Alexandr Kazda 1 Department of Algebra, Charles University, Prague, Czech Republic

Transcription

1 #A3 INTEGERS 9 (009), 6-79 CONVERGENCE IN MÖBIUS NUMBER SYSTEMS Alexandr Kazda Department of Algera, Charles University, Prague, Czech Repulic alexak@atrey.karlin.mff.cuni.cz Received: 0/30/08, Accepted: /9/09 Astract The Möius numer systems use sequences of Möius transformations to represent the extended real line or, equivalently, the unit complex circle. An infinite sequence of Möius transformations represents a point x on the circle if and only if the transformations, in the limit, take the uniform measure on the circle to the Dirac measure centered at the point x. We present new characterizations of this convergence. Moreover, we show how to improve a known result that guarantees the existence of Möius numer systems for some Möius iterative systems. As Möius numer systems use sushifts instead of the whole symolic space, we can ask what is the language complexity of these sushifts. We offer (under some assumptions) a sufficient and necessary condition for a numer system to e sofic.. Introduction The theory of Möius numer systems was introduced in []. A Möius numer system assigns numers to sequences of Möius transformations otained y composing a finite starting set of Möius transformations. Möius numer systems display complicated dynamical properties and have connections to other kinds of numer representation systems. In particular, Möius numer systems can generalize continued fractions (see [3]). As Möius transformations are ijective on the complex sphere, we cannot use the contraction theorem of Barnsley to define convergence of a series of maps to a point. Instead, in [], convergence of sequences of Möius transformations to points was defined using convergence of measures. An infinite sequence of Möius transformations represents a point x on the circle if and only if the images of the uniform measure on the circle converge to the Dirac measure centered on the point x. There are multiple equivalent ways to state this convergence, including the convergence of the images of point 0. We present several new equivalent definitions of convergence at the end of Section 3. The existence of the numer system is already known for a considerale class of sets of Möius transformations (see [3], Theorem 9). In Section 4, we improve this The research was supported y the Czech Science Foundation research project GAČR 0/09/0854.

2 INTEGERS: 9 (009) 6 result y slightly weakening the assumptions of the existence theorem. As Möius numer systems are essentially sushifts, we can ask what the language of a given system is. Because these systems can e descried in terms of point sets on the unit circle instead of the usual language-theoretic ways, we need tools for investigating the complexity of sushifts involved. In Section 5, we present one such tool for some Möius numer systems, a sufficient and necessary condition for eing sofic, similar to the classic Myhill Nerode theorem for regular languages.. Preliminaries Denote y T the unit circle and y D the closed unit disc in the complex plane. For x, y T denote y (x, y) the counterclockwise interval from x to y. To make notation more convenient, we define the sum x+l for x T and l R y arg(x+l) = arg x+l modulo π. Let A e a finite alphaet. Denote y A the monoid of all finite words on A with the operation of concatenation and y A ω the set of all one-sided infinite words. Let v denote the length of the word v. We use the notations w = w 0 w w and w [i,j] = w i w i+ w j. Let v A e a word of length n. Then we write [v] = {w A ω : w [0,n ] = v} and call the resulting suset of A ω the cylinder of v. Let X e a metric space. We denote y ρ the metric function of X, y Int(V ) interior of the set V and y B r (x) the open all of radius r centered at x. The diameter of a nonempty set V, denoted y diam(v ), is the supremum of {ρ(x, y) : x, y V }. If I is a finite union of intervals on T, denote y I the total length of I. We equip C with the metric ρ(x, y) = x y and T with the circle distance metric (i.e., metric measuring distances along the circle). The shift space A ω of one-sided infinite words comes equipped with the metric ρ(u, v) = max ( { k : u k v k } {0} ). It is easy to see that the topology of A ω is the product topology. A sushift Σ A ω is a set that is oth topologically closed and invariant (i.e., σ(σ) Σ) under the shift map σ(w) i = w i+. The language of a sushift L(Σ) is the set of all words v A such that there exists w Σ and indices i and j satisfying v = w [i,j]. See [5] for details. Unlike in [], we will only consider symolic representations of T. The extended real line R { } is homeomorphic to the unit complex circle using the map u : T R { }, u(z) = iz +. z i Therefore, as long as we are not interested in arithmetics, representing T is equivalent to representing the extended real line. A Möius transformation (MT for short) of the complex sphere C = C { } is

3 INTEGERS: 9 (009) 63 any map of the form F (x) = ax + cx + d where (a, ), (c, d) are linearly independent vectors from C. We can associate with F the matrix A F = ( ) a c d normalized y det F =. It is well-known (see []) that MTs take circles and lines to circles and lines (possily turning a circle into a line or vice versa). Also, Möius transformations form a group under addition and composing MTs corresponds to multiplying their respective matrices: A F G = A F A G. In the following we identify F with A F. By default, we will consider Möius transformations that map D onto D. It is straightforward to prove that these are precisely the transformations of the form ( ) α β F = β α with the normalizing condition α β =. These transformations not only map T onto T, ut they also preserve orientation of intervals on the circle (clockwise versus counterclockwise). Denote y µ the uniform measure on T such that µ(t) =. If ν is a measure on T and F : T T an MT, we define the measure F ν y F ν(e) = ν(f (E)) for all measurale sets E on T. The Dirac measure centered at point x is the measure δ x (E) = {, if x E 0, otherwise for any E measurale suset of T. Definition. Let {F i } i=0 e a sequence of Möius transformations. We say that the sequence {F i } i= represents the point x T if and only if lim i F i µ = δ x, where the convergence of measures is taken in the weak topology, i.e., ν i ν if and only if for all f : T R continuous it is fdν i fdν. In [], Lemma 4 gives a sufficient condition for a sequence of MTs to represent a point. Note that the original statement of this Lemma in [] contains a mistake which we correct here. Lemma. Let {M n } n= e a sequence of Möius transformations. Assume that there exists t T and c > 0 such that for each open interval I with t I we have lim inf n M nµ(i) > c. Then lim n (M n µ)(i) = and lim n M n µ = δ t.

4 INTEGERS: 9 (009) 64 Note that the word open was missing in the original statement of Lemma. It is sufficient to consider only open intervals, while the equality lim n (M n µ)(i) = may fail when t is an endpoint of I: Consider the sequence of transformations {M n } n= where M n (z) = (n+ n)z+(n n) (n n)z+(n+ n). Then M nµ δ ut the interval I = [, ] is invariant under all M n and so lim n (M n µ)(i) = µ(i) =. We will need a slightly modified Proposition 6 from []: Lemma 3. Let {F i } i=0 e a sequence of Möius transformations. Then {F i} i=0 represents x T if and only if lim i F i (0) = x. 3. General Properties of Möius Transformations We will now show several useful properties of circle-preserving MTs, as well as equivalent descriptions of what it means for a sequence of MTs to represent a point on T. Definition 4. A Möius transformation of the form R α (z) = e iα z is called a rotation. A contraction to is a transformation of the form ( ( ) ( )) C r = r + r r ( ) ( r ) r r +, where r. r Oviously, C is the identity map and the identity transformation is oth a contraction to and a rotation. Oserve that any contraction to fixes the points ±. For r >, C r acts on T y making all points (with the exception of ) flow towards and so the sequence {C n } n= represents the point. Lemma 5. Let F e a Möius transformation. Then there exist two rotations R φ, R φ and C r, a contraction to, such that F = R φ C r R φ. Moreover, if F is not a rotation then R φ, R φ, C r are uniquely determined y F. Proof. We want to satisfy the equation ( ) ( α β = F = R β α φ C r R φ = ( ) r + r e i φ +φ ( ) r r e i φ φ r ( r ( r r + r ) e i φ φ ) e i φ +φ ( ) ( ) Choose r so that r + r = α. It is easy to see that we will then have r r = β. Now choose the parameters of the rotations φ and φ so that we satisfy the conditions φ + φ = arg α and φ φ = arg β and we are done. Oviously, as long as β 0, this linear system has a unique solution. Because β = 0 if and only if F is a rotation, the uniqueness follows. ).

5 INTEGERS: 9 (009) 65 Denote y F (x) the modulus of the derivative of F at x. Direct calculation gives us that F (x) = βx + α Definition 6. Let F e a Möius transformation. Then, inspired y [] and [], we define the four point sets U = {x T : F (x) < }, V = {x T : (F ) (x) > }, C = {x C : F (x) }, D = {x C : (F ) (x) }. Call U the contraction interval of F, V the expansion interval of F and C resp. D the expansion sets of F resp. F. Oviously, U = T \ C and V = T Int(D). By Lemma 5 we have that for every F there exist φ, φ and r such that F = R φ C r R φ. As R φ =, we have F (x) = C r (R φ (x)) and F (x) = Cr (R φ (x)). Because the sets U and C are defined using F (x) = C r (R φ (x)), the value of r determines the shapes and sizes of U and C while φ rotates U and C clockwise around the point 0. Similarly, the shapes of V and D depend on r while φ determines positions of V and D, rotating them (counterclockwise) around 0. Lemma 7. Let F e a Möius transformation that is not a rotation. following hold: Then the. F (C \ C) = Int D.. F (U) = V. 3. C and D are circles with the same radius β and centers c, d such that c = d = β V < π. 5. U + V = π. 6. If x y are points in V then I, the shorter of the two intervals joining x, y, lies in V. Proof. Part () follows from the formula for the derivative of composite function. We can write F (F (z)) = F (z) and so F (z) < if and only if F (F (z)) >. Part () is a consequence of () and the definition of U and V.

6 INTEGERS: 9 (009) 66 U c e + C 0 V d e D Figure : The geometry of C and D. We prove (3) y direct calculation. We have F (x) = x C if and only if x + α β < β. βx+α and therefore This is the equation of a circle with the center c = α β and radius β. Also, it is c = α + β β = = β β +. The case of F is similar. Elementary analysis of the situation yields (4) and (5), with (6) eing a direct consequence of (4) (see Figure 3). Remark 8. Oserve that the triangles 0de + and 0de in Figure 3 are right y Pythagoras ( theorem. ) Also, we can compute that the length of V is equal to β arccos and the distance of d from V is + β. Therefore, + β the size of D, length of V and the distance of d and T are all decreasing functions of β. Theorem 9. Let {F n } n= e a sequence of MTs. Denote y V n the expansion interval of Fn, D n the expansion set of Fn and d n the center of D n. Then the following statements are equivalent:

7 INTEGERS: 9 (009) 67 () The sequence {F n } n= represents x T. () lim n V n = {x} (3) lim n d n = x (4) lim n D n = {x} (5) For all K Int(D) compact we have lim n F n(k) = {x}. (6) For all z Int(D) we have lim n F n(z) = x. (7) There exists z Int(D) such that lim n F n(z) = x. In (), (4), and (5), we take convergence in the Hausdorff metric on the space of nonempty compact susets of T, C and D, respectively. In particular, E n {x} if and only if for every ε > 0 there exists n 0 such that n > n 0 we have E n B ε (x). Proof. Assume (). To prove (), we show that the length V n, as well as the distance ρ(x, V n ) (measured along T), go to zero. Consider the interval I ε = (x ε, x + ε) of T. It is easy to see that we have lim n Fn (I ε ) = π. Therefore, there exists n 0 such that Fn (I ε ) > for all n > n 0. If now ε < and n > n 0, then Fn (I ε ) > I and so there exists z I ε such that Fn (z) >. Therefore z I Vn, so ρ(x, D n ) ε. Now assume that there exists r > 0 such that for any n 0 there exists n > n 0 with V n > r. For any interval I we have the inequality F n (I) = F (I V n ) + F (I \ V n ) U n + I = π V n + I, n n as the part of I inside V n will expand at most to the length U n and the part of I outside V n will only get shorter. Now consider I r. If V n > r then Fn (I r ) < π r+r = π r. Therefore, it cannot e true that F n (I r ) π, a contradiction. An elementary examination of the geometry of V n and D n shows that (), (3) and (4) are all equivalent. The distance of d n from V n is an increasing function of V n and ρ(d n, V n ) 0 as V n 0, proving (3). In a similar way, the diameter of D n goes to zero when ρ(d n, T) 0. Denote y C n the expansion set of Fn and assume (4). Recall that the diameter of C n is always equal to the diameter of D n. This means that diam(c n ) 0 and so for any compact K Int(D) there exists n 0 such that K C n = holds whenever n > n 0. Then F n (K) D n and so F n (K) {x}, proving (5). As {z} is a compact set, (6) easily follows from (5). Also, statement (7) is an ovious consequence of (6). It remains to prove (7) (). Consider the circle-preserving MT M that sends 0 to z. Then the sequence {F n M} represents the point x y Lemma 3. Therefore,

8 INTEGERS: 9 (009) 68 if I is an open interval on T containing x, then lim (F n M) (I) π. It remains to see that (F n M) (I) = M Fn (I) m Fn (I) where m > 0 is the minimum of M (y) for y T. Therefore, {F n } n=0 represents x y Lemma and we are done. Remark 0. To see that we cannot expect to have convergence for all z T, let {m n } n= e a sequence of real numers such that the set E = {z : z = R mn ( ) for infinitely many n} is dense in T. Consider the sequence of transformations F n = C n R mn. Oviously, {F n } n= represents the point, ut the set {x T : lim n F n (x) = } contains the whole E and therefore is dense in T. Corollary. Let {F k } k= e a sequence of Möius transformations such that for some x 0 T we have lim k (F k ) (x 0 ) =. Then {F k } k= represents x 0. Proof. As (F k ) (x 0 ) = β k x 0 + α k, we have α k x β 0. But this means precisely that d k x 0 and so {F k } must k represent x 0 y part () of Theorem Möius Numer Systems In an attempt to put Möius numer systems in a more astract context we give one possile definition of symolic representation. Afterwards, we prove Theorem, which strengthens part of Theorem 9 of []. Definition. Let X e a compact topological space. A mapping P : L(Σ) X together with a sushift Σ A ω form a symolic representation of X if the following are true:. P (v) is a closed suset of X for each v L(Σ).. i= P (w [0,i]) is a singleton for every w Σ. Denote the element of this set y p(w). 3. The map p : Σ X is surjective. The aove definition is reminiscent of the definition of a generating cover from [4]. The following lemma is an exercise in the use of compactness.

9 INTEGERS: 9 (009) 69 Lemma 3. Let P e a symolic representation of X. Then p is continuous in Σ. Proof. Given an open set U X we want to show that p (U) is open in Σ. Let w p (U) and assume that for all k there exists w such that w [0,k] = w [0,k] and p(w ) U. Then we must have k i= P (w [0,i]) U c and from compactness of X we otain that i= P (w [0,i]) U c. But then i= P (w [0,i]) cannot e a singleton as p(w) U c, a contradiction. In the rest of this section, let us have a set {F a : a A} of MTs where A is a finite alphaet. Such a set is called a Möius iterative system. Assign to each word v A the transformation F v = F v0 F v F vn where n = v. Denote y V v the expansion interval of Fv. For a A, lael e + a the counterclockwise and e a the clockwise endpoint of the interval V a (that is, in our notation we have V a = (e a, e + a )). Definition 4. The map Φ assigns to each w A ω the point x T such that {F w[0,n) } n= represents x. If {F w[0,n) } n= does not represent any point in T, let Φ(w) e undefined. Denote the domain of Φ y X F. Definition 5. The sushift Σ A ω is a Möius numer system if and only if Σ X F, Φ(Σ) = T and Φ Σ is continuous. It is easy to oserve that if Φ(w) = x then Φ(σ(w)) = F w 0 (x). We will use this simple property later. Originally, the following theorem comes from [3]. Here, we slightly modified it so that it refers to T instead of the extended real line. It gives some sufficient and some necessary conditions for a Möius iterative system to represent T. Theorem 6. (Theorem 9, [3]). Let F : A + T T e a Möius iterative system.. If u A + V u T, then Φ(X F ) T.. If {V u : u A + } is a cover of T, then Φ(X F ) = T and there exists a sushift Σ X F on which Φ is continuous and Φ(Σ) = T. Oserve that (y compactness of T) if {V u : u A + } covers T, then there exists a finite B with {V : B} covering T. We show that in part two of Theorem 6 it suffices to demand that there exists a finite B A such that the set of closures {V : B} covers T. First, we prove several technical lemmas needed to nail down Theorem. The following lemma is stated in a different form in [] as Lemma. Lemma 7. Let {F a : a A} e a Möius iterative system and B A e finite. Let L e the length of the longest of intervals in {V : B}. Then there exists an increasing continuous function ψ : [0, L] R such that ψ(0) = 0, ψ(l) > l for

10 INTEGERS: 9 (009) 70 x k+ x k I k+ I k V y k+ y k U Figure : A twist l > 0 and if I is an interval and B a word such that I V then F (I) ψ( I ). Proof. Thanks to Lemma 5 we can without loss of generality assume that all F are contractions to. Let ψ(l) = inf{ F (I) : I V and I = l}. By analyzing contractions, it is easy to see that ψ is increasing, continuous and ψ(l) > l for l > 0. Lemma 8. Let {F a : a A} e a Möius iterative system. Let x, y e points and w A ω a word such that for all k we have Fw (x), Fw (y) V wk. Then x = y or there exists a k 0 such that F w[k,k+] is a rotation for all k > k 0. Proof. Denote x k = Fw (x), y k = Fw (y). Assume that x y. Then for each k there exists the closed interval I k with endpoints x k, y k such that I k V wk. Moreover, I k > 0 for all k, as Möius transformations are ijective on T. Oserve first that the sequence { I k } k=0 is a nondecreasing one. For any particular k we have two possiilities: Either Fw k (I k ) = I k+ and therefore I k+ ψ( I k ) > I k (y Lemma 7), or I k+ = T \ Fw k (I k ); see Figure. In the second case, T \ U wk I k+ and so I k+ V wk I k. Call the second case a twist. Assume first that the numer of twists is infinite. Whenever a twist happens for some k, we have V wk+ I k+ V wk. There must exist a k 0 such that I k is constant for all k > k 0. Otherwise, the letter set {w k : k = 0,,,... } would need to contain infinitely many letters, as I k+ > V wk means we cannot use the letter w k again. Therefore, I k+ = V wk for all ut finitely many twists and the finiteness of A gives us I k = I l for all k, l sufficiently large. Assume that k > k 0. Simple case analysis shows that I k = I k+ only when x k, y k are the endpoints of V wk and the transition is a twist. Therefore, x k+, y k+ are also endpoints of V wk+ and I k+ = I k. By applying R, the rotation sending I k+ to I k, we otain that R Fw [k,k+] has two fixed points x k, y k and that

11 INTEGERS: 9 (009) 7 (R Fw [k,k+] ) (x k ) = (R Fw [k,k+] ) (y k ) =. This can happen only when R Fw [k,k+] = id. Therefore, Fw [k,k+] is a rotation for all k > k 0 and we are done. It remains to investigate the case when the numer of twists is finite, i.e., Fw k (I k ) = I k+ for all k k 0. By Lemma 7 we otain I k+ ψ( I k ) for all k k 0 and therefore I k ψ k k0 ( I k0 ) for all k k 0. Let l = I k0 so that if x y, then l > 0. Consider the sequence {ψ k (l)} k=0. Assume that ψk (l) L for all k. Then this sequence is increasing and ounded and therefore it has a limit. But the only fixed point of ψ is 0, a contradiction. Therefore, there always exists a k such that ψ k (l) > L. Then I k+k0 cannot possily fit into any of the intervals V, a contradiction with x k, y k V wk. This means that we must have x = y. The following easy oservation is going to e useful when constructing our numer system. Lemma 9. Let I,..., I k e open intervals on T. Then x k i= I i if and only if oth of the following conditions hold: () x k i= I i. () If x is an endpoint of I i and I j, then it is either a counterclockwise endpoint of oth intervals or a clockwise endpoint of oth intervals. Proof. Oviously, if x k i= I i then x k i= I i. Were x a clockwise endpoint of I i and counterclockwise endpoint of I j then there would exist a neighorhood E of x such that E I i I j = and so x does not elong to k i= I i. In the other direction, assume that x has all the required properties. Then a simple case analysis shows that for any neighorhood E of x we have E k i= I i and so x k i= I i. Assume that {F : B} is a Möius iterative system. Given (x, w) T B ω, we use the shorthand notation x i = Fw [0,i) (x). Define X T B ω to e the set of all pairs (x, w) such that: () For all i =,,... we have x i V wi. () For no i it is true that x i = e + w i and x i+ = e w i+, neither it is true that x i = e w i and x i+ = e + w i+. Note that the second condition says that endpoints cannot alternate : If x i, x i+ are endpoints of their respective intervals then they are oth of the same type (clockwise or counterclockwise). The situation is illustrated in Figure 3.

12 INTEGERS: 9 (009) 7 x i = F w[0,i ) (x) F w i (V wi ) V wi V wi x i+ = F w [0,i) (x) Figure 3: The situation in part () We show that the set X is actually a closed suspace of T B ω. Using Lemma 9, it is easy to see that (x, w) X if and only if x F w[0,i) (V wi ) F w[0,i+) (V wi+ ) for all i Denote P (v) = n i=0 F v [0,i) (V vi ) F v[0,i+) (V vi+ ) for v B such that v = n >. For formal reasons, let P () = V for B and P (λ) = T where λ is the empty word. Rewriting the aove oservation, (x, w) X if and only if x i=0 P (w [0,i)), and therefore X = k=0 v =k P (v) [v]. As the set v =k P (v) [v] is closed for every k, we have that X is a closed (and hence compact) suset of T B ω. Oservation 0. If (x, w) X, then F w [k,k+] is not a rotation for any k. Proof. From () and () we otain that the open set W = V wk F wk (V wk+ ) is nonempty. Now it is easy to see that Fw [k,k+] is an expansion on W and therefore Fw [k,k+] cannot e a rotation. We are now ready to improve Theorem 6. Theorem. Let {F a : a A} e a Möius iterative system. Assume that there exists a finite suset B of A + such that {V : B} covers T. Then there exists a sushift Σ A ω that, together with the iterative system {F a : a A}, forms a Möius numer system. Proof. Take B as our new alphaet. We will show that there exists Σ B ω with the required properties. Once we have Σ, we can define the sushift Σ A ω as

13 INTEGERS: 9 (009) 73 Σ = d i=0 σd (Σ) where d = max{ : B}. It is easy to see that if Σ is a Möius numer system then Σ is also a Möius numer system. Consider the set X T B ω introduced aove. We have already shown that X is closed and therefore compact. Taking projections π, π of X to first and second elements, we otain the set of points π (X) T and the set of words π (X) B ω. To conclude our proof, we verify that: (i) π (X) = T. (ii) π (X) = Σ is a sushift of B ω. (iii) For (x, w) X we have Φ(w) = x. (iv) The map P (v) together with Σ forms a symolic representation of T. (v) Φ Σ is continuous. (i) This follows from the fact that the system {V : B} covers T. Given x T we can construct w B ω with (x, w) X y induction. (ii) As X is compact and π is continuous, Σ is compact. To show σ-invariance, consider (x, w) X. Then it is easy to see that (Fw 0 (x), σ(w)) X. (iii) Let (x, w) X. Recall the notation x i = Fw [0,i) (x). We will divide the proof into several cases. Assume first that x i is an endpoint of V wi for all i. Then either x i is always the clockwise endpoint or always the counterclockwise endpoint of V wi. Assume without loss of generality that x i = e + i and denote l = min{ V : B}. We will use Lemma. Let I V w0 e a nondegenerate closed interval with x as the counterclockwise endpoint. We want to show that lim inf i Fw [0,i) (I) l, that is, there exists j such that i > j Fw [0,i) (I) l. This will e enough to prove Φ(w) = x since for every J open interval, x J, there exists a suitale I J. Oserve first that if Fw [0,i) (I) l for some i then Fw [0,i) (I) V wi l and, ecause F w i expands all intervals in V wi, we have F w [0,i+) (I) > l. All we have to show is that there exists i such that the length Fw [0,i) (I) is at least l. But this follows from Lemma 8; all we have to do is choose x y I and oserve that is not a rotation for any k y Oservation 0. Therefore, in this case we have Φ(w) = x. If there exists k such that x i is an endpoint of V wi for all i k then Φ(σ k (w)) = F w [k,k+] Fw (x) y the aove argument, and so Φ(w) = x, which was to e proven. It remains to consider the case when x i is not endpoint for infinitely many values of i. We want to show that then lim i (Fw [0,i) ) (x) =. This will e enough to finish the proof, thanks to Lemma. First of all, note that there exists a numer ξ > 0 with the following property: For all, the intervals ( F (e ), F (e + ξ) + ξ) and ( F (e + ξ) ξ, F (e + )) do not contain any endpoint e c, e + c of any interval V c (where c B, see Figure 4). The existence of ξ follows from the fact that the set of all endpoints of all intervals

14 INTEGERS: 9 (009) 74 e + e + ξ No endpoints here. V F (e + ξ) ξ F (e + ξ) F (e + ) Figure 4: The condition for ξ {V c : c B} is finite. We find such a ξ for every B and then choose the minimum ξ as the gloal ξ. Oserve that (Fw [0,i) ) i (x) = k=0 (Fw k ) i (Fw (x)) = k=0 (F w k ) (x k ). ( ) As x k V wk, we have that (Fw k ) (x k ). Moreover, there exists a δ > 0 such that (F ) (y) > + δ whenever y [e + ξ, e+ ξ]. Therefore, the only way the limit of the product ( ) can e finite is when there exists an i 0 such that ρ(x i, {e w i, e + w i }) < ξ for all i i 0. But this leads to a contradiction: Assume (without loss of generality) ρ(x i0, e + w i0 ) < ξ. Then we have ρ(x i0+, e w ) ξ (ecause e i0 + w [x i0 + i 0+, x i0++ ξ) y the choice of ξ) and so ρ(x i0+, e + w i0 ) < ξ. By induction, it is then true that + ρ(x i, e + w i ) < ξ for all i i 0 and we otain that Fw (e+ [i0,i) w i0 ) V wi (ecause y the choice of ξ, we have e + w i0 (F + w (e+ [i0,i) w i0 ), x i0+], see Figure 5), which is a contradiction with Lemma 8 and Oservation 0. (iv) All we need to show is that P (w) = i=0 P (w [0,i)) is a singleton for every w Σ. We know that when (x, w) X then x P (w). Let y P (w). Then for all k we have Fw ({x, y}) V wk and applying Lemma 8 together with Oservation 0 we otain x = y, which is what we needed.

15 INTEGERS: 9 (009) 75 e + w i+0 x i0 V wi0 F w i0 (x i0 ) = x i0+ ξ F w i0 (e + w i0 ) e + w i0 + Figure 5: The situation rought aout y the properties of ξ (v) Follows from the previous point and Lemma 3. From the proof of Theorem, we otain that there are alternative sushifts that also produce a Möius numer system. Assume that we replace all the sets V, B with some smaller open intervals W, B satisfying W V and B W = T. Then we can define a Möius numer system derived from the set X T A ω. Let (x, w) X if and only if: (3) For all i =,,... we have x i W wi. (4) There are no i, j indices such that x i is the counterclockwise endpoint of W wi and x j is the clockwise endpoint of W wj. Note that the new conditions are at least as restrictive as the original conditions () and (), therefore X X. The set X is closed, which follows from the oservation that (x, w) X if and only if and therefore k k, x F w[0,i) (W wi ) i=0 n X = F v[0,i) (W vi ) [v]. n=0 v =n i=0

16 INTEGERS: 9 (009) 76 To see that π (X ) = T, consider x T. Then there exists w 0 with x W w0 such that x is not the clockwise endpoint of W w0. Similarly, there always is a set W w such that Fw 0 (x) W w and Fw 0 (x) is not the clockwise endpoint of W w. We continue in this manner, producing w, w 3,... until we have the whole word w. By the same arguments as in proof of Theorem, we then otain that Σ = π (X ) is a sushift. And X X gives us that Φ Σ is continuous and Φ(w) = x for (x, w) X. We conclude that Σ together with {F : B} forms a Möius numer system. Corollary. Let {W : B} e a set of open intervals such that W V for every and B W = T. Let Σ W B ω e a sushift such that w Σ W if and only if k, k F w[0,i) (W wi ). i=0 Then Σ W together with {F : B} is a Möius numer system. Proof. It suffices to show that Σ W = Σ. Let w B ω and denote E k = k i=0 F w [0,i) (W wi ). By compactness of T, w Σ W if and only if there exists x k=0 E k. But from Lemma 9, we otain that x k=0 E k if and only if (x, w) X. So w Σ W x, (x, w) X, which means that Σ W = π (X ) = Σ. 5. Sofic Representations A sushift Σ is called sofic if and only if the language of Σ is regular (recognizale y a finite automaton). Papers [] and [3] contain several examples of such sushifts. In the following, we offer a tool for testing whether a Möius numer system is sofic. The tool is far from universal, ut should cover most interesting systems. Define the follower set of a word v as F v = {w Σ : vw Σ} and denote the image of F v y Z v = Φ(F v ). Oservation 3. Let Σ e a Möius numer system such that Σ is sofic. Then {Z v : v A } is a finite set. Proof. It is a well-known fact that a sushift is sofic if and only if {F v : v A } is finite (see [5]). Therefore, if Σ is sofic then {Z v : v A } = {Φ(F v ) : v A } is finite.

17 INTEGERS: 9 (009) 77 We state the converse statement only for one special class of Möius numer systems. We shall see in a moment that the shift from Theorem is precisely such a shift. Theorem 4. Let t e a positive integer. Let X T A ω e a set such that for all x T, v A with v t we have the equivalence [ ( )] w A ω, (x, vw) X k = 0,,..., v t, τ v [k,k+t), Fv (x) where τ is some predicate (the same condition for all k). Assume moreover that Φ(w) = x for all (x, w) X. Then L(π (X)) is regular if and only if {Z v : v A } is finite. Proof. The only if direction of the claim was proved in Oservation 3. To prove the other direction, assume that X satisfies all the requirements. We construct a finite automaton recognizing the language of Σ. The states of our automaton will e T v = {(Z v, v [ v t, v ) ) : v A }. The a transitions are of the form T v Tva. Designate (Z λ, λ) as the initial state and let all states e accepting except for states of the form (, u). As v L(Σ) Z v, our automaton indeed recognizes L(Σ). To complete the proof, we need to show that the automaton is defined correctly, i.e., if T v = T u then T va = T ua. We do this y showing that x Z va if and only if x Fa (Z v ) & τ(v [ v + t, v ) a, x). This will e enough, as then T va is a function of T v. We can assume that v > t, as we can add all the finitely many short words as special states. By rewriting the conditions, we otain: Z v = { x : w, (F v (x), vw) X }, { ( )} Z v = x : k = 0,,..., v t, τ v [k,k+t), Fv (F v (x)), { ( )} Z va = x : k = 0,,..., v t +, τ (va) [k,k+t), Fva (F v F a (x)). A simple inspection of the conditions for x Z va shows that we can write them in the form: Z va = { x : F a (x) Z v & τ ( va [ v + t, v +), F va (F v F a (x)) )}, Z va = { x : F a (x) Z v & τ ( v [ v + t, v ) a, x )}. Therefore, Z va indeed depends only on Z v and the word v [ v + t, v ) a. As a sample application of this theorem, we prove a criterion for soficity of the Möius numer system introduced in Theorem.

18 INTEGERS: 9 (009) 78 Corollary 5. Assume that Σ is the Möius numer system set up in Theorem. Then the shift Σ is sofic if and only if {Fv (P (v)) : v A } is a finite set. Proof. We can assume that B = A, as the transition from sushift of B ω to a sushift of A ω does not affect soficity due to B eing finite. We know that (x, w) X if and only if x i=0 P (w [0,i)). Moreover, given v, we can find for each x P (v) a w such that (x, vw) X ecause {V a : a A} covers T. Therefore Z v = Fv (P (v)). Setting t = and τ(u, y) equal to y P (u) satisfies the requirements of Theorem 4 and so Σ is sofic if and only if {Z v : v A } = {Fv (P (v)) : v A } is finite. Recall that after Theorem, we have introduced an additional Möius numer system. This shift Σ would require extra effort, due to the nonlocal character of (3) and (4). On the other hand, it is easy to see that if we take some system of open intervals {V : B} such that V V and {V : B} covers T and write conditions ( ) and ( ) with V replaced y V then the proof of Theorem as well as of Corollary 5 goes veratim. Therefore, one strategy to produce sofic Möius numer systems could e to find some cover of T, say {V : B}, and then try to adjust the sets {V : B} to otain a system such that {Z v : v A } is finite and so the resulting shift is sofic. At present, we are not ale to show any concrete examples of sofic systems otained in this way. 6. Conclusions In the whole paper we have uilt upon the results of [] and [3]. In particular, Theorem narrows the gap etween the Möius iterative systems that admit a Möius numer system and the iterative systems that do not. However, there are quite a few open prolems, practical as well as theoretical, in this area. Most oviously, we would like to have a sufficient and necessary condition for the existence of a Möius numer system for a given iterative system. Having such a characterization, we would like to know if it is effective, i.e., given a Möius iterative system with coefficients in some suitale computale field, can we decide whether there is a corresponding Möius numer system? For manipulation with numer systems, it would e nice to have a sofic Möius numer system. There are several known examples of such systems, ut we do not know whether existence of a Möius numer system for a given iterative system implies the existence of a sofic system or not. Neither can we at the moment tell if a given set of MTs admits a sofic Möius numer system. We have offered a tool to ring light to the sofic prolem in Theorem 4. Unfortunately, our result does not work for all systems and requires us to generate possily

19 INTEGERS: 9 (009) 79 infinitely many sets Z v (although the generation process eventually stops if the system in question is sofic). A large part the complexity of these prolems seems to come not from the numer systems themselves ut from the fact that we don t properly understand how do large numers of MTs compose (or, equivalently, how long sequences of matrices multiply). This suggests that maye the way forward lies in studying the limits of products of matrices. Acknowledgements. The author thanks Petr Kůrka for support and stimulating discussions. References [] Svetlana Katok. Fuchsian Groups. The University of Chicago Press, Chicago and London, 99. [] Petr Kůrka. A symolic representation of the real Möius group. Nonlinearity, :63 63, 008. [3] Petr Kůrka. Möius numer systems with sofic sushifts. Nonlinearity, (): , 009. [4] Petr Kůrka. Topological and symolic dynamics. Société Mathématique de France, Paris, 003. [5] Brian Marcus and Douglas Lind. Introduction to Symolic Dynamics and Coding. Camridge University Press, Camridge, 995.