A class of noninjective circle transformations without periodic points ...

Lithuanian Mathematical Journal, Vol. 51, No. 1, January, 2011, pp. 3–12

A CLASS OF NONINJECTIVE CIRCLE TRANSFORMATIONS WITHOUT PERIODIC POINTS Vytautas Kazakeviˇcius Faculty of Mathematics and Informatics, Vilnius University, Naugarduko 24, LT-03225 Vilnius, Lithuania (e-mail: [email protected]) Received October 8, 2009

Abstract. We construct a class of noninjective aperiodic circle transformations. As a by-product, we establish some new properties of Mersenne numbers. MSC: 37E10 Keywords: circle transformation, periodic points, Mersenne numbers

1

INTRODUCTION

Let S denote the interval [0; 1) endowed with the metric   d(x, y) = min |x − y|, 1 − |x − y| . Then S is a compact metric space homeomorphic to the unit circle in R2 . Let C(S, S) denote the set of all continuous functions from S to S . We call any f ∈ C(S, S) a circle transformation. For f ∈ C(S, S), by f n we denote the iterations of f . By Per(f ) we denote the set of all periodic points of f :    Per(f ) = x ∈ S  ∃n  1, f n x = x . If Per(f ) = ∅, the transformation f is called aperiodic. Furstenberg [1] proved that each aperiodic homeomorphism on S is strictly ergodic, i.e., there exists a unique probability μ such that μf −1 = μ (here μf −1 denotes the image of μ with respect to f : (μf −1 )(A) = μ(f −1 A) for Borel subsets A). Our initial aim was to extend this result to arbitrary transformations (not necessary homeomorphisms). However, such an extension makes sense only if there exists at least one noninjective aperiodic transformation: if no such transformation exists, the assertion that each aperiodic transformation is strictly ergodic is equivalent to the result of Furstenberg. It turned out that construction of such transformation is a surprisingly difficult problem. In this paper, we consider the transformations f defined by  α for 0  x < 1/2, fx = α  2x for 1/2  x < 1, c 2011 Springer Science+Business Media, Inc. 0363-1672/11/5101-0003 

3

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V. Kazakeviˇcius

where α ∈ S is an irrational number, and  denotes addition modulo 1. Obviously, f ∈ C(S, S) and is not a homeomorphism. The question is whether f is aperiodic for some α. Our main result (as usual, {x} denotes the fractional part of x) is the following: Theorem 1. f is aperiodic if and only if {(2n − 1)α} > 1/2 for all n  1. There exists a continuum of α with this property. Clearly, if f is aperiodic, then α > 1/2, and also f n α > 1/2 for all n  1; a straightforward induction shows that, in this case, f n α = {(2n+1 − 1)α}. This proves the necessity. To prove the sufficiency (this is done in Section 4), we need some auxiliary results: in Section 2, we prove a useful combinatorial lemma, and in Section 3, we use the lemma to establish a result about orbits under the map x → 2x (mod 2n − 1). Finally, in Section 5, we show that there exists a continuum of α with the desired property. In what follows, Mn = 2n − 1 denotes the nth Mersenne number, N the set of all positive integers, x the integer part of x, gcd(a, b) the greatest common divisor of a and b, gcd(A) the greatest common divisor of the set A, #A the number of elements in A, and a | b means that “a divides b.” We also set    Orbn (r) = 2i r (mod Mn )  i  1 . 2 A COMBINATORIAL LEMMA Let A = {a, b} be an alphabet of two elements called letters. By m-word in A we call any string of letters x = (xi | i ∈ Zm ) = (x0 , . . . , xm−1 ) indexed by elements of the cyclic additive group Zm = {0, 1, . . . , m − 1}. The set of all m-words is denoted by Am . The number m is called the length of x and is denoted by |x|. The only element of A0 is the empty string also called the empty word. A letter is identified to the corresponding 1-word. If x = (x0 , . . . , xm−1 ) ∈ Am and y = (y0 , . . . , yn−1 ) ∈ An , then xy denotes the concatenated word (x0 , . . . , xm−1 , y0 , . . . , yn−1 ) ∈ Am+n . Concatenation is an associative operation in the set of all words ∗ A = m0 Am , and the empty word is a neutral element with respect to this operation. Moreover, each x ∈ Am can be written as the product x0 · · · xm−1 , where x0 , . . . , xm−1 are the letters of x interpreted as words from A1 . By xk we denote the concatenation of k instances of x. More precisely, x0 is the empty word, and k x = xk−1 x for k  1. A word x is called periodic if x = y k for some y ∈ A∗ and k  2 (the empty word is thus periodic). Now let  be any order on A. It induces a partial ordering on A∗ , x < y meaning that there exists an index k < min(|x|, |y|) such that xi = yi for i < k and xk < yk . As usual, x  y means that “x < y or x = y .” So by construction each set Am is totally ordered. Moreover, x < y implies that xx < yy  for any words x and y  . The words of different length can be incomparable: for example, x and xy are incomparable if y is not empty. In the sequel, we assume that a < b. An m-word x is called maximal if xq xq+1 · · · xq+m−1  x for all q ∈ Zm . For example, any word of the form bk am−k is maximal (and aperiodic if 0 < k < m). If x is maximal and aperiodic, then x0 = b and xm−1 = a. Lemma 1. If m  2 and x is a maximal aperiodic m-word, then there exists s ∈ Zm such that xs = a, xs+m−1 = b and bxs+1 · · · xs+m−2 a  x. Proof. Since x is maximal and aperiodic, x0 = b and xm−1 = a. Consider first the case where xm−1 is the only a in x, i.e., x = bm−1 a. Then taking s = m − 1, we get xs · · · xs+m−1 = abm−1 ,

Hence, in this case, the lemma is proved.

bxs+1 · · · xs+m−2 a = bbm−2 a = x.

5

Noninjective aperiodic circle transformation

If m = 2, there are no more cases. Therefore, in what follows, we assume that m  3, and we can suppose that the lemma is proved for all n < m and that there are at least two a in x. Then let x = u0 · · · un−1 , where n  2 and uj = bqj a for some qj  0. By the maximality of x, q0  qj for all j . Suppose that q0  2 and qj  q0 − 2 for some j , and let j be the first such index. Let s be the index in x of the last letter of uj−1 . Then xs = a and xs+m−1 = xs−1 = b. Moreover, the word bxs+1 · · · xs+m−2 a begins with buj = bqj +1 a; therefore, it is less than x which begins with bq0 a. Hence, the lemma is proved in the considered case. It remains to consider the case where each qj equals either q0 or q0 − 1. Let U = {u, v} be an ordered alphabet of two letters with u < v . Let y → y˜ be the homomorphism from U ∗ to A∗ defined by u ˜ = bq0 −1 a q 0 and v˜ = b a. In the remaining case, x = y˜ with y = y0 · · · yn−1 , where yj = u if qj = q0 − 1 and yj = v if qj = q0 . Obviously, y is aperiodic and maximal. Since n < m, we can use the inductive assumption and find a t such that yt = u, yt+n−1 = v and vyt+1 · · · yt+n−2 u  y.

Then ut =

bq0 −1 a,

ut−1 = ut+n−1 =

bq 0 a

and

bq0 aut+1 · · · ut+n−2 bq0 −1 a  x.

If we replace the first letter of the word in the left-hand side by a and the last letter by b, we get the word abq0 −1 aut+1 · · · ut+n−2 bq0 = xs · · · xs+m−1 ;

here s is the index in x of the last letter of ut−1 . Hence, xs = a, xs+m−1 = b and bxs+1 · · · xs+m−2 a  x.

3



THE MERSENNE NUMBERS

This section contains four lemmas, more or less related to Mersenne numbers. The first one is a classical result (see [2, Chapter 4, Exercise 38] and [4, Section 4.3.2, Exercise 6a]); so we omit the proof. Lemma 2. (1) If m, n ∈ N, then gcd(Mm , Mn ) = Md , where d = gcd(m, n). (2) If k and l are nonnegative integers and s ∈ N, then Mk ≡ M l

(mod Ms )

⇐⇒

k≡l

(mod s).

Lemma 3. If k, l, m  0 and aMk + bMl = Mm ,

(3.1)

for some a, b ∈ N, then m  k + l. Proof. Let A denote the set of all triples (k, l, m) of nonnegative integers satisfying (3.1) and such that k + l > m. We have to prove that A = ∅. Suppose the contrary and define    m = min m  (k, l, m ) ∈ A for some k, l . Let (k, l, m) ∈ A. Since Mm  aMk  Mk , we get m  k and analogously m  l. Let, for example, k  l. Then, by (3.1), −a − b ≡ −1 Lith. Math. J., 51(1):3–12, 2011.



 mod 2k ,

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V. Kazakeviˇcius

i.e., a + b = 1 + c2k for some integer c  1. This yields 

 1 + c2k Mk + b(Ml − Mk ) = Mm ,   c2k Mk + b 2l − 2k = 2m − 2k , cMk + bMl−k = Mm−k .

Moreover, k + (l − k) = l > m − k , and, therefore, (k, l − k, m − k) ∈ A. By the minimality of m, k = 0. Then Mm = bMl  Ml , i.e., m  l = l + k , a contradiction.

Lemma 4. If r ∈ ZMn and m = #Orbn (r), then m ∈ A, and m divides any m ∈ A, where A is the set of all Mn m | n such that M | r. Hence, m = gcd A and m = min A. m Proof. For i  0, denote ri = 2i r (mod Mn ). Clearly, m is the least positive integer with the following property: rm = ri for some 0  i < m. Then 2m r ≡ 2i r

(mod Mn ),

2m−i r ≡ r

(mod Mn ),

i.e., rm−i = r0 . By the minimality of m, i = 0. Hence, m is the least positive integer such that 2m r ≡ r (mod Mn ) or, equivalently, Mm r ≡ 0 (mod Mn ). So, for some integer q , Mm r = qMn .

(3.2)

Let d = gcd(m, n); then by Lemma 2, Md = gcd(Mm , Mn ). Denote a = Mm /Md and b = Mn /Md . Then gcd(a, b) = 1 and ar = qb. Hence, a | q . Denote q  = q/a. Then r = q  b, and, therefore, Md r = q  Mn . By Mn Mn the minimality of m, d = m. Hence, m | n. Moreover, (3.2) implies that r = q M , i.e., M | r. m m Mn Mn Mn    Let m | n and Mm | r. Then, for some integers q, q , r = q Mm = q Mm , i.e., qMm = q  Mm .

Denote d = gcd(m, m ). By Lemma 2, Md = gcd(Mm , Mm ); therefore, q

Mm  Mm = q Md Md

and gcd(Mm /Md , Mm /Md ) = 1. This yields q = uMm /Md and q  = uMm /Md for some integer u. Hence, r = uMn /Md , i.e., Md r ≡ 0 (mod Mn ). By the minimality of m, m = d. Hence, m | m .

Remark 1. Let dr = #Orbn (r). By Lemma 4, for all m | n, dr | m

and, therefore,

⇐⇒

 d|m

r≡0

ν d = Mm ;

Mn mod , Mm

Noninjective aperiodic circle transformation

7

here νd denotes the number of all r ∈ ZMn with dr = d. This yields  m

μ Md = ν m , d d|m

where μ is the Möbius function. Moreover, νm equals m times the number of orbits consisting of m elements. Hence, for all m,  m

μ Md ≡ 0 (mod m). d d|m

We thus restated (for a = 2) an old result of K.F. Gauss, who proved (see [6, p. 191]) that  m

μ ad ≡ 0 (mod m) d d|m

for all a, m with gcd(a, m) = 1. Lemma 5. If #Orbn (r) = m, then max Orbn (r) − min Orbn (r) 

 Mn  m−1 2 −1 . Mm

Proof. The inequality is obvious for m = 1. If n = 1, then also m = 1. Therefore, it suffices to consider the case m, n  2. First, we represent any r in ZMn by the binary word r0 · · · rn−1 such that r = r0 · 2n−1 + · · · + rn−1 is the binary expansion of r. Notice that this representation preserves the order and that 2r (mod Mn ) is represented by the word r1 · · · rn−1 r0 . Therefore, if x = max Orbn (r), then its representation x0 · · · xn−1 is maximal. If x0 · · · xn−1 = (x0 · · · xd−1 )k , then d | n and   Mn 2n − 1 =y x = y 2n−d + 2n−2d + · · · + 1 = y d Md 2 −1 with y = x0 · 2d−1 + · · · + xd−1 . By Lemma 3, m | d. Therefore, if m = n, then the word x0 · · · xn−1 is aperiodic. By Lemma 1, there exists s, 0  s  n − 1, such that xs = 0, xs+n−1 = 1, and 1xs+1 · · · xs+n−2 0  x0 · · · xn−1 . Going back to the corresponding integers, we get 2n−1 − 1 = 2n−1 +

n−2 

xs+i 2n−1−i −

n−2 

i=1

xs+i 2n−1−i − 1

i=1

 x − min Orbn (r), as expected. So, the lemma is proved in the case m = n. Mn If m < n, then, by Lemma 3, m | n, and c = M divides r, i.e., r = cr for some r ∈ ZMm . Moreover, m Mn   m if d | m and M Md | r , then d | n and Md | cr = r ; therefore, d = m. This means, again by Lemma 3, that  #Orbm (r ) = m. It is easily checked that Orbn (cr ) = c Orbm (r );

therefore, the general case is reduced to the case just proved.

Lith. Math. J., 51(1):3–12, 2011.

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4 AN APERIODICITY CRITERIUM In this section, we prove the “if” part of the first statement of Theorem 1. Suppose that  n   1 2 −1 α > 2 for all n  1. This immediately yields    f n x = 2n − 1 α

(4.1)

(4.2)

for all 0  x  1/2 and n  1. We have to show that f is aperiodic. Suppose that, on the contrary, f n x = x for some n  1 and x ∈ S . Then, for all i  0, f n f i x = f i f n x = f i x,

and, therefore, 1 f ix > . 2 Indeed, if f i x  1/2, (4.2) and (4.1) yield a contradiction: f ix = f nf ix =



(4.3)

  1 2n − 1 α > . 2

From (4.3) we derive that, for all i  0,

  f i x = 2i − 1 α  2i x.

By (4.4) and x = f n x, we have

(4.4)

  x ≡ 2n − 1 α + 2n x (mod 1),  n    2 − 1 x = − 2n − 1 α + a, a x = −α + n 2 −1

for some a ∈ Z. Then, by (4.3)–(4.4),

 −α +

This yields, for all i  0, 1 α− < 2

2i a 2n − 1





2i a 2n − 1

1 > . 2

< α.

(4.5)

Let 2i a = qi (2n − 1) + ri for some integers qi and 0  ri < 2n − 1. Then (4.5) means that, for all i  0, α−

1 ri < n 1/2}. Proof. (⊃) Let x ∈ / A. Since   A= In (ε1 , . . . , εn ) = ε∈E n1





In (ε1 , . . . , εn ) =

ε∈{0,1}∞ n1





In (ε1 , . . . , εn ),

n1 ε1 ,...,εn ∈{0,1}

there exists n such that x is not contained in any In (ε1 , . . . , εn ). Then either x  1/2 or   x ∈ In−1 (ε1 , . . . , εn−1 ) \ In (ε1 , . . . , εn−1 , 0) ∪ In (ε1 , . . . , εn−1 , 1) for some ε1 , . . . , εn−1 ∈ {0, 1}. In the first case, {(21 − 1)x} = x  1/2, and in the second, (2l + 1)2p−1 + q (2l + 1)2p−1 + q + 1/2  x  2k+p − 1 2k+p − 1

for some l, k, q, p; therefore, {(2k+p − 1)x}  1/2. (⊂) Suppose that, on the contrary, {(2n − 1)x}  1/2 for some x ∈ A and n  1. Let k  1 be the least integer for which there exists x in A such that {(2k − 1)x}  1/2, and let x be any element of A such that {(2k − 1)x}  1/2. Since x ∈ (1/2; 1), we have that k  2. Denote l = (2k − 1)x; then l + 1/2 l