Notes on Piecewise Isometries

NOTES ON PIECEWISE ISOMETRIES

AREK GOETZ

Abstract. This document contains selected notes used in the third part of the Dynamical Systems. They will evolve and will be updated after almost every lecture.

Date: May 17, 2003. 1991 Mathematics Subject Classiﬁcation. Primary 58F03; Secondary 22C05. 1 2 AREK GOETZ

1. Rotations

Deﬁnition 1. A (planar) isometry is a map R : C → C which preserves distances, that is such that for all z1,z2 ∈ C, |z1 − z2| = |Rz1 − Rz2|.

All non-identity planar isometries can be divided into categories according to the number of fixed points that they have. Maps with no fixed points are translations, those with exactly one fixed points are rotations and those with more than one fixed point are reflections. As the following propositions state, all planar piecewise isometries are conveniently de- scribed using complex notation. Proposition 1. Any rotation or translation can be written as R(z)=ρz + t where ρ = eIα, 0 ≤ α<2π.

The map R first rotates a point z by angle α, and then it translates the result ρz by t.If ρ = 1 that is when α = 0, the map R is a translation. Otherwise, if ρ =6 1, the map R,a proper rotation, has exactly one fixed point, z0 = t/(1 − ρ). Proposition 2. Any reflection can be written as R(z)=ρz where ρ = eIα, 0 ≤ α<2π and z denotes a conjugation of z.

The conjugation z flips a point z with respect the real axis. Basic properties of the dynamical systems of rotations are well understood. There are two distinct cases, (a) rational rotations, and (b) irrational rotations. A rotation R is rational if α/π ∈ Q. Otherwise we say that R is irrational. Iterates of all points under rational rotations are periodic. Orbits of all points other than the fixed point are infinite under irrational rotations. These orbits are dense in a circle. Homework Problems Exercise 1. Show that the map R : C → C, R(z)=ρz + t where ρ, t ∈ C, |ρ| = 1 is an isometry. Exercise 2. Show that the map R : C → C, R(z)=ρz, |ρ| = 1 is an isometry. Exercise 3. Show that for an irrational rotation R, all z ∈ C either z is fixed or R or the orbit of z is infinite. R α π k . C Exercise 4. Let be a rotation by the angle =2 m Show that All points in are periodic of period m. Show that ρ = eiα satisfies ρm = 1, that is ρ is a root of unity. Exercise 5. Show that under a proper rotation R the orbit of any point lies on a circle centered at the fixed point z0 = t/(1 − ρ)ofR. Exercise 6. Show that a rotation R(z)=ρz + t,(|ρ|6= 1) is conjugate to a rotation 0 R (z)=ρz via the translation h(z)=z − z0, C −−−R→ C     hy yh

0 C −−−R → C, where z0 = t/(1 − ρ) is the fixed point for R. Exercise 7. Show that for the map R(z)=ρz + t, the set of fixed points is the a line L with slope 1/2 arg(ρ). Then show that R(z) reflects z with respect to L. NOTES ON PIECEWISE ISOMETRIES 3

Exercise 8. Show that if an orbit of a point z under an isometry R is unbounded, that is, if for all M, there is an iterate T k(z), |T k(z)| >M, then R must be a translation. Exercise 9. Prove that an isometry is a bijection (1-1 and onto).

iπ Computer Exercise 10. Graph the ﬁrst 100 iterates of 1 under the rotation R(z)=e k where k = 3+1/2. Then graph the ﬁrst 100 iterates of 1 under the rotation R(z) for k 1 =3+2+1/3 . What can you say about the orbit behavior in each case.

2. Piecewise Rotations

In the previous section, we saw that a Euclidean rotation T in the plane has a rela- tively simple dynamics. For every point x other than the center of rotation T the orbit {x, T x, T (Tx),...} is a ﬁnite set or it is a dense subset of a circle when T . This depends on whether T is a rotation by an angle commensurate with the full angle. Composing two or more rotations in the plane leads to complicated and fascinating dynamical systems. These systems are also surprising, since they can have self-similar structures, attractors, repellers, which are usually observed in systems whose generating map has either contacting or expanding properties. The dynamics of maps changes dramatically if we introduce discontinuities in the generating maps. In this section we will introduce a class of maps which are rotations when restricted to subsets of the domain.

Deﬁnition 2. A map T : X → X is called a piecewise rotation with atoms P = {Pj}j∈N if

Tx = ρjx + tj ∈ Pj for some complex numbers: sj and ρj such that |ρj| = 1 for all j ∈ N, and ρj =6 1 for at least one j. If all ρj are roots of unity, then T is called a rational piecewise rotation.

Frequently, we will refer to sets Pi as atoms.

Figure 1. An example of a piecewise rotation with two atoms. The angle α = π/5. The right ﬁgure represents a partition of space 4oca into sets that follow the same pattern of visits to the atoms. Iterates of each colored pentagon never get broken apart by the discontinuity.

In Figure 1, we illustrated one of the most simple examples of a piecewise rotation T defined on a triangle X = 4oac. Associated with this map are two subdividing X triangles, P0 and P1 on which T acts as a rotation. On P0, T rotates P0 by π − π/5 about S0.On P1, T is a rotation by π/5 about S1. The center S0 is the center of the circle subscribed in P0 and S1 is the center of the circle circumscribed in P1. The right figure is a partition of triangle X into sets that follow the same periodic pattern of visits to the atoms P0 and P1. It is surprising that this mosaic appears to be fractal. In order to study it we now define a convenient and widely used in dynamical system tool called symbolic encoding. 4 AREK GOETZ

Deﬁnition 3. Coding. The partition P of X associated to a piecewise rotation T gives rise to a natural one-sided coding map i : X → Ω=N N for T . The map i encodes the forward orbit of a point by recording the indices of atoms visited by the orbit, that is i(x)=w0w1 ..., T jx ∈ P where wj .

Deﬁnition 4. Cells. A set of all points following the same coding ω ∈ Ω under T will be called a cell.

All cells and are convex provided that the atoms {Pi} are convex (Goetz 2000). Piecewise rotations are natural two dimensional generalizations of well studied interval exchange transformations studied in (Arnoux, Ornstein & Weiss 1985, Boshernitzan 1988, Boshernitzan 1985, Katok 1980, Masur 1982, Keane 1977, Keynes & Newton 1976, Veech 1987, Veech 1982, Veech 1978) and interval translation maps (Boshernitzan & Kornfeld 1995, Troubetzkoy & Schmeling 2000). Invertible piecewise rotations (Adler, Kitchens & Tresser 2001, Ashwin & Fu 2001, Ashwin & Fu 2002, Haller 1981, Gutkin & Haydn 1997, Kahgng 2002, Poggiaspalla 2002) (those that preserve Lebesgue measure) are closely related to the theory of dual billiards (Tabachnikov 1995) and Hamiltonian systems (Scott, Holmes & Milburn 2001). A somewhat unusual application is found outside of mathematics, in electrical engineer- ing, in particular in the theory of digital ﬁlters (Ashwin 1996, Ashwin, Chambers & Petrov 1997, Chua & Lin 1988, Chua & Lin 1990, Davies 1992, Kocarev, Wu & Chua 1996, OgorzaÃlek 1992). Digital ﬁlters are algorithms widely incorporated in electronic components in con- temporary electronics devices such as cellular phones, radio devices and voice and image recognition systems. The most basic examples of piecewise isometries are exchanges of intervals (Arnoux et al. 1985, Boshernitzan 1988, Boshernitzan 1985, Katok 1980, Masur 1982, Keane 1977, Keynes & Newton 1976, Veech 1987, Veech 1982, Veech 1978).

Definition 2 (Interval Exchange Transformation.) A piecewise isometry T : X → X is called a interval exchange transformation with atoms if X is an left open interval, T is Lebesgue measure preserving, T restricted to each atom is a translation. Homework Problems Exercise 11. Determine all fixed points for the map T illustrated in Figure 1. Exercise 12. Show that there are no periodic points of period 2 for the map T illustrated in Figure 1. Exercise 13. Find a periodic point of period 3 for the map T illustrated in Figure 1. ρ 2π i 2π j ∈{ , ,..., } jπ i jπ Exercise 14. Let = cos 5 + sin 5 . For each 1 3 9 , write down cos 5 + sin 5 as a monomial expression in ρ. ρ 2π i 2π Exercise 15. Let = cos 5 + sin 5 be the first fifth root of unity. In Figure 1 let 3 the common vertices of the triangles P0 and P1 be o =0,b = −1. Show that a = ρ 2 3 and c = ρ + ρ + ρ . Then show that the map T rotates triangle P0 by (π − π/5) and then 2 3 translates it by c = ρ+ρ +ρ . Finally show that the triangle P1 is rotated by (π/5−π) under T , and then it is translated by a = ρ3 (Hint: Exercise 14 and also 1 + ρ + ρ2 + ρ3 + ρ4 = 0). *Exercise 16. Show that for the map T illustrated in Figure 1, the centers of rotations S / − ρ − ρ2 ρ3 S / ρ ρ2 ρ3 ρ 2π i 2π 0 =1 5( 2+ +2 ) and 1 =1 5(1 + 2 +3 +4 ) where = cos 5 + sin 5 . NOTES ON PIECEWISE ISOMETRIES 5

Figure 2. An example of a piecewise rotation with two triangles based on the angle π/7.

*Exercise 17. In a similar manner as illustrated in Exercise 15, use the following ﬁgure in order to determine the vertices of the triangles P0 and P1 as polynomial expressions in ρ ρ 2π i 2π where = cos 7 + sin 7 . *Exercise 18. Using the notation from problem 17, show that the centers of rotations 2 3 4 5 2 3 4 5 S0 =(−1 − 2 ρ − 3 ρ − 4 ρ +2ρ − 6 ρ )/7, and S1 =(5+3ρ + ρ +6ρ +4ρ +2ρ )/7.

3. Rational polynomial expressions involving roots of unity

In this section we illustrate an algebraic tool which allows us to precisely and conveniently deﬁne rational piecewise rotations.

A piecewise rotation T will be completely defined if we list and identify all atoms {Pi} and the corresponding rotations {Ri}. In order to define polygons, we need to specify vertices of each of the polygons. A vertex is a complex number. As illustrated in Exercise ?? all vertices of the map defined in section 2 in Example 1, can be written as rational polynomial expressions in ρ where ρ5 = 1, that is ρ is a fifth root of unity. In all other examples, this will be the case as well, that is for some positive integer number p, all vertices will be elements of the set:

p−1 p Q[ρ]={a0 + a1ρ + ···+ ap−1ρ | ρ =1,ai ∈ Q} The atoms in Example 1 from section 2 can be then deﬁned as: 4 3 P0 =[0, −1 − ρ , −1],P1 =[0, −1,ρ ].

Let us note that all rational polynomial expressions in ρ are in Q[ρ]. This is because ρp = 1. For any integer m we can write m = kp + r where k, r ∈ Z and r is the remainder in division of m by r. Then any higher power monomials in ρ, ρm, where m = kp + r where can be replaced by ρr. ρm = ρkp+r =(ρp)kρr = ρr. The numbers {0,ρ,...,ρp−1} are all p-th roots of unity. Using an algebraic notion of a vector space, we can say that Q[ρ] is a called a vector space of spanned by p-th roots of unity over rational numbers. Note that Q[ρ] is closed under addition and subtraction. The set Q[ρ] is also closed under multiplication because as we pointed out all rational polynomial expressions in ρ are in Q[ρ]. 6 AREK GOETZ

For example, if ρ5 = 1, then (1+2ρ3)(2 − ρ4)=2+4ρ3 − ρ4 − 2ρ7 =2+4ρ3 − ρ4 − 2ρ2 ∈ Q[ρ].

It is a remarkable fact that Q[ρ] is also closed under division, that is for all a, b ∈ Q[ρ], b =0,6 a/b ∈ Q[ρ]. A set that contains zero and one and that is closed under addition, subtraction, multiplication and division is called a ﬁeld. In Algebraic Number Theory, the set Q[ρ] is referred to as a cyclotomic ﬁeld.

4. First return maps

Figure 3. The return orbit of triangle P1 to itself.

As we saw in Figure 3, all pieces of the bottom atom P1 come back quickly. No piece was stuck forever in the top triangle. Let us explain why this is so. Let us consider an orbit of a polygon K ⊂ P0

K −→R0 TK −→R1 T 2K −→R1 T 3K −→R1 T 4K −→R1 T 5K (1) ∩∩∩∩∩∩ P0 P1 P1 P1 P1 P1

This is because if the orbit of a polygon K ⊂ P0 remained P1 for more than five iterations, 6 5 then the first iterate of K, TK = R0K would be the same as its sixth iterate T K = R1(R0K) 5 since R1 is the identity map. However, this would violate that T is one-to-one since R0K 5 4 would be the T -image of K as well as the image of T K = R1(R0K). The argument above works only for a case of a map that has two atoms and in order for the orbit to close itself we used that R0 was rational. In general, it is not true that all points must come back to the original atom. This motivates the following general definition. Definition 5. Return map. Let f : X → X and U ⊂ X. Let ∆ be a subset of U be such that for all x ∈ ∆, there is n such that f n(x) ∈ U. Given an x ∈ ∆, let τ(x) be the smallest n τ(x) n such that f (x) ∈ ∆. The first return map of f to U is defined as fU = f x.

While we cannot guarantee that all points return, if a map preserves area (note that the our example, T25 preserves area), then the following famous theorem guarantees that the set of points that never return in negligible. Theorem 19 (Poincar´e Recurrence). Let f : X → X be a measure preserving map and let U ⊂ X. Then the return map fU is deﬁned for almost all x ∈ U.

Homework Problems

Exercise 20. Let f : R → R,f(x)=x + 1. What is the return map f[0,1]? NOTES ON PIECEWISE ISOMETRIES 7

Exercise 21. Let f : C → C be a rotation on the unit circle C = {z||z| =1}. Do all points in the upper semi-circle return back to it? Let the angle of rotation be α. What is the maximum return time to the upper semi-circle?

Exercise 22. Let f : R → R,f(x)=2x. Determine the return map f[0,1] and its domain.

Exercise 23. Find the similarity that is the conjugation between the map T25 illustrated in Figure 1 and its ﬁrst return map to the bottom triangle P1. (See Figure 3).

5. Substitutions

The periodic blocks of period pentagons in our main example, can be obtained by starting with 0, and then applying the following rule: 0 −→ 1 σ : 1 −→ 001. For example, the coding of the largest pentagon is 000 .... The coding of the second largest pentagon is 111 .... The coding of the smaller pentagon is 001 001 001 .... Exercise 24. Determine the length p(n) of the period of a sequence obtained from 0 by a recursive application of the substitution n times. Hint: Let p0(n) and p1(n) be the number n of zeros in the sequence σ (0). Find recursive relations for p0(n) and p1(n). Exercise 25. Let the substitution σ act on the collection of inﬁnite binary sequences Σ = {w0w1 ...,| wj ∈{0, 1}}. What are the ﬁxed points for σ?

6. Noninvertible piecewise rotations and attractors

In the previous sections we considered maps that rearranged atoms without overlaps. These maps were except for the boundaries of the domains one-to-one and onto. In this section, we will brieﬂy study the maps for which atoms overlap after the iteration of the map.

References

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