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The Fibonacci Unimodal Map.

arXiv:math/9201291v1 [math.DS] 12 Aug 1991The Fibonacci Unimodal Map.Mikhail Lyubich and John MilnorMathematics Department and IMS, SUNY Stony BrookTo Yuri Lyubich on his 60th birthday.§1. Introduction.The Fibonacci recurrence of the critical orbit appeared in the work of Branner and Hubbard on complexcubic polynomials [BH, §12], and in Yoccoz’s work [Y] on quadratic ones, as the “worst” pattern of recurrence.On the other hand, a real quadratic Fibonacci map f was suggested by Hofbauer and Keller [HK] asa possible candidate for a map having a “wild” attractor (because the ω -limit set of the critical pointpossesses all known topological properties of wild attractors [BL2] ).

Also, Shibayama [Sh] has describedthis real Fibonacci map as the limit of a sequence of quadratic maps with attracting orbit whose period isa Fibonacci number.This paper will study topological, geometrical and measure-theoretical properties of the real Fibonaccimap. Our goal was to figure out if this type of recurrence really gives any pathological examples and tocompare it with the infinitely renormalizable patterns of recurrence studied by Sullivan [S].

It turns out thatthe situation can be understood completely and is of quite regular nature. In particular, any Fibonacci map(with negative Schwarzian and non-degenerate critical point) has an absolutely continuous invariant measure(so, we deal with a “regular” type of chaotic dynamics).

It turns out also that geometrical properties of theclosure of the critical orbit are quite different from those of the Feigenbaum map: its Hausdorffdimensionis equal to zero and its geometry is not rigid but depends on one parameter.Branner and Hubbard introduce the concept of a tableau in order to describe recurrence of criticalorbits.Their “Fibonacci tableau” is a basic example, which corresponds to one particularly close andregular pattern of recurrence. If a complex quadratic map z 7→z2 + c realizes this Fibonacci tableau, thenthe orbit0 = z0 7→z1 7→z2 7→· · ·of the critical point returns closer to zero (in a certain invariant sense) after each Fibonacci number ofiterations.

In the real case, it follows that|z1| > |z2| > |z3| > |z5| > |z8| > |z13| > · · · .In §2 we will prove that a real quadratic map is uniquely defined by the last property. More precisely weprove the following.

We denote the Fibonacci numbers byu(1) = 1 , u(2) = 2 , . .

. ,withu(n + 1) = u(n) + u(n −1) .Theorem1.1.Thereisoneandonlyonerealquadraticmapoftheformfc(x) = x2 + cwith the property that the critical orbit0 = x0 7→x1 7→· · ·has closest re-currence at the Fibonacci values, so that|x1| > |x2| > |x3| > |x5| > · · · , with x4 < 0 .

† Thekneading invariant for this uniquely defined map fc can be described by the conditions thatxu(n) < 0forn ≡0 , 1mod 4xu(n) > 0forn ≡2 , 3mod 4 ,and thatsgn(xi) = sgn(xi−u(n))foru(n) < i < u(n + 1) .Infactnumericalcomputationshowsthatc=−1.8705286321646448888906 · · ·.The associated topological entropy ish = log 1.7292119317 · · · .For a fairly general unimodal map f with this same kneading data, we prove the following.LetO = {x0 , x1 , . .

.} ⊂R be the critical orbit.† We conjecture that this condition on x4 is automatically satisfied.1

Theorem 1.2. If f is C2 - smooth with non-flat critical point, and with kneading data as above,then:1.

The closure O of the critical orbit is a Cantor set, with the xi , i ≥1, as the end points ofthe complementary intervals.2. The map f from O onto itself is one-to-one except that the critical point has two pre-images.This map f|O is minimal, and is uniquely ergodic with entropy zero.

It is semi-conjugate tothe golden rotationt 7→t −(√5 −1)/2(mod 1)of the circle R/Z .The proof, in §3, will give an explicit description of the ordering of this critical orbit closure. It will alsoshow that it is canonically homeomorphic to the set of all infinite sequences (a1 , a2 , .

. . ) of zeros and oneswith no two consecutive ones, or to the set of all finite or infinite “Fibonacci sums”.

(Compare 2.3 and 3.3. )Theorem 1.3.

If f is C2 -smooth with non-degenerate critical point then:1. The ratio of xu(n) to xu(n−1) decreases exponentially, withλn ≡|xu(n)|/|xu(n−1)| ∼a/2n/3asn →∞for some constant a > 0 .2.

The critical orbit closure O has Hausdorffdimension zero and the Liapunov exponent at thecritical value is equal to zero.3. Any two Fibonacci maps with the same parameter a are smoothly conjugate on O .4.

If the Schwarzian derivative is negative, then f has a unique absolutely continuous invariantmeasure, with support equal to the entire closed interval [x1 , x2] , and with positive entropy .Remark 1. Uniqueness and other properties of an absolutely continuous invariant measure hold auto-matically (see [BL2].) Existence we will derive from the Nowicki-van Strien “series” condition [NvS].Remark 2.

Unlike the Feigenbaum map, the geometry of O goes down to zero under renormalization,and is not rigid but depends on the parameter a . (We can effectively vary this parameter).Remark 3.

It is essential here that the critical point be non-degenerate ( f ′′(0) ̸= 0 ). We hope toshow in a later paper that, for example, a Fibonacci map of the form f(x) = x4 + c has completely differentbehavior, with bounded geometry and with no absolutely continuous invariant measure.Let us describe the structure of the proof of the last theorem, which is somewhat complicated.

In §4we get some a priori bounds on the ratios λn . In §5 we prove the Theorem assuming that inf λn = 0 .

Inorder to verify this assumption we introduce in §6 an appropriate notion of renormalization so that infinitelyrenormalizable maps are exactly Fibonacci ones. Applying Sullivan’s ideas [S] to our case we prove that ifgeometry of O is bounded from below then there is a sequence of renormalizations converging to a mapwhich can be analytically continued in a quite big domain of the complex plane.In §7 we discuss polynomial-like maps, in an appropriate generalized sense.

A version of the Douady-Hubbard theorem is valid in this situation: any cubic-like map is quasi-conformally conjugate to a cubicpolynomial with one escaping critical point. It follows that all real cubic-like Fibonacci maps are quasi-symmetrically conjugate.

So, any example of a cubic-like Fibonacci map with unbounded geometry showsthat all of them have unbounded geometry. Finally, we renormalize a quadratic-like Fibonacci map into acubic-like one which completes the proof for the polynomial-like case.In the last §8 we show that the limits of the renormalizations of a smooth Fibonacci map are actuallypolynomial-like which completes the proof of the Theorem.Remark 4.

The Fibonacci recurrence is a well-known phenomenon for monotone maps of the circlewith golden rotation number. The scaling laws in this situation were studied by Herman (at least implicitly),by Swiatek [Sw1] (smooth homeomorhisms with critical points) and by Tangerman and Veerman [TV] (mapswith flat spots).

In the two former cases one has bounded geometry, in the latter the geometry goes down tozero in the similar manner as in our example. Such circle maps are explicitely related to certain unimodalmaps of the interval which are different from ours but also have a sort of Fibonacci recurrence ; see [PTT].2

The notation f n will always be used for the n -fold iterate of f .Acknowledgement. We want to thank Branner, Douady, Sullivan, and Tresser, for helpful conversations.We also profited from the discussions with the participants of the Stony Brook dynamical systems seminar,particularly: Brucks, Yu.Lyubich, Shishikura, Tangerman and Veerman.3

§2. Kneading.Letf:I→Ibeaunimodalmapwithminimumatx=0 .Asusual,let0 = x0 7→x1 7→· · ·be the critical orbit, and letu(1) = 1 , u(2) = 2 , u(3) = 3 , u(4) = 5 , .

. .be the Fibonacci numbers.

In order to avoid the hypothesis that f is an even function, we will use thenotation x 7→x′ for the order reversing involution, defined on some suitable subinterval of I , which satisfiesf(x′) = f(x) . Let ∥x∥be the larger of x and x′ .Definition.

We will say that f is a Fibonacci map if ∥xu(n)∥> ∥xu(n+1)∥for n ≥1 , so that∥x1∥> ∥x2∥> ∥x3∥> ∥x5∥> ∥x8∥> ∥x13∥> · · · ,(2 −1)and if x4 < 0 .Lemma 2.1. The map f is a Fibonacci map if and only if the signs of the successive images xiare given bysgn(xj) = sgn(xj−u(n))foru(n) < j < u(n + 1) ,with(2 −2)sgn(xu(n)) = (−1)(n+1)(n+2)/2 .

(2 −3)Remark1.Someconditionsuchasx4<0isneededinordertoavoidtheuninteresting casex1 < 0

Thus this particular case can never occur for a quadratic map. )Remark 2.We can describe these conditions in different language as follows.If we assume thatx1 < 0 < x2 , then Conditions (2-2) and (2-3) are completely equivalent to the statement that the intervalbetween 0 and xu(n) is mapped homeomorphically by the iterate f ◦i for 0 ≤i ≤u(n −1) , but is notmapped homeomorphically by f ◦u(n−1)+1 .

The condition that some large iterate of f restricted to aninterval [a, b] is a homeomorphism is an invariant way of specifying that a is very close to b . Thus Lemma2.1 can be thought of as giving an invariant description of just how close xu(n) is to the critical point.Remark 3.

The Branner-Hubbard description of f would be rather different. Following Yoccoz, theycut the interval not at the critical point, but rather at the interior fixed point α < 0 .

In terms of theresulting partition of the interval, the appropriate description of the critical orbit is that the two images xiand xi+u(n) lie on the same side of α for i < u(n+1)−2 , but on opposite sides of α for i = u(n+1)−2 .Proof of 2.1. If (2-2) and (2-3) are satisfied, then according to Remark 2 above, we see that thesuccessive images xu(n) are closer and closer to zero.

Since x4 < 0 , it follows that f is a Fibonacci map.Conversely, the proof that every Fibonacci map satisfies (2-2) and (2-3) will be by induction on n , usingthe following induction hypothesis.Hypothesis Hn . For i in the range 0 < i < u(n) with i ̸= u(n −1) , the points xi have signas specified in Conditions (2 −2) and (2 −3) above, and furthermore ∥xi∥> ∥xu(n−1)∥.The following elementary observation will be used over and over.

For any unimodal map with minimumat x0 = 0 :if∥xp∥< ∥xq∥thenxp+1 < xq+1 .To start the induction, we must show that every Fibonacci map satisfies H4 . Since ∥x1∥> ∥x2∥> ∥x3∥by definition, we need only show thatx1 , x4 < 0 < x2 ,and that∥x4∥> ∥x3∥.Note first that the ∥xi∥must all be distinct.

For otherwise the critical orbit would have only finitely manydistinct elements. We have assumed that x4 < 0 .

If 0 < x1 then we see inductively that 0 < x1 < x2 < · · · ,4

which contradicts our hypothesis. Similarly, if x2 < 0 hence x1 < x2 < 0 , then we see inductively thatx1 < x3 < x5 < · · · < x6 < x4 < x2 < 0 ,which contradicts our hypothesis.

Finally, suppose that ∥x4∥< ∥x3∥. Applying the map f , we see thatx5 < x4 < 0 , and applying f again we see that x5 < x6 .

Since ∥x5∥< ∥x3∥by hypothesis, hencex6 < x4 , we have x5 < x6 < x4 < 0 and a similar inductive argument shows that x5 < x7 < x9 < · · ·

Since 0 < ∥xu(n)∥< ∥xu(n−1)∥, we havex1 < x1+u(n) < x1+u(n−1) .Now xi and xi+u(n−1) have the same sign for 0 < i < u(n −2) by Hn . Hence it follows by inductionon i that xi+u(n) lies between them, and hence also has the same sign, for i in this range.

Since bothxi and xi+u(n−1) have absolute value greater than ∥xu(n−1)∥by Hn , it follows also that ∥xi+u(n)∥>∥xu(n−1)∥> ∥xu(n)∥, for i in this range. For i = u(n −2) , this argument proves that xu(n−2)+u(n) liesbetween xu(n−2) and xu(n) , but does not determine its sign.

However, it does follows that0 < ∥xu(n−2)+u(n)∥< ∥xu(n−2)∥,hencex1 < x1+u(n−2)+u(n) < x1+u(n−2) .Nowasimilarinductiveargumentshowsthatxi+u(n−2)+u(n)liesbetweenxiandxi+u(n−2) , and hence has the required sign, for 0 < i < u(n−3) . Furthermore, this shows that ∥xi+u(n−2)+u(n)∥>∥xu(n−1)∥>∥xu(n)∥foriinthisrange.Inthelimitingcasei=u(n −3) ,thisargumentprovesthatxi+u(n−2)+u(n)=xu(n+1)liesbetweenxu(n−3)andxi+u(n−2)=xu(n−1) ,butdoesnotdetermineitssign.However,since∥xu(n+1)∥< ∥xu(n−1)∥< ∥xu(n−3)∥, this proves that xu(n−3) and xu(n−1) have opposite sign, so thatxu(n−1) also has the required sign.

Thus, we have almost proved Hn+1 . The only missing pieces of infor-mation are the sign and magnitude of xi for i = u(n −2) + u(n) .We must prove that ∥xu(n−2)+u(n)∥> ∥xu(n)∥.

But if ∥xu(n−2)+u(n)∥< ∥xu(n)∥thenx1 < x1+u(n−2)+u(n) < x1+u(n) .This is impossible. For a similar inductive argument would show that xi+u(n−2)+u(n) must be between xiand xi+u(n) for 0 < i ≤u(n −2) .

In particular, taking i = u(n −3) it would follow that xu(n+1) mustbe between xu(n−3) and xu(n−3)+u(n) . By the part of Hn+1 which has already been proved, these twohave the same sign, and it would follow that ∥xu(n+1)∥> ∥xu(n)∥, which contradicts our hypothesis.

Thus∥xu(n−2)+u(n)∥> ∥xu(n)∥.Now recall that xu(n−2)+u(n) is known to lie between xu(n−2) and xu(n) . Since ∥xu(n−2)+u(n)∥>∥xu(n)∥, it follows easily that xu(n−2)+u(n) has the same sign as xu(n−2) .

This completes the proof thatHn ⇒Hn+1 . ⊔⊓To show that this result is not vacuous, we must prove the following.Lemma 2.2.

Fibonacci maps exist.We will outline two different proofs. The proof below is an immediate application of the formal machineryof kneading theory, as developed in [MT].

An alternative proof, which is more direct and gives a more explicitdescription of the critical orbit, will be given in Lemma 3.1. Both proofs will make use of the following.Definition 2.3.

By a Fibonacci sum we will mean a finite or infinite formal sumµ = u(n1) + u(n2) + u(n3) + · · ·of non-consecutive Fibonacci numbers. That is, we always assume that ni+1 ≥ni + 2 , with n1 ≥1 .

Itis not difficult to check that every positive integer has a unique expression as a finite Fibonacci sum. As anexample, the difference u(n) −1 can be expressed asu(n) −1 =(u(1) + u(3) + u(5) + · · · + u(n −1)for n even ,u(2) + u(4) + u(6) + · · · + u(n −1)for n odd .

(2 −4)(For infinite Fibonacci sums, compare the proof of Lemma 3.2. )5

As in [MT], we describe the kneading invariant of a unimodal map f by a formal power series D(t) =1 + ǫ1t+ ǫ2t2 + · · · , where each coefficient ǫn is equal to +1 or −1 according as the function x 7→|f ◦n(x)|has a local minimum or local maximum at the origin. Since the xi are non-zero for i > 0 , we can checkinductively thatǫn = sgn(x1x2 · · · xn) .

(2 −5)Such a kneading invariant is admissible (ie., actually occurs) if and only if the inequality∞X0ǫi ti ≤∞X0(ǫmǫm+i) ti(2 −6)is satisfied for every m ≥1 . Here, by definition, an inequality P aiti < P biti between formal power seriesmeans that the first difference bi−ai which is non-zero is actually positive.

Thus, for each m we require thatthesmallestiforwhichǫm+i̸=ǫm ǫi(if any such exist) must satisfy ǫi = −1 .In the case of a Fibonacci map, it follows inductively from (2-2), (2-3) and (2-5) that we must haveǫu(n) = −1 for every Fibonacci number u(n) .In fact, according to (2-5), ǫu(n+1) is equal to ǫu(n)multiplied by the sign of the product xu(n)+1xu(n)+2 · · · xu(n+1) . This coincides with sgnx1x2 · · · xu(n−1)=ǫu(n−1) = −1 except that the very last factor xu(n−1) has the wrong sign.

Thus it follows inductively thatǫu(n) = ǫu(n+1) = −1 for all n . In other words, each map x 7→|f ◦u(n)(x)| must have a local maximum atx = 0 .

For a k -fold Fibonacci summ = u(n1) + · · · + u(nk) ,where alwaysn1 ≥1 and ni+1 ≥ni + 2 ,(2 −7)Equations (2-2) and (2-5) imply that ǫm is equal to the product ǫu(n1)+···+u(nk−1)ǫu(nk) . Hence it followsinductively that ǫm = (−1)k .

Thus, in order to prove 2.2 we need only show that the formal power seriesP ǫmtm , with ǫm defined by this equation, satisfies Condition (2-6). That is, for each fixed m the smallesti with ǫm+i ̸= ǫmǫi must satisfy ǫi = −1 .

However, if we express m as a Fibonacci sum as above, thenit is not hard to show that the smallest i with ǫm+i ̸= ǫmǫi is either i = u(n1 −1) or i = u(n1) or (inthe special case n1 = 1 ) i = 2 . Since ǫi = −1 in each of these cases, the required inequality (2-6) follows.This completes the proof of 2.2.

⊔⊓Proof of Theorem 1.1. Since any unimodal kneading invariant which is admissible can be realizedby a quadratic map, we can certainly find at least one quadratic map fc which realizes the given kneadinginvariant.

(See for example [MT].) But for any real quadratic map fc which is not infinitely renormalizableand has no attracting periodic orbit, Yoccoz has recently shown that the constant c is uniquely determinedby its kneading invariant.

(This is an immediate corollary of his much more general result about complexquadratic parameter space.) Since it is easy to check that a quadratic Fibonacci map is not renormalizableand has no attracting periodic orbit, this proves 1.1.

⊔⊓§3. The critical orbit.Out of the kneading data, it is not difficult to determine the precise ordering of the points xm in thecritical orbit.

We can describe the resulting ordering by a fairly concrete model as follows. The constructionwill provide an alternative proof of 2.2.Choose a parameter 0 < t < 1 −t2 , or in other words0 < t < (√5 −1)/2 = .61803 · · · ,for example t = 12 .

Now for each integer m ≥0 , expressed as a Fibonacci sum (2-7), define a real numberym by the formulaym = ±tn1 −tn2 + −· · · ± tnk,where the initial sign is to be −1 for n1 ≡0 , 1 (mod 4) , and +1 for n1 ≡2 , 3 (mod 4) , as in (2-3)above. Thus the initial term ±tn1 is the dominant one, and subsequent terms alternate in sign, decreasingby a factor of t2 or more at each step since ni+1 ≥ni + 2 .6

Remark 3.1. More precisely, this ordering can be described as follows.

For Fibonacci sums m withdifferent dominant terms, the order of the ym is determined by the rules:y1+··· < y5+··· < y8+··· < y34+··· < · · · < 0 < · · · < y21+··· < y13+··· < y3+··· < y2+··· .Here, in each case, the dots in the subscript stand for higher terms, which may be zero, for an arbitraryFibonacci sum. For two Fibonacci sums which have the same leading summands u(n1) + · · · + u(nk) butdiffer at the (k+1) -st summand, the relative order is determined as follows.

Setting s = u(n1)+· · ·+u(nk) ,we have|ys| > · · · > |ys+u(nk+5)+···| > |ys+u(nk+4)+···| > |ys+u(nk+3)+···| > |ys+u(nk+2)+···|if k is odd; and the same but with all inequalities reversed if k ≥2 is even. Here all of these points ys+···have the same sign, depending only on the leading summand n1 , as described above.We claim that the resulting ordering of the ym is precisely the required ordering of the points xm inthe critical orbit.

More precisely, we will prove the following.Lemma3.2.Thecorrespondenceym7→ym+1isunimodal,thatis,itismonotoneincreasingonthesetofymforwhichym≥0 ,butmonotonedecreasing for ym ≤0 . Furthermore, this correspondence is uniformly continuous.

Thus, if weextend linearly over each gap between the ym , then we obtain a continuous unimodal map F fromthe interval [y1 , y2] to itself, satisfying the Fibonacci condition thaty1 < y′2 < y′3 < y5 < y8 < y′13 < · · · < 0 ,where ym = F m(0) . ( Here, as in §2, we use the notation y 7→y′ for the orientation reversinginvolution of the subinterval [y′2 , y2] which satisfies the condition that F(y′) = F(y) .

)Proof. It is convenient to divide the various ym into intervals An , n ≥0 , which are ordered accordingto the following pattern:A2 < A6 < A10 < · · · < A8 < A4 < A0≤A1 < A5 < A9 < · · · < A11 < A7 < A3 .

(Here the two sequences {A2n} and {A2n+1} converge towards the two pre-images of zero. Compare 3.6.

)LetA0=[y5 , 0]betheclosedintervalcontainingallyu(n)+···withn ≡0 , 1 (mod 4) , n ≥4 , and also containing the limit point zero. Here, as above, the notation u(n) + · · ·stands for an arbitrary Fibonacci sum with leading term u(n) .

Similarly, let A1 = [0 , y3] be the interval con-taining all yu(n)+··· with n ≡2 , 3 (mod 4) , n ≥3 , together with the limit point zero. For n ≥2 even, letAnbethesmallestintervalcontaining all ym with m of the form u(1) + u(3) + · · · + u(n −1) + (higher terms) , where the higherterms if any must start with u(n + 2) or higher.

Using the identity (2-4), it follows easily that An is equalto the closed interval spanned by the two points yu(n)−1 and yu(n)+u(n+2)−1 . Here the relative order of thesetwo endpoints depends on whether n is congruent to 0 or 2 modulo 4.

Similarly, for n ≥3 odd, we defineAn to be the smallest interval containing all ym with m of the form u(2)+u(4)+· · ·+u(n−1)+(higher) ,where again the higher summands if any must start with u(n + 2) or higher. Again using the identity (2-4),we see that this interval An is again spanned by the points yu(n)−1 and yu(n)+u(n+2)−1 , where the relativeorder of the two end points depends on whether n is congruent to 1 or 3 modulo 4.It is not difficult to show that every ym with m > 0 belongs to exactly one of these intervals, and thatthese points are ordered according to the pattern described above.

For ym ∈An a brief computation showsthat the map ym 7→ym+1 is linear with slope (−1)n−1 . In particular, it is either order preserving or order re-versingaccordingasAn ⊂[0, y2] or An ⊂[y1 , 0] .

If we extend this map to be linear in the gap between An and An+4 ,then computation shows that the slope in this gap takes the value∆F(x)∆x= (−1)n−1 tn −tn+2 −tn+4tn+1 −tn+2 −tn+3for n > 0 . This is independent of n except for sign.

For n = 0 it takes a different value, but still withthe appropriate (negative) sign. As an example, for t = 12 this gap slope is equal to ± 112for n > 0 , and7

is −65for n = 0 . In this way, we obtain the required explicit unimodal map F which realizes the givenkneading data.

This 3.2, and completes the alternate proof of 2.2. ⊔⊓Lemma 3.2.

If the Fibonacci map f has no “homtervals” within the interval [x1 , x2] , that is,if the pre-critical points are everywhere dense, then f restricted to this interval is topologicallyconjugate to this model map F .The proof is straightforward. ⊔⊓Remark.

By definition, a homterval is a subinterval of I which is mapped homeomorphically by alliterates of f . A wandering interval is a homterval which is not contained in the basin of attraction forany periodic orbit.

According to Guckenheimer [G1], a unimodal map has no wandering intervals within[x1 , x2] provided that it has negative Schwarzian, with non-flat critical point. According to de Melo andvan Strien [MS], it has no wandering intervals provided that it is sufficiently smooth, with non-flat criticalpoint.

(See also Blokh and Lyubich [L], [BL1]. )Lemma 3.4.

More generally, if a Fibonacci map has no wandering intervals, then its critical orbitclosure O is a Cantor set, homeomorphic to the corresponding critical orbit closure for the modelmap F . In particular, this Cantor set is canonically homeomorphic to the set of all finite or infiniteFibonacci sums, suitably topologized.Proofof3.4.TheappropriatetopologyforthesetofallfiniteorinfiniteFibonacci sums can be described as follows.

Let Σ be the “Fibonacci shift”, consisting of all sequences(a1 , a2 , . .

.) of zeros and ones with no two consecutive ones.

In other words, Σ is a one-sided subshiftof finite type corresponding to the matrix T =1110. (The name is suggested since the number ofcylinders in Σ of length n is equal to u(n + 1) .) This set Σ is topologized as a subset of the infiniteCartesian product {0, 1} × {0, 1} × · · · .

Each sequence {an} ∈Σ determines an associated Fibonacci sumµ = P anu(n) , and we give the set consisting of all Fibonacci sums the corresponding compact topology.It is easy to check that the correspondence m 7→xm , where m ranges over positive integers expressed asfinite Fibonacci sums, extends uniquely to a homeomorphism µ 7→xµ ∈O , where now µ ranges over finiteor infinite Fibonacci sums. Further details of the proof are straightforward.

⊔⊓Remark 3.5. It is sometimes convenient to partially order the Cantor set Σ using lexicographical orderfrom the right.

Thus two sequences of zeros and ones, with no two consecutive ones, are comparable wheneverthey are eventually equal, or in other words have the same tail. In terms of this ordering, the map from Σto itself which corresponds to the map f|O can be described as the immediate successor function, whichcarries each such sequence to the next largest sequence with the same tail (such a transformation is calledan adic shift, compare [V]).

However, there are two exceptional sequences which are maximal, and hencehave no successor, immediate or otherwise, namely the two sequences (1, 0, 1, 0, . .

.) and (0, 1, 0, 1, .

. .

)corresponding to the Fibonacci sums1 + 3 + 8 + · · ·and2 + 5 + 13 + · · ·respectively. These both mapto the zero sequence.

(Compare (2-4) in §2. )Corollary 3.6.

The mapping f from the Cantor set O onto itself is one-to-one except that thepoint zero has two different pre-images, corresponding to the infinite Fibonacci sums u(1) + u(3) +u(5) + · · · and u(2) + u(4) + u(6) + · · · .The proof is straightforward. ⊔⊓Here is a more explicit description of this Cantor set as a subset of the real line.

For each n ≥1 letIn ⊂R be the smallest closed interval containing all of the points xu(q) with q ≥n . Thus In is a closedneighborhood of the origin.

One end point of this interval is xu(n) and the other end point is either xu(n+1)or xu(n+2) according as n is odd or even. Note that the map f folds In over onto the closed interval[x1 , xu(n)+1] , which in turn maps onto the closed interval [xu(n)+2 , x2] provided that n ≥3 .

For eachk ≥0 , we will use the notation Ink for the image f k(In) . According to §2, this image Ink is disjoint fromthe origin for 1 ≤k < u(n −1) , but contains the origin for k = u(n −1) .

However, Inu(n−1) contains asmaller interval In+1u(n−1) which again is disjoint from the origin. It will be convenient to use the notationJn = In+1u(n−1) ,and more generallyJnk = f k(Jn) = In+1k+u(n−1) .8

Note in particular that Jnu(n−2) = In+1u(n) .Definition. Let M n be the u(n) -fold unionM n =[0≤k

The u(n) closed intervalsIn0 , In1 , . .

. , Inu(n−1)−1andJn0 , .

. .

, Jnu(n−2)−1are pairwise disjoint. Denoting their union by M n as above, the M n form a nested sequence ofclosed sets M 1 ⊃M 2 ⊃M 3 ⊃· · · with intersection equal to the Cantor set O .Proof.

We will show by induction on n that the u(n) subintervals of M n are pairwise disjoint, thatthe M n are nested, and that each M n contains the critical orbit closure. The idea of the proof is toshow that, as we pass from M n to M n+1 , each of the u(n −1) intervals Ink ⊂M n will be replaced bytwo subintervals In+1kand Jn+1kin M n+1 , while each of the u(n −2) intervals Jnk = In+1k+u(n−1) remainsunchanged.To start the induction, it is trivially true that M 1 = [x1 , x2] contains the critical orbit closure.

Thefirst step in the induction is to note that each In contains In+1 and Jn+1 as disjoint subsets. For exampleif n ≡3 (mod 4) then these two subinterval ofIn = [xu(n+1) , xu(n)]are situated as follows:xu(n+1) xu(n+2) 0xu(n+3)xu(n)+u(n+2)xu(n)In+1Jn+1Figure 1.

The interval In in the case n ≡3 (mod 4) .The picture for n ≡1 (mod 4) is a mirror image, and the pictures for n ≡0 , 2 (mod 4) are quite similar.Note that the map f u(n) folds the subinterval In+2 ⊂In+1 over onto Jn+1 , while the map f u(n−1) carriesJn+1backontoaneighborhoodoftheorigin,spannedbythetwopointsxu(n+1)andxu(n+3) .Inthecasen≡3(mod 4)asillustrated,In+2istheinterval[xu(n+2) , xu(n+3)] ,whiletheimagef ◦u(n−1)(Jn+1)= [xu(n+1) , xu(n+3)]coincides with the interval In+1 .It follows easily from 3.1 and 3.4 that the two subintervalsIn+1 , Jn+1 ⊂Inare indeed disjoint, and together contain all of the points of O ∩In . For 1 ≤k < u(n −1) , a similar9

argument shows that the two subintervalsIn+1k, Jn+1k⊂Inkare disjoint, and together contain all of the points of O ∩Ink . This completes the induction, and shows thatM 1 ⊃M 2 ⊃M 3 ⊃· · · ⊃O .Since each endpoint of each subinterval of M n belongs to the orbit O , using the hypothesis that there areno wandering intervals we see easily thatT M nis equal toO .

⊔⊓Using the sets M n one can give another description of the above correspondence between O and Σ(see 3.4). Given x ∈O , let M n(x) be an interval of the set M n containing x .

Then set an = 0 ifM n(x) = Ink , and an = 1 if M n(x) = In+1k(for appropriate k ’s). One can check that {an} ∈Σ is thesequence corresponding to x ∈O .In what follows we will use the notation M na1...an for the interval of M n corresponding to the cylinder[a1...an] ⊂Σ .Lemma 3.8.

Still assuming that there are no wandering intervals, the points xi ,i ≥1 are theendpointsofthecomplementaryintervalsforthecriticalorbit closure O ⊂R .More explicitly, the Cantor set O can be obtained from the closed in-terval [x1 , x2] by removing a dense collection of disjoint open subintervals (xp , xq) as follows. Ifone of p , q is a Fibonacci sum of the formu(n1) + · · · + u(nk−1) + u(nk) + u(nk + 2)with k ≥2 , then the other is equal tou(n1) + · · · + u(nk−1) + u(nk + 1) .Ontheotherhand,ifoneisu(n) + u(n + 2),thentheotheriseitheru(n + 1)oru(n + 3)according as n is even or odd.As an example, the first seven open subintervals to be removed are as follows, in their natural order:(x9 , x19) ∪(x6 , x12) ∪(x4 , x5) ∪(x18 , x8) ∪(x13 , x11) ∪(x3 , x7) ∪(x20 , x10) .In other words, the Cantor set O is contained in the disjoint union[x1 , x9] ∪[x19 , x6] ∪[x12 , x4] ∪[x5 , x18] ∪[x8 , x13] ∪[x11 , x3] ∪[x7 , x20] ∪[x10 , x2](which coincides with the closed set M 5 ).

The proof of this statement is a straightforward consequence ofthe ordering of the points in the critical orbit, as described above. ⊔⊓We can obtain a different model for this critical orbit closure as follows.

Letγ = (1 −√5)/2 = −.61803 · · · ,so thatγ=γ2 −1 .To each finite or infinite Fibonacci sumµ=u(n1) + u(n2) + · · · ,let us assign the real numberφ(xµ) = γ (γn1 + γn2 + · · · )modulo one.Lemma 3.9. The resulting map φ from the critical orbit closure O onto the circle R/Z is one-to-one except at the countably many iterated pre-images of zero.

It semi-conjugates the map f|Oonto the golden rotationt 7→t + γ (mod 1) .Proof. It is easy to check that φ is well defined and continuous.

Note that the identity u(n−1)+u(n) =u(n + 1) corresponds to the identity γn−1 + γn = γn+1 . Using this fact, it is not difficult to check therequired identityφ(f(xµ)) = φ(xµ+1) ≡φ(xµ) + γ(mod 1) .Thus the image is a compact subset of the circle, invariant under the golden rotation, and hence is equal tothe entire circle.

Now consider any Fibonacci sum with leading term u(n) . A brief computation shows thatthe corresponding imageφ(xu(n)+···) = γn+1 + · · ·10

lies somewhere betweenγn+1 + γn+3 + γn+5 + · · · = γn+1/(1 −γ2) = −γnandγn+1 + γn+4 + γn+6 + · · · = γn+1 −γn+3 = −γn+2 .Thus, depending on the leading summand, the image φ(xµ) lies in one of the non-overlapping intervals[−γ2, −γ4] ∪[−γ4, −γ6] ∪[−γ6, −γ8] ∪· · · ∪{0} ∪· · · ∪[−γ7, −γ5] ∪[−γ5, −γ3] ∪[−γ3, −γ] ,having total length −γ −(−γ2) = 1 . Hence the value φ(xµ) ∈R/Z determines the leading summand u(n)uniquely, except in countably many cases which can be explicitly described.

For two Fibonacci sums withthe same leading term, a similar argument shows that the value φ(xµ) determines the second term uniquely,again with the exception of countably many cases which can be explicitly described; and a similar argumentapplies to higher terms. ⊔⊓Corollary 3.10.

With hypotheses as above, the map f|O is minimal, that is every orbit is dense,and has topological entropy zero. Furthermore this map is uniquely ergodic, that is it has one andonly one invariant probabiltity measure.Proof.

This follows easily from the corresponding assertion for an irrational rotation of the circle. ⊔⊓Combining 3.4–3.10, this evidently completes the proof of Theorem 1.2.

⊔⊓§4. A priori bounds.In the following two sections we assume that f : [−1, 1] →[−1, 1] is a C2 -smooth unimodal map withnon-degenerate minimum point 0, and normalized by the condition f(−1) = f(1) = 1 (which does notrestrict the generality).

Denote this class of maps by U , and let us discuss topology on this space.We will mainly be interested in the subspace U0 ⊂U consisting of those f for which f is an evenfunction, f(−x) = f(x) . We will first discuss the differentiability conditions and topology on this subspace,and then generalize to the full space U .

If f is even, then we can write it uniquely asf(x) = Ax1 ◦g ◦Qwhere Q is the squaring map ξ 7→ξ2 , g is some orientation preserving diffeomorphism of [0, 1] , and Ax1is the orientation preserving affine map which carries [0, 1] onto [x1, 1] , where x1 = f(0) is the criticalvalue.Now the Ck -topology on U0 , k ≤2 , comes from the Ck -topology on the space of diffeomorphisms g, together with the line topology on the range of the parameter x1 . Let ∥f∥denote the maximum of theC2−norms for g, g−1 which is a continuous functional in C2 -topology on our space.To obtain a corresponding topology of the full space U we need one extra step.

Let x 7→x′ be theorientation reversing diffeomorphism of T which satisfies f(x) = f(x′) . This involution is certainly C2 -smooth.

Consider a map B : x 7→(x −x′)/2 . Evidently f can be expressed as a function of (x −x′)2/4 ,so that we have a presentation f(x) = Ax1 ◦g ◦Q ◦B instead of the above one.

Now we must incorporatethe Ck topology on the involution as part of our topology. In practice, it is easiest simply to carry out thissymmetrizing change of coordinate x 7→(x −x′)/2 in the beginning, and thereafter to deal only with evenmaps f .

Moreover, we can also assume without loss of generality that f is purely quadratic x 7→x2 −cnear 0 (since any f ∈U is C2 -conjugate to such one).Denote by F the subspace of Fibonacci maps f ∈U .The following notations will be kept throughout the paper:dn = |xu(n)| ,λn = dn/dn−1 .11

The goal of this section is to obtain some a priori estimates for the λn (compare [G2], [L], [MMSS], [BL3],[M], [S],...). The proofs are based upon the Schwarz lemma and the Koebe Principle stated in the Appendix.First let us introduce a convenient terminology and notations.

A family of intervals G = {Gi}ni=0 iscalled a chain of intervals if Gi is a component of f −1Gi+1 for i = 0, 1, ..., n −1 . The chain is calledmonotone if all maps f : Gi →Gi+1 are homeomorphisms.For a given interval G and a point x such that f nx ∈G one can construct a chain G0, G1, ..., Gn ≡Gpulling G back along the n -orbit of x .

This construction is an efficient tool in one dimensional dynamicsbecause it is often possible to estimate the distortion of f n along chains of intervals (see [L] , [S] ).For a family of intervals G = {Gi} denote by |G| = P |Gi| the total length of intervals Gi and bymult G the maximal intersection multiplicity of intervals Gi , that is the maximum number of Gi havingnon-vacuous intersection.Let us consider now the pull-backHn+1 = {Hn+1m}u(n)−1m=0,Hn+10≡Hn+1 ⊃In+11(4 −1)of the interval T n−2 along the orbit {f mIn+11}u(n)−1m=0. The following two topological lemmas easily followfrom the above combinatorics.Lemma 4.1.

The chain Hn+1 is monotone (so that f n monotonously maps Hn+1 onto T n−2 ).Let us consider any interval I = Ilk , l ∈{n, n + 1} , of the family M n different from In0 , In1 , In2 . Definean interval F ≡Fn(I) ⊃I as follows(i) If I ̸= Jn then F is the convex hull of two neighbors of I in the family M n ;(ii) If I = Jn then F is the half of the interval T n−2 containing I .Now consider the pull-back G = {Gi}ki=0 of F ≡Gk along the k -orbit of Il0 .Lemma 4.2 Under the above circumstances1.

{Gi}ki=1 is a monotone chain of intervals;2. G0 ⊂T l−1 .Lemma 4.3.The intersection multiplicities of the above chains G and Hn+1 are uniformlybounded:mult G ≤8andmult Hn+1 ≤8.Proof.

If t intervals of the chain {Gi}ki=1 have a common point, then there is an interval Gi amongthem containing at least (t −1)/2 intervals Ns of the (k −1) -orbit of Il1. Since f k−i|Gi is monotone,f k−iNs belongs to the (u(l −1) −1) -orbit of Il1 .

But Gk contains at most three intervals of this orbit.Hence t ≤7 .The argument for H is similar, and we omit it. ⊔⊓Now we have enough topological information for getting a priori bounds.Lemma 4.4.supn λnλn+1 < 1.Proof.

Choose the smallest interval I among [0, xu(n)] and Ilk ∈M n with k > 0 . It is easy toanalyse the cases I = [0, xu(n)] or I = Ink for k = 1, 2.

So, we restrict ourselves to other cases, and thenthe interval F is well-defined. Moreover, the Poincar´e length [I : F] does not exceed log 4 .It follows from Lemmas 4.2.1, 4.3 and the Schwarz lemma that the Poincar´e length [Il1 : G1] is uniformlybounded (by a constant depending on ∥f ∥).

Since f is quadratic (and hence quasi-symmetric) near thecritical point, the ratio|G0||G0∖T l|can be estimated through [Il1 : G1] , and hence the ratio |T l|/|G0| is bounded away from 1.By Lemma 4.2.2, G0 ⊂T l−1 , so λl ≤|T l|/|G0| . It remains to mention that λl is equal to either λnor λn+1 .

⊔⊓Lemma 4.5.11−λ2n+1 ≤1+λnλn−11−λnλn−12(1 + O(|Hn+1|)).12

Proof. Applying the Schwarz lemma to the monotone mapf u(n)−1 : (Hn+1, In+11) →(T n−2, In0 )we get[In+11: Hn+1] ≤[T n : T n−2] + O(|Hn+1|) = 2 log 1 + λnλn−11 −λnλn−1+ O(|Hn+1|).

(4 −2)Let G be the component of f −1Hn+1 containing 0, µ = |T n+1|/|G|. The calculation for the quadraticmap shows thatlog11 −µ2 ≤[In+11: Hn+1].

(4 −3)Furthermore, since f u(n) is not unimodal on T n , G ⊂T n. Hence λn+1 ≤µ. The last estimate togetherwith (4-2) and (4-3) yield the required.

⊔⊓From Lemmas 4.4 and 4.5 we get immediately an a priori bound of λn :Lemma 4.6.supn λn < 1.Lemma 4.7. Let Ln be the gap between T n and Jn .

Thensupn|Ln||xu(n)| < 1.Proof. Because of Lemma 4.6, it is enough to show that the gap L is not too small as compared withJn .

Let N be a monotonicity interval of f u(n−2) adjacent to Jn on its outer side. Consider the mapf u(n−2)|L ∪Jn ∪N and apply to it the Schwarz lemma taking into account Lemmas 4.1 and 4.6.

⊔⊓Now we can prove that the Lebesgue measure of M n and Hn go down exponentially fast (compare[G2], [BL3], [MMSS]). Let [[α , β]] denote the smallest closed interval containing both α and β (similarly,((α , β)) will denote the smallest open interval containing α and β ).Lemma 4.8.

There exist constants C > 0 and q < 1 such that|Hn| ≤Cqnand|M n| ≤Cqn.Hence, the Lebesgue measure of ω(0) is equal to zero.Remark. The last statement is a corollary of more general results [BL2], [M].Proof.

By Lemma 4.7, density of M n+1 in In0is bounded away from 1. Consider now an intervalInl ∈M n, l > 0.

It follows from Lemmas 4.1 and 4.6 that the mapf u(n−1)−l : Inl →[[xu(n−1), xu(n+1)]]has bounded distortion. But this map carries M n+1 ∩Inlinto In+10∪Jn.

By Lemma 4.7, density of thelatter set in [[xu(n−1), xu(n+1)]] is bounded away from 1. Hence density of M n+1 in Inlis bounded awayfrom 1 as well.

So, there is a q < 1 such thatλ(M n+1) ≤qλ(∪u(n−1)−1l=0Inl ) + λ(∪u(n)−1l=u(n−1)In+1l).Applying this estimate twice we getλ(M n+2) ≤qλ(M n),and we are done with M n .Now consider a pair Hn+1 ⊃Hn+2 and apply f u(n)−1. Then Hn+1 is mapped onto T n−2, whileHn+2 is mapped into T n−1 (since f u(n−1) is monotone on its image).

By Lemma 4.6 and the Schwarzlemma, the density of f mHn+2 in f mHn+1 is bouded away from 1 for m = 0, ..., u(n) −1. Furthermore,f u(n)+mHn+2 ⊂In−1m,m = 1, ..., u(n −1).13

Cosequently, for some q1 < 1 we have|Hn+2| ≤q1|Hn+1| + |M n−1| + |M n−2|,and the required follows. ⊔⊓Lemma 4.9.

(i). There is a q < 1 such that λ2n+1 = O(λnλn−1 + qn).

( ii). λ2n+1 = O|Jn||T n−1|.Proof.

The point (i) follows from Lemmas 4.5 and 4.8. To prove (ii), consider f u(n−1) : In+1 →Jnand apply the Schwarz lemma.

⊔⊓Remark 4.10. All constants in the above estimates depend only on ∥f ∥.

Moreover, they are uniformover the maps with negative Schwarzian derivative (since the Schwarz lemma and the Koebe Principle areuniform over this class). Finally, all estimates are asymptotically uniform over the whole class U (”beauestimates”, see Sullivan [S]).

For example, Lemma 4.6 can be improved in such a way:lim supn→∞λn ≤C < 1for an absolute constant C .§5 Scaling, characteristic exponent and Hausdorffdimension.In this section we will prove Theorem 1.3 assuming that there is a good enough a priori bound of λn .It follows that the Theorem holds for an open set of Fibonacci maps invariant under quasi-symmetricalconjugacy.Let q < 1 be the constant from Lemma 4.8 , σn = maxn−1≤i≤n+1(λi, λi+1) .Lemma 5.1. For any x ∈In+11dnd2n+1(1 + O(σn + qn))−1 ≤|(f u(n)−1)′(x)| ≤dnd2n+1(1 + O(σn + qn)).Proof.

Let us apply the Koebe Principle to the mapf u(n)−1 : (Hn+1, In+1) →(T n−2, T n)taking into account Lemma 4.8:|(f u(n)−1)′(x)||(f u(n)−1)′(y)| = 1 + O(λnλn−1 + qn),x, y ∈In+11.Besides,dnd2n+1≤|In||In+11| ≤(1 + λn+1λn+2) dnd2n+1,and the Lemma follows. ⊔⊓Lemma 5.2.

There is a ρ = ρ(∥f ∥) and L = L(∥f ∥) ∈N such that if λl < ρ for some l ≥Lthen λn exponentially decrease. For maps with non-positive Schwarzian derivative one can chooseL = 1 and uniform ρ .Proof.

Let n be so large that f(x) is a quadratic map in the neighborhood Tn−1 . Then by the chainrule,|(f u(n)−1)′(x1)| = |(f u(n−1)−1)′(x1)| · 2dn−1|(f u(n−2)−1)′(xu(n−1)+1)|.

(5 −1)By Lemma 5.1,dn−1d2n· 2dn−1dn−2d2n−1≤dnd2n+1(1 + O(σn−2 + σn−1 + σn + qn)). (5 −2)14

It follows from Lemma 4.5 that λk keep to be small for k = n −2, ..., n + 1 , once λn−3 becomes small forbig enough n . Hence, by (5-2)λ2n+1 ≤γλnλn−1(5 −3)for some γ < 1 .

Setting Λn = max(λn, λn−1), we get from (5-3) thatΛn+1 ≤√γΛn. (5 −4)So, once λn become small, they start exponentially decrease.

It follows that they exponentially decreaseforever.The final remark: since the constants in the Schwarz Lemma and the Koebe Principle depend only on∥f ∥, the constants ρ and L depend only on this data as well . Moreover, all estimates are uniform inthe case of negative Schwarzian derivative.

⊔⊓Recall that a one dimensional homeomorphism h is called quasi-symmetric if any two adjacent com-mensurable intervals I and J are mapped into commensurable ones:|I||J| ≤K ⇒|fI||fJ| ≤γ(K).Denote by F0 the set of Fibonacci maps for which inf λn = 0 .Lemma 5.3.1. The set F0 is invariant under quasi-symmetrical conjugacy.2.

The set F0 is C0−open in the C2 -balls B(r) of the space F .Proof. The first point is clear from the definitions .

The second one follows from the fact that theconstants in the previous lemma are uniform over B(r) . ⊔⊓Let us write αn ∼βn if | log (αn/βn)| exponentially decrease, and αn ≍βn if it is bounded.The next lemma gives the asymptotical formula of Theorem 1.3.1 for the subclassF0 (compareTangerman and Veerman [TV]).Lemma 5.4.

For any f ∈F0 the following asymptotical formulas hold:1. λn+1 ∼λn/3√2.2. λn ∼a2−n/3.3.

dn ∼(1/2)n2/6+βn+γfor some constants a > 0, β and γ . Moreover| log(a/λ0)| ≤R(∥f ∥) ,and the constant R is uniform over maps with negative Schwarzian derivative.Proof.

Since λn exponentially decrease, Lemma 5.1 yields for x ∈In+11|(f u(n)−1)′(x)| ∼dn/d2n+1. (5 −5)Substituting this into the recurrent equation (5-1), we getλ2n+1 ∼12λnλn−1.

(5 −6)Setting sn = log(λn/λn−1) , we have from the last formulasn+1 = −12sn −12 log 2 + O(qn)with q < 1 . It yieldssn = −13 log 2 + O(ρn)(5 −7)with ρ = max(1/2, q) which proves the first point of the lemma.Setting now c = 13 log 2, νn = log λn + cn we get from (5-7)νn+1 = νn + O(ρn).15

So, there is a limitlim νn ≡log a = ν0 + O(1),with exponential convergence and the constant depending only on ∥f ∥and uniform over maps withnegative Schwarzian. Equivalentlya ≡lim λnenc ≍λ0.It proves the second point together with the last remark.

The reader can easily derive the third point fromthe second one. ⊔⊓Let us estimate now the ratio of any two intervals M ns1...sn ⊂M n−1s1...sn−1 .

The previous lemma gives theasymptotics for the ratio λn ≡|M n0...0|/|M n−10...0 | . Besides, M ns1...10 = M n−1s1...1 .

Other cases are covered bythe following lemma.Lemma 5.5. For f ∈F0 the following scaling laws hold:|M n0...01||M n−10...0 |≡|Jn||In−1| ∼a222(n+1)/3 .If [s1...sn−1] ̸= [0...0] then|M ns1...sn−11||M n−1s1...sn−1|∼a222(n−1)/3 .and|M ns1...sn−10||M n−1s1...sn−1|∼a222(n−2)/3where a is the constant from Lemma 5.4.

All asymptotics are uniformly exponential.Proof. Let us consider a chain of two maps(In−1, Jn)→(In−11, Jn1 )→(In−2, In).ff u(n−2)−1Note that by Lemma 5.4 |In| ∼|xu(n)| .

Setting rn = |Jn|/|In−1| we get|fJn||fIn−1| ∼1 −(1 −rn)2 ∼2rn.On the other hand, f u(n−2)−1 has an exponentially small distortion on In−11, and hence2rn ∼|In||In−2| ∼λnλn−1 ∼a22(2n−1)/3 ,and the first asymptotical formula is proved.In order to get the others, consider the mapf k : M n−1s1...sn−1 →In−2for an appropriate k . It carries M ns1...sn−10 into Jn−1 and M ns1...sn−11 into In with exponentially smalldistortion.

It yields the required. ⊔⊓Now we can prove the next piece of Theorem 1.3 for f ∈F0Lemma 5.6.

For f ∈F0 the critical orbit closure O has Hausdorffdimension 0.Proof. Let us consider covering of O by the intervals M ns1...sn .

By the above two lemmas, the lengths oftheseintervalsdecreaseuniformlysuperexponential( O(qn) for any q ∈(0, 1) ), while their number increases exponentially ( ≤2n ). Let γ = −log 2/ log q,lγbe the Hausdorffmeasure on O of exponent γ .

Thenlγ(O) ≤C2nqnγ ≤C.Hence, dim O ≤γ , and γ is arbitrary small positive number. ⊔⊓16

Now we are going to show that the geometry of the set O is completely determined by only oneparameter a from Lemma 5.4. Let f and g be two Fibonacci maps,φ : O(f) →O(g)be the natural topological conjugacy.

Let us say that φ is smooth if for any x ∈O there existlim |φ(x) −φ(y)||x −y|̸= 0as y →x along O(f) , and this limit depends continuously on x .Lemma 5.7.If two Fibonacci maps f and g in F0 have the same parameter a then theconjugacy φ is smooth on O(f) .Proof. Indeed, it follows from Lemmas 5.4 and 5.5 that for any Fibonacci sequence s = s0s1... thereis a limitlimn→∞|M ns1...sn(f)||M ns1...sn(g)|depending continuously on s .

⊔⊓Lemma 5.8. Let f ∈F0,n = [s1...sk] be the Fibonacci expantion of n .

Then|(f n)′(x1)| ∼223Pmsm+γ Psm+δfor some constants γ and δ .Proof. Let mi be the places where smi = 1 .

Decompose n -orbit of x1 into the parts of length u(mi) .By (5-5) it gives the factorization of the derivative into factors of order ∼2λ−2mi+1 . Now Lemma 5.4 impliesthe required asymptotics.

⊔⊓Clearly, it follows from the last lemma that the growth of the n -fold derivative at x1 is subexpo-nential.The maximal growth of order exp κ(log n)2 (which is faster than any power nγ ) is attainedat noments u(m) −1 .However, at the next moments n = u(m) the derivative drops to nγwithγ = 2 log 2/3 log(√5+12) < 1. These oscillations are balanced in a “convergent way” .Lemma 5.9.

The series∞Xn=11|(f n)′(x1)|αis convergent for any α > 0 .Proof. By the last lemma , this series has a majorant of the following form:Xsm∈{0,1}2−Pkm=1(am+b)sm =∞Ym=1(1 +12am+b ) < ∞.⊔⊓This Lemma and the Nowicki-van Strien Theorem [NvS] imply the existence of an absolutely continuousinvariant measure for f ∈F0 .

So, Theorem 1.3 is proved for the subclass F0 .17

§6. Real renormalizations.Now we need another class of maps on which we can define a renormalization in such a way that theFibonacci maps can be exactly characterized as infinitely renormalizable.

LetJ = [a, b] ,T = [α, β] , where −1 < a < b < α < β < 1 ,Dom(f) = J ∪T ,and letf : Dom(f) →[−1, 1] be a C2 -smooth map such that (see Figure 2) :(i) f|J is a diffeomorphism from J onto [−1, 1] , which may be either orientation preserving or orientationreversing. (ii) f|T is a unimodal map from T into [−1, 1] with non-degenerate minimum point, and with f(∂T ) = 1 .Let us denote the space of all such maps by A .

Since we don’t specify whether f|J preserves or reversesorientation, A can be decomposed into the union of two connected components A+ and A−, where “+”corresponds to the case of orientation preserving f|J .TJFigure 2, Graph of a function in A+0 .Now suppose that some map f ∈A , with critical point x0 ∈T , satisfies the conditions that the criticalvalue x1 = f(x0) lies in J , and the its image x2 = f(x1) lies back in T . Then we will be interested intwo segments of the first return map from T to itself, as follows.

There is an interval T1 around the critcalpoint which is mapped unimodally by f 2 into T , with both endpoints of T1 mapping to one endpoint ofT . Further, there is a disjoint interval J1 ⊂T which maps diffeomorphically onto T under the map fitself.

Here we choose J1 to the left of T1 if f|J preserves orientation, or to the right of T1 if f|J reversesorientation (so that J1 lies on the same side of 0 as x2 ). The resulting map V f : J1 ∪T1 →T , affinelyconjugated (rescaled) so that T is replaced by the original interval [−1, 1] , is the required renormalizationRf (there is choice of two rescalings ; select that one which makes the critical point to be minimum point ).This renormalization interchanges the two spaces A+ and A−.

If f is n -fold renormalizable then Rnfcomes as rescaling of a map V nf ≡fn , the restriction of appropriate iterates of f to the union of twoappropriate intervals, Tn and Jn .Let T+ and T−be the semi-intervals on which 0 divides T . The kneading sequence of f ∈A is thesequence of symbols Un ∈{T+, T−, J} such that xn ≡f n0 ∈Un .

Two maps f ∈A+ (or A−) withoutlimit cycles are topologically conjugate if and only if they have the same kneading sequence (compare [MT]).In terms of kneading sequences the above renormalization can be described in the following way. Therenormalizable kneading sequences start with JTs, s ∈{+, −} .

To write its renormalization do the followingoperations moving along the sequence:( i). When you see J , cross it;( ii).

When you see TsJ, s ∈{+, −} , change Ts for Tks provided f ∈Ak, k ∈{+, −} . ( iii).

When you see TsTr , change the first Ts for J .Let us say that a map f ∈A+ is a Fibonacci map if it has the following kneading sequence:fib+ = J|T−|T+|JT+|JT−T−|JT−T+JT−|...(In order to write the block from u(n)+ 1 to u(n+ 1) repeat the beginning of the sequence till the momentu(n −1) , and then change the last symbol Ts for the “opposite” one, T−s ). Denote this class of maps byF+ .

Similarly, the kneading sequence of a map f ∈F−is produced by the same rule but with different18

initial:fib−= J|T+|T+|JT−|JT+T−|JT+T+JT+|...A class F of Fibonacci maps is defined as F+ ∪F−. One can also describe this class by the followingproperties: x1 ∈J, and f u(n−1) is well-defined and monotonous on the interval [[0, xu(n)]] , andf u(n−1)((0, xu(n))) ≡((xu(n−1), xu(n+1))) ∋0.

(6 −0)If we want to emphasize that f ∈A then we say that f has type (2,1). In the unimodal case we saythat f is of type (2) (see the next section for more general discussion).

As in the unimodal case, we will usethe notations T n and Jn for the intervals [[xu(n), x′u(n)]] and [[xu(n−1), xu(n−1)+u(n+1)]] correspondingly(don’t confuse with Tn and Jn introduced above).Lemma 6.1. A map f ∈A is infinitely renormalizable if and only if it is a Fibonacci map :f ∈F .

In this case the following inclusions hold:T n+2 ⊂Tn ⊂T n+1(6 −1)Jn+2 ⊂Jn. (6 −2)Proof.

Let f ∈A be infinitely renormalizable. Then one can check by induction thatfn|Tn = f u(n+1)andfn|Jn = f u(n).

(6 −3)Since fn−1 is renormalizable,xu(n) = fn−1(0) ∈Jn−1andxu(n+1) = fn(0) ∈Tn−1.Hence, xu(n+1) lies closer to 0 than xu(n), n = 1, 2, ...Let us study now the combinatorics of several first iterates of 0. Since f is renormalizable,T 2 ≡[[x2, x′2]] ⊂T ⊂[x1, x′1] ≡T 1.

(6 −4)Furthemore, x3 = f2(0) ∈T1 ; hence x4 = fx3 ∈J. So,J2 ≡[x1, x4] ⊂J.

(6 −5)Consider now the following map σ : N →N of the set of natural numbers: if m = P u(li) is theFibonacci expantion of m then σ(m) = P u(li + 1) ( σ is induced by the shift on the space of Fibonacciexpantions). Then we have the following rule:(fn)m(0) = xσn(m).

(6 −6)So, if we have a combinatorial property of several points xm then repalcing f by fn we immediately get thesame property of points xσnm (provided f is infinitely renormalizable). In particular we can replace pontsx1, x2, x4in(6-4)and(6-5)byxu(n+1), xu(n+2), xu(n+1)+u(n+3) .

Then we obtain the required properties (6-1) and (6-2).Let us show now that x1 and x2 lie on the same side of 0 for f ∈A+ , and they lie on the oppositesides of 0 for f ∈A−. Indeed, otherwise consider f|[x1, x4] and conclude that x5 lies farther from 0 thanx2 .Changing f for f1 we get the same statement for the points x2 and x3 .

Since the renormalizationinterchanges A+ and A−, we conclude that ((x1, x3)) ∋0 . Replacing f by fn−2 we get (6-0).Finally, since x2 ∈T, f|[0, x2] is well-defined and monotone.

Replacing it again by fn−2 we concludethat f u(n−1)|[0, xu(n)] is well-defined and monotone. So, f is a Fibonacci map.Vice versa, let fibsn, s ∈{+, −} be the initial parts of length u(n) of the kneading sequences fibs .Then one can easily check by induction that the renormalization turns fibsn into fib−sn−1 .

So, it interchangesfibs and fib−s which certainly implies that both sequences are infinitely renormalizable. ⊔⊓Now let us briefly discuss topology on the space A (compare §4).

We can restrict ouselves to thesubspace A0 ⊂A consisting of those f for which f|T is an even function, f(−x) = f(x) . Then we canwrite f|T uniquely asf(x) = Ax1 ◦fT ◦Q ◦AT19

where AT is the orientation preserving linear map which carries T onto [−1, 1] , Q is the squaring mapξ 7→ξ2 , fT is some orientation preserving diffeomorphism of [0, 1] , and Ax1 is the orientation preservingaffine map which carries [0, 1] onto [x1, 1] , where x1 = f(0) is the critical value. Similarly, we can writef|J as fJ ◦AJ where AJ is the orientation preserving affine map from J onto [−1, 1] , and where fJ isa diffeomorphism of [−1, 1] .Now we suppose that both fJ and fT are C2 -smooth.

The Ck -topology on A0, k ≤2, comes fromthe Ck -topology on the space of diffeomorphisms fT and fJ , together with the Euclidian topology on thefinite dimensional space of parameters a, b, α, β, x1 . Let ∥f∥denote the maximum of the C2−norms forfJ, f −1Jand fT , f −1Twhich is a continuous functional on our space.We can assume without loss of generality that the original map f is quadratic near 0 (though thisproperty is not preserved under renormalization).

Let us remark also that clearly all estimates of §§4,5 holdnot only for unimodal maps but in the class A as well.Lemma 6.2. The norms ∥Rnf∥are uniformly bounded.Proof.

By (6-3), fn|Tn = f u(n+1) which can be decomposed as a quadratic map and the diffeomorphismf u(n+1)−1 : Hn+2 →T n−1(6 −6)(see Lemma 4.1). On the other hand,f u(n+1)−1(fTn) = fnTn ⊂Tn−1 ⊂T n(6 −7)(the last inclusion is by (6-1)).

It follows from (6-6), (6-7) and a priori bounds proven in §4 that f u(n+1)−1|fTnhas bounded distortion. By rescaling we getlog(Rnf)′T (x)(Rnf)′T (y) = O(|x −y|)for any x, y ∈[0, 1].

This implies(Rnf)′′T(Rnf)′T = O(1).Because of bounded distortion, the derivative (Rnf)′T is uniformly bounded from below and above, andthe boundedness property for the second derivaty (Rnf)′′T follows. The same argument applies to (Rnf)Jand to the inverse maps.

⊔⊓Corollary. If inf λn > 0 then there is a C1 -convergent sequence of renormalizations Rni →g ∈A .Proof.It follows from the assumtion and inclusions (6-1) that the ratio |Tn| : |Tn−1| is boundedaway from 0.

Moreover, Lemma 4.9 and (6-2) imply the same for the ratio |Jn| : |Tn−1| . Now one canplay the “distortion game” in manner of §4 to check that three complementary gaps (that is, componentsof Tn−1∖(Tn ∪Jn) ) are also commensurable with Tn−1 .

After rescaling we conclude that the domainsDom(Rnf) don’t degenerate, so we can select a convergent sequence Dom(Rnif) . Then by the last lemma,families of diffeomorpfisms {(Rnif)T } and {(Rnif)J} are C1 -precompact, and we can extract from themconvergent subsequences as well.

⊔⊓For an interval I ⊂R denote by P(I) the plane slitted along two rays:P(I) = C∖(R∖I).Let us introduce now a subspace E ⊂A consisting of maps f : T ∪J →[−1, 1] with the followingproperty: The map f −1T: [0, 1] →[0, 1] can be analytically continued to a map P[0, 1] →P[0, 1] , andf −1 : [−1, 1] →J can be analytically continued to a map P[−1, 1] →P(J) .Lemma 6.3. Let Rnif →g in C1 -topology.

Then the limiting function g belongs to the classE .Proof. The map (Rnf)−1Tcan be written as long compositions of type h1 ◦q1 ◦... ◦hk ◦qk where hiare diffeomorphisms between apropriate intervals with a small total distortion while qi are square root maps(we reserve this term for affine conjugates to the standard square root).

Such a map can be rewritten asHn ◦Qn where the distortion of Hn does not exceed the total distortion of hi, i = 1, ..., n , and Qn is thecomposition of Qi renormalized by appropriate M¨obius maps (see [S] , [Sw2]). The maps Qn analytically20

map P[−1, 1] into itself, and hence form a normal family. So, we can select a convergent sequence Qn →Qwith Q to be a self-map of P[−1, 1] .

On the other hand, Hn →id in C1 -topology. So, (gT )−1 = Q .

Inthe same way we can treat gJ . ⊔⊓Correspondence between Fibonacci maps of classes U and A−.

We are going to describe aneasy surgery interchanging these classes without touching the critical orbit. It will follow that any resultabout the critical orbit established in one of the classes immediately yields the same statement in the otherclass.Let f ∈U be a unimodal Fibonacci map.

Let us restrict it onto the union of two disjoint intervalsI2 ∪J2 ≡[x5, x2] ∪[x1, x4]. (6 −8)Then let us embed these intervals into disjoint intervals T and J correspondingly, and continue f to amap of class A−defined on T ∪J .Vice versa, given a Fibonacci map g ∈A−, we can also restrict it onto the union (6-8), and thencontinue to a unimodal map of class U .

This is possible since g(x5) ≡x6 < x5 ≡g(x4) .Since orb (0) ⊂I2 ∪J2 , the above surgeries keep the critical orbit untouched.§7. Polynomial-like maps.Now we are going to show that all polynomial-like maps f ∈A−(or A+ ) are quasi-symmetricallyconjugate.

It is convenient to introduce more general terminology.Consider k + 1 topological disks Ui and V bounded by piecewise smooth curves, and such that clUiare disjoint and contained in V . Let us say thatf : ∪Ui →Vis a polynomial-like map of type (n1, ..., nk) if f|Ui is a branched covering of degree ni ; d = P ni is calledthe degree of f .

Note that polynomial-like maps of type (d) are exactly polynomial- like maps in the senseof Douady and Hubbard [DH].Lemma 7.1. Any polynomial-like map f : U1 ∪U2 →Vof type (2, 1) is quasi-conformallyconjugate to a cubic polynomial with at least one escaping critical point.Proof.

Consider an “eight-like” neighborhood N of U1 ∪U2 and smoothly continue f there so thatf becomes a double covering on the annulus around U1 and a diffeomorphism on the annulus around U2 ,and both annuli are mapped on the same annulus around V , see Figure 3.Then continue f to a slightly bigger domain so that it turns into a three sheeted smooth covering of atopological disk over a bigger disk. Now use the Douady-Hubbard surgery [DH] in order to quasi-conformallyconjugate this map to a cubic polynomial.

⊔⊓VU1U2Figure 3.Lemma 7.2. Any polynomial-like map f ∈A−is quasi-symmetrically conjugate to a real cubicpolynomial with one escaping critical point.Proof.

For f ∈A−one can carry out the above construction in an R -symmetrical way. ⊔⊓21

Lemma 7.3. All Fibonacci real cubic polynomials are quasi-symmetrically conjugate.Proof.

Consider a locus F+ of real cubic polynomials z 7→z3 −3a2z + b for which the critical point ais a preimage of the left fixed point (it is equivalent to b = 2a3 −2a ) and a < 1/3 . By Branner and Douady[BD] , there is a natural one-to-one correspondence between F+ and the 1/2-locus of quadratic polynomialsz 7→z2 −c with −2 ≤c < −3/4.

Hence, in F+ there is only one Fibonacci map (Theorem 1.1 ). On theother hand, conjugacy classes of cubic maps with escaping critical point a ( which means b < 2a3 −2a )are in one-to-one correspondence with F+ as well: go toward the curve b = 2a3 −2a along external rays(this argument is due to Douady).

⊔⊓From the last two lemmas we have immediateCorollary 1.All polynomial-like Fibonacci maps f ∈F−are quasi-symmetrically conjugate.Corollary 2. Either all polynomial-like Fibonacci maps belong to the set F0 or to its complement.Proof.

For maps f ∈F−it follows from the last Corollary and Lemma 5.3. For maps f ∈F+ justobserve that it belongs to F0 or its complement together with the renormalization.

⊔⊓Now we will give an example of a polynomial-like map belongning to F0 which will yield that allFibonacci polynomial-like maps belong to F0 .Example. Consider disjoint union of two intervals I = [−1, λ] and J = [−c, −c + qλ2] with positivec, q, λ , c is big, λ is small.

Let f|I be a quadratic map x 7→qx2 −c , while f|J be linear x 7→αx + b.Let us adjust parameters α, b, c, q, λ in such a way that0 7→−c →−1 7→λ 7→−c + qλ2 7→v ∈[0, λ].It yield the relationsq = c + λ ∼c, α =1 + v(c + λ)λ2 ∼1cλ2 , b = αc −1 ∼1λ2(7.1)It remains three free parameters c, λ and v . Let us show that for c2λ2 < 1 this map is cubic-like.

To thisend consider a disk D = {z : |z| < 2} . On its boundary ∂D our map acts asf(z) = c(z2 −1) + λz2 ∼c(z2 −1).Hence,3c < |f(z)| < 5cforz ∈∂D(7.2).Consider a disk V = {z : |z| < 2c} and its inverse image U1 (under the quadratic map.) By (7.2), U1 ⊂Dand f : U1 →V is a quadratic-like map.

Moreover, U1 ⊃[−1, 1] since f[−1, 1] = [−c, λ] ⊂V.Furthermore, consider the preimage U2 of V under the linear map z →αz + b . It is a disk containingJ of radius2c/α ∼2c2λ2 < 2(by (7.1)).

Hence, for big enough c the closure of this disk is contained in V and does not intersect cl U1 .So, f : U1 ∪U2 →V is a polynomial-like map.Now one can adjust v to get a Fibonacci map. Since f has non-positive Schwarzian derivative, itbelongs to F0 provided λ is sufficiently small (Lemma 5.2).

⊔⊓Renormalization of a quadratic-like Fibonacci map. This procedure associate to a quadratic-likeFibonacci map (of type (2)) a cubic-like Fibonacci map (of type (2,1)).

It will complete the proof of Theorem1.3 for quadratic-like Fibonacci maps (in particular, for the quadratic polynomial). We can restrict ourselvesto the case of the quadratic Fibonacci polynomial.

Now let us consider the beginning of the Yoccoz partitionconstruction (see [H] ). Draw a curve S consisting of two external rays through the fixed point α and anequipotential level γ .

We will obtain two pieces of level 0, W 0 (containing 0) and W 01 (containing x1 ).Define pieces of level n as n -fold preimages of the pieces of level 0. Denote by W n(x) the piece of level ncontaining x , set W n ≡W n(0) .

Let us consider the piece V ≡W 4 ⊃T 4 satisfying the property thatcl W 4 ⊂W 3. (7.3)22

Define a piece U1 ≡W 9 ⊃T 5 as the pull-back of Vof order 5, and U2 ⊃J5 as the pull-back of Voforder 3. One can check that cl U1 and cl U2 are pairwise disjoint and are contained in V(it is a formalcorollary from (7.3)).

So, the map g defined as f 5|U1 and f 3|U2 is polynomial-like of type (2,1). ⊔⊓Remark.

The above construction actually can be applied to any non-infinitely renormalizable ”persis-tently recurrent” quadratic polynomial (see [L2]).Geometry of ω(c) is not rigid. We would like to show that parameter a can really be changed inclass U , so the geometry of ω(c) is not rigid.

The above Example provides us with a Fibonacci map ofclass A with arbitrary small λ0 = 1/c . By Lemma 5.4, parameter a ≍λ0 is getting arbitrary small aswell.

Renormalizing f if necessary we obtain a Fibonacci map of class A−with arbitrary small a . Nowthe surgery of §6 turns this map into a unimodal Fibonacci map with the same parameter a .Remark.

Actually, in order to vary parameter a in class A it is enough to observe that the renor-malization turns a into a/3√2 .§8. Polynomial-like property of analytic Fibonacci maps.In this section we will prove that analytic Fibonacci maps f ∈E become polynomial-like after apropriaterenormalization.

Together with the results of the previous two sections it will complete the proof of Theorem1.3.For an interval I ⊂R denote by by D(I) the Eucledian disk based upon I as the diameter.Lemma 8.1 (see [S]). Let φ : P(I) →P(J) be an analytic map which maps I diffeomorphicallyonto J .

Then φD(I) ⊂D(J).Proof. The interval I is a Poincar´e geodesic in P(I) , and the disk D(I) is its Poincar´e neighborhood(of radius independent of I ).

Since φ contracts the Poincar´e metric, we have the required. ⊔⊓Lemma 8.2.

Let f ∈E be an analytic Fibonacci map. Given n , consider a disk V = D(Tn)and its pull-backs U1 ⊃Tn+1 and U2 ⊃Jn+1 of order u(n + 2) and u(n + 1) correspondingly.Then cl Ui are disjoint and are contained in V .Proof.

Let Tn = [[tn, t′n]] with tn being closer to xu(n+2) .The branch φ : V →U2 of f −u(n+1) satisfies the asumptions of Lemma 8.1, and hence U2 ⊂D(Jn+1) .By the same reason, fU1 ⊂D(Q) where Q ≡[b, a] ∋x1 is the monotone pull-back of Tn of orderu(n + 2) −1 ( b < x1 is the preimage of tn ).Now let Xn−1 be the component of Tn−1∖Tn adjacent to tn . Since P |Xn| < ∞, we can select suchan n that|Xn| < |Xn−1|(8 −1).By Lemma 4.1, the map f u(n+2)−1 has a monotone continuation beyond the point b to the interval Wwhich is mapped onto Xn−1 .

So, we have three interval mapf u(n+2)−1 : W ∪[b, x1] ∪[x1, a] →Xn−1 ∪[[tn, xu(n+2)]] ∪[[xu(n+2), t′n]]. (8 −2)Let q = |xu(n+2)| : |tn| .

Applying the Schwarz lemma to (8-2) taking into account (8-1) we getlog a −ba −x1≤log 2 + log21 + q ,so thatx1 −ba −x1≤3 −q1 + q . (8 −3)Nowletustakethef -preimageofD(Q) .Sincef −1isjustasquarerootψ : ζ 7→√ζ −x1 on D(Q) , this preimage is contained in a domain based upon Tn+1 with atitudeh = |tn+1|rx1 −ba −x1≤|tn+1|r3 −q1 + q ≤|tn+1|/q < tn.23

Moreover,thisdomainiscontainedinthediskcenteredatzeroofradiusmax(tn+1, h) < tn . So, cl U1 ⊂V .Let us show now that cl U1 ∩cl U2 = ∅.

If a −x1 ≥x1 −b then ψD(Q) ⊂D(Tn+1) , and the statementfollows. Assume that x1 −b > a −x1 .

Then one can check the following elementary fact about the squareroot map: ψD[b, a] is convex if and only if x1 −b ≤3(a −x1) . By (8-3), the last estimate holds, so ψD(Q)is convex.

Hence, ψD(Q) ∩D(Jn+1) = ∅, and we are done. ⊔⊓Appendix.

Schwarz Lemma and Koebe Principle.We refer the reader to [Y], [G2], [Sw1-2], [MS] and [S] for the following technical background.Let us consider four points a < b < c < d and two nested intervals L = [a, d] and H = [b, c] . ThePoincar´e length of H in L is the logarithm of an appropriate cross-ratio:[H : L] = log (d −b)(c −a)(d −c)(b −a).Let g : (L, H) →(L′, H′) be a C3 diffeomorphism,Sg = g′′′g′ −32g′′g′2be its Schwarzian derivative.Schwarz Lemma.

If g has non-negative Schwarzian derivative then it contracts Poincar´e length:[H′ : L′] ≤[H : L].Koebe Principle. Let g has non-negative Schwarzian derivative.

If [H : L] ≤ℓtheng′(x)g′(y) ≤K(ℓ)for any x, y ∈H . Moreover, K(ℓ) = 1 + O(ℓ) as ℓ→0 .One can essentially extend the range of applications of these results combining the Schwarzian derivativecondition on some intervals with bounded non-linearity on others.

Let us consider a chain of (closed) intervaldiffeomorphismsI1 →J1 →... →In →Jnwhere gi : Ii →Ji have negative Schwarzian derivative while hi : Ji →Ii+1 are just C2 smooth. SetF = hn ◦gn ◦... ◦h1 ◦g1 .

Let Gi ⊂intIi and Hi ⊂intJi be closed subintervals related by diffeomorphisms.Denotebyhthefamilyofmapshi ,byIthefamilyofintervalsIietc.Let∥hi∥= max |h′′(x)/h′(x)| , ∥h∥= max ∥hi∥be the maximal nonlinearityof h , |I| = P |Ii| be thetotal length of I .Schwarz Lemma (smooth version). Expantion of the Poincar´e length by the map F is con-trolled by h in the following manner[Hn : Jn] ≤[Gn : In] + O(|J|)with the constant depending on ∥h∥.Koebe Principle (smooth version).

Distortion of F|G1 can be estimated as follows:F ′(x)F ′(y) ≤K(ℓ; |h|, |J|)where K = 1 + O(ℓ+ |J|) as |J|, ℓ→0 with the constant depending on |h| .References.24

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