2 e8 Γ17

a ˜24b24

b

f 0

e

j0

Γ6

45 i29

a ˜29

Γ23

c0 Γ 7 i0

1321

cc

A6

c14 c22

A0

Γ24 b14 e33 i17 a ˜1 b1 c1

33

d8

e

B e

i0 g0 c0 k 0 h0 d0 e0

i23 e39 8

7

d

a ˜15 d e

i24 c29 e40c9 b9

c

2

a ˜

11

b11 c11

c25 e36

e3 Γ27

4

i2a ˜27 18 e29c

a ˜ d1212 e

c4 4 b

a

b0

e21 i5

Tˆ3 →

b19 a ˜19 Γ12

d52 h80 b77

d0

6

e46 f 61

g0 c0

Figure A.1. The first return map Tˆ3 and large parts of R, λ =

B

b37

f0 0

i30 j1

i20 d328 b

A4

c32

i13

16

h49 a ˜30 d21 30 e46 k 30 f c

d 44

i27

a ˜2

A

b

g 43

b

19

b2 e43

a ˜

e19

4

d10

2

c1616

35

B

1

Γ9 Γ20

41c30

i12 e28

a ˜26 i1 b26

1

b15 c15e34 c23 i18

Γ19

8

Γ25b35 e10 Γ4

b32 17 a ˜9 17 d

i25

→

A1

a ˜8

Γ2

46

c28 c8

b

b42

0

b0

1

7

Γ

e38 b7 a ˜7

Γ29 d5d15 Γ0 e5 b40 e6 6 e15 b30 d d16 b31 e16 b41Γ1 B0

d0 0 e

a ˜14

31

i22

a ˜ 0 b0

i16

a ˜13 b13 e32

d20 i15

e

b

j0 a ˜

e

Γ11

i4 Γ22 c34

20

a ˜12b12e c20 c12

i28

˜5 c5 a 5 b38 b d13 B7

13

˜18 b18a

c27c7

A5

Γ28

e22 i6

B5

b45

11 11 e a ˜3 d Γ5

b27

2

21

e30 c19

e13

e

d2

c33

Γ10 ˜ c a b17 3 i a ˜2814 i

19

b36

3

17

35

Γ26

Γ3 e9

17

b10c10

33

h

0

e

0

g 74 a31 c66

√ 3. (ℓk stands for T k (Dℓ ).)

√ Appendix A. The map Tˆ for λ = 3. √ As the scaling domain D is very small in case λ = 3, the determination of Tˆ is done in several steps. Figure A.1 shows the action of Tˆ3 , which is the first return map on the domain √ √ √ √ √ √ √ √ {x > 4 3−6, 2y < 3x+2− 3, y > 3x+1− 3}∪{x < 1, y > 12 3x−20, y > 3x+6 3−11},


34

H 67 Z1

k 11

H 18

E 31 E 40

E 3 H 52 H 3 ∆10

f 25 E 57 h28 E 13

E 66

10 h18f 10g

h46

E 22 33 H

9 ∆39 h37 h H 82 H 43 k 26 g 39 ∆74 H 97

∆17 ∆25 ∆60 Z4 17 10 g 25 g H 59 H f 17 H 21 E 49

∆46

g3

H 89 g 41

k 19

h30 E 59 50 18 85 55 H 23 H 101 f 0 f E H 11 2 k 29 H H h 100 Z5 H 75 H6 H 72 H H 86 26 42 42 60 H g ∆ H ∆043 g 0 15 13 E 13 ∆32 f 4 16 ∆77 ∆ E 45 k ∆29 ∆ H 56 E 33 k 13 H 22 g 20 g 29 H7 E 68 E 5 h f 29 71 ∆14 ∆56 k0 0 H 28 45 48 21∆64 ∆ H 24 h h g 3 E ∆ 15 39 24 54 14 k h H 11 H f h H 104 73 H 5 f 12 44 g 14 46k 17 H H 35 E 3 E 84 g 28 57 f H H 102 ∆ ∆21 g 1220 ∆63 g 30 ∆301 ∆12 55 g 22 65 f1 g ∆ k 28 ∆41 ∆ ∆ ∆53

H 99 ∆76

∆67g

32

∆70

∆20 ∆49 g 6 ∆6

g 35 93g 7 H ∆35 H 78

k 22

f7

∆50

∆736 ∆

g 1 87 ∆44 H Z2

∆22 ∆71 g 36

H 19

H 68 k 12 E 41

∆15 ∆69 H 57 ∆51 92 H 94 H0 f 26 34 H 8 29 g 8 23 g H 49H 79 k 32 H 6 43 107 H f H h E 58 E f 28 h5H 39h15 f 15 k 1 h29 h19 H 48 H 30 k 14 k 21 f8 15 0 18h33 1 g 27 g 42 6 E E 47 34 58 43 h h h ∆ H 77 ∆ E 19 h 47 14 E 71 f 11 E4 H 1 ∆8 E18 ∆62 E 62 E 28 E h34 23 31 k H 25 E h14 H 534 11 g g 19 g 28 38 H H H g H 40 50 E 63 h16 h44 H E 23 h4 h32 27 ∆66 ∆54 74 5 67 E 19 37 31 H ∆ E 61 80 ∆ 48 ∆ 72 ∆ 23 11 E H 1 ∆ ∆ ∆ ∆ E 29E h38 E 17 E9 H 44 h10 H 34 g 5H 106 g 37 47 103 31 H 95 H E 10 H 91 f 5 E8 E 20 Hk 24H ∆2 ∆68 H 83 40 E 70 E 53 24E 37 52 h41 ∆ h E 54 g9 ∆ g 33 h7 E 36 9 h13 k 20 f 21k 8 22 h25 f 2 k 27 23 35 f h h 7 52 64 k H 14 f E 38 E H 37 E 26 E H 41 g 2 g 40 17 H 64 H 76 3 31 h 9 98 20 45 h45 H 2 k H f 15 ∆33 H 27 h h ∆ 88 H 75 9 55 30 H 63 35 ∆ E E 60 E H 51 ∆ E 7 E 6 H 13 H H 65f 23 E 11E 2 k 26 h E 16 h22 H 62 H 16 Z9 Z3 8 E 51 Z ∆16 ∆4 k 10 E 39 21 H 81 Z 10 69 f 19 38 26 H ∆ ∆ 66 E H 105H 46 E 69 9 7 H H Z H 32 H 20 H 12 Z 11 40 H 58 h8 E 34 5 H 61 H 17 f 24 E 56 65 59 47 16 4 h 73 61 ∆ ∆ f f 12 k 38 26 E ∆ ∆ h E 25 g4 g 16 h36 k 25g k 13g k3

H 70

Z0

∆27

h27E 12

H 42

H 96

∆24 g 24 E 48 k 2 f 27 E 42 g 18 ∆18

H 90

H 36

E 6 h21

Figure A.2. Trajectories of long lines in R and Df , Dg , Dh , Dk , λ = stands for T k (Dℓ ).)

Z6

√ 3. (ℓk

on sets Da , . . . , Dh . To this end, we first determine the trajectory of sets Da˜ , Db , . . . , Dk , which partition a symmetric version of this domain. Figure A.1 shows the trajectory of the open sets Da˜ , Db , Dc , Dd , De , Di , Dj , Figure A.2 completes the picture with the trajectories of the lines Df , Dg , Dh , Dk . All points which are not on these trajectories are periodic. ¿From the symmetric first return map, it is easy to determine Tˆ3 . Next, we consider the first return map on √ √ √ √ √ √ {(x, y) : 2y < 3x + 2 − 3, 2x < 3y + 2 − 3, x ≥ 30 3 − 51 or y ≥ 30 3 − 51} in Figures A.3 and A.4, partitioned into open sets Dk , Dl1 , Dl2 , Dm ˜ , Dn ˜ and lines Dl3 , Do , Dp˜, Dq˜. √ √ ˆ ¿From this√map, we easily √ obtain the first√return map T4 on {(x, y) : 2y < 3x + 2 − 3, 2x < √ 3y + 2 − 3, x ≥ 30 3 − 51 and y ≥ 30 3 − 51}, which is partitioned into the sets Dl , . . . , Dq . Observe that the return time on Dl is not constant since the trajectories of the three parts Dl1 , Dl2 , Dl3 are different. This implies that the return times on D0 , D1 and D0 are not constant.


35

k7 k0

l113

k2

n ˜ 33 l214 n ˜

19

n ˜

Θ5

l15

n ˜ 12 n ˜ 35

m ˜ 12 m ˜ 00 n ˜

n ˜2

n ˜ 16

n ˜ 37 n ˜ 14

n ˜5 m ˜ 540 n ˜

l12

l10 m ˜2

l25

l22

l20

47

n ˜ 26 n ˜ 21

Θ0

Θ2

n ˜ 30

l27 n ˜ 23 n ˜ 28

l17 n ˜7 n ˜ 42 l29 m ˜7

k

m ˜ 10

k9 l210

l110

n ˜ 10

l115

n ˜ 24

n ˜ 45

n ˜ 29

n ˜8

Θ3

n ˜ 4914 m ˜

n ˜ 38

n ˜ 15

l18

Θ1

m ˜3 n ˜

31

n ˜ 17

n ˜

l11

l13 l23

m ˜ n ˜ 48

k1

n ˜ 27

n ˜1 n ˜ 36 m ˜ 1 13 n ˜

l21

l28

˜ m ˜6 m Θ6

n ˜6

l14 n ˜4

n ˜ 41

13

k4 l24

n ˜ 22

3

k3

n ˜ 43 8 m ˜

l216

l111 l212

→

n ˜ 44

n ˜9 9 m ˜

l19

5

l211

n ˜ 18

n ˜ 32

4

n ˜ 39

Θ4

l213 l112

k8

l114

l16

l215 n ˜

34

n ˜

20

l26 n ˜

25

Figure A.3. An intermediate first return map, λ =

n ˜ n ˜

˜ 11 m

11

46

k6

√ 3. (ℓk stands for Tˆ3k (Dℓ ).)


36

o7 l310

Λ33 Λ

Λ

19

47

12

q˜

Λ

q˜5

47

q˜

q˜0

o0

I0 l30 I 2

0 q˜35 Λ0 p˜

Λ35

p˜5

l35

12

p˜19

Λ14

Λ5

q˜2

o2 l32

p˜2

16 Λ2 Λ

q˜37 Λ37

40

q˜

26 Λ40 Λ Λ21 p˜12

Λ30 Λ28

p˜7 Λ7 42 Λ 14 p˜ q˜742 q˜ K

Λ23

p˜16

9 Λ p˜ Λ 44 l37 q˜9 q˜ 10 p˜ q˜10 10 Λ24 Λ q˜45 9

0

o5

o

p˜

l312

Λ

Λ q˜49 q˜14

Λ

Λ15 K1

q˜3

Λ

3

p˜ 17

p˜8

49

Λ38 q˜38 3

43 Λ29 Λ q˜43 q˜8 Λ8

p˜21

17

Λ31

p˜15 9

Λ45

l38

→

44

l33

l31

o3

o1 I1

p˜1 Λ1 Λ36

q˜136 q˜

q˜13

Λ

l311 Λ

p˜20 34

p˜6 l36

Λ20

Figure A.4. The trajectory of the lines, λ =

Λ

o4 l34 p˜13

Λ39 p˜11

25

Λ11

Λ18

p˜4 Λ32

Λ41 4 4 q˜41 q˜ Λ 39 q˜6 q˜

Λ6

Λ48 o8

Λ27

13

q˜48

22

p˜18 Λ46 q˜46

l39 o6

q˜11



37

(1, 1)

Tˆ4 →

k0

k 473 m2284 q 4067

m0 0 0 o0 n q p0

p2647 4059 n o501 l{655,787}

l0 √ √ (30 3 − 51, 30 3 − 51)

Figure A.5. The map Tˆ4 , λ =

√ 3. (ℓk stands for T k (Dℓ ).)

Finally, we consider in Figure A.6 the first return map on √ √ √ √ √ √ {(x, y) : 2y < 3x + 2 − 3, 2x < 3y + 2 − 3, x ≥ 72 − 41 3 or y ≥ 72 − 41 3}, partitioned into sets D0 , D˜ , . . . , D˜ , D˜ , D˜ , D˜ , from which it is easy to deduce Tˆ on D. 1

9

0

1

3

√ Appendix B. The map Tˆ for λ = − 3.

√ For λ = − 3, we consider in Figure B.1 the first return map on √ √ √ √ {(x, y) ∈ [0, 1)2 : 2x + 3y > 3 3 − 2 or 3x + 2y > 3 3 − 2}, partitioned into sets Da , . . . , Di . Figure B.2 provides the first return map Tˆ on D. Again, all points in R are periodic.


38

00

03

55

50

˜ 15

˜4 1 ˜ 83

˜5 3

˜ 60 ˜ 44Ξ4

˜ 20 ˜ 23

˜9 ˜7 0 ˜ 32 7˜ 47 93 ˜ 72 Ξ7 ˜

˜0 ˜ 90 0 0 ˜ Ξ0 4 ˜ 70 ˜ 30 ˜ 75

˜ 62 ˜4 54 1 Π3

02

Ξ5˜ 45

˜ 10

˜ 63

→

˜ 42 Ξ2

˜5 0 O3

˜ 80 ˜ 0 1 ˜0 3

DM

Ξ˜33 4 ˜ 92

˜ 22

Ξ6 ˜ 46 Ξ1 8 ˜ 0 ˜6 ˜1 7˜1 4 ˜ 82 3 ˜4 ˜ 31 7˜ 1 0 ˜1 ˜3 3 1 ˜ 21 81 ˜1 ˜ 1

˜ 91

P1 ˜2 P 0 ˜ 31 3 ˜2 ˜1 ˜7 3 6 0 O5 0 2 ˜ O0 Π0

˜1 1 51

52 ˜ 12

Π1

Π2

˜ 3O 1 0

O 20 ˜4

DN

01

53

˜ 13 O 46 ˜ 0

Figure A.6. Almost the map Tˆ, λ =



f 11

a11 g 13 f 13 E1

f 19 g 19 a17

g

11 f

d12

22

a9

E4 a a

e

b3

g8

a21

d6

e

g5

f5

g 16

a22g

6

c4

f4

g 15

3

E0

i5

g9

e

a18

g 20 a7 20 f

g g 21

d9

10

d1

b1

g f 10

i3

1

c1

f 14

a

a1 14

a12

19

c2 d10

∆2 18g 18a16

d2

∆1

a15

g 17

f2

a20 g

22

e5

5

7

d

d

d0 e

1

8 f 21a

c3

f

g0 f 0

E2

3

f7 g7 a5

g1 i f1

E3 b2

DΓ

a0

h0

15

e4

f

i0

1

DB

g4

d3

b0

f9

d4

g3

h

a13

d8

d0

DA

∆0

f a2

e

i6 24

a14

∆3

d11

e

0

c0

f8 g 23

i2

i4

c5

3

f 16

6

7

39

a0

0

e2

h

b0

2

g

f

6 g 6 a4 E

5a

10g 12f 12

f 23

a23 g 25

b i7

c0

2

f 17

4

c6

→ i0

0 g0 f

h0

d13 e

8

√ Figure B.1. A first return map and large parts of R, λ = − 3. (ℓk stands for T k (Dℓ ).)


40

14

20b 20a 0 5 5268 60 10

I4 54 24a

I0

I 47

˜ 30

DH

581 281 a 482

059

281 b

Tˆ →

5349 I 53

686 I 57

DZ

132

00

64 24b 5272

˜ 3183

40

138

142 5353 285 b486

5343 275 b

585 285 a

476

275 a 575

10

5278 2b

610

210 a 510 I 10

110 5293 225 625 b 225 a 125 25 40 5 I

535 663 235 a 436

I 25

227 a 527

Θ0

˜ 3147

˜ 371

Θ51

5336 68 2b

023

˜ 323

469

˜ 3104 Θ76 ˜ 361

250 b

Θ86 217 5285 b 617

650

Θ33

217 a 517

423

036

˜ 379 Θ94 268 a

568 ˜ 3114 Θ43

Θ8

˜ 3160

˜ 3122

˜ 336

Θ41 ˜ 369

5318 ˜ 3112 Θ84

260 260 a 560 5328 b 61 117 4 I 17

58

5326 2b

I 32

√ Figure B.2. The map Tˆ and small parts of R, λ = − 3. (ℓk stands for T k (Dℓ ).)