On invariant measures of "satellite" infinitely renormalizable quadratic polynomials
Abstract.
Let be an infinitely renormalizable quadratic polynomial and be the intersection of forward orbits of "small" Julia sets of its simple renormalizations. We prove that if admits an infinite sequence of satellite renormalizations, then every invariant measure of is supported on the postcritical set and has zero Lyapunov exponent. Coupled with [14], this implies that the Lyapunov exponent of such at is equal to zero, which answers partly a question posed by Weixiao Shen.
1. Introduction
We consider the dynamics of a quadratic polynomial. Up to a linear change of coordinates, has the form for some . In this paper, which is the sequel of [9], we assume that is infinitely-renormalizable. Moreover, in the main results we assume that has infinitely many "satellite renormalizations", see e.g. [19], or below for definitions. Dynamics, geometry and topology of such system can be very non-trivial, in particular, due to the fact that different renormalization levels are largely independent.
Historically, the first example of infinitely-renormalizsable one-dimensional map was, probably, the Feigenbaum period-doubling quadratic polynomial , where [6]. The Julia set of is locally connected [7] as it follows from so-called "complex bounds", a compactness property of renormalizations. This is a key tool since [28], in particular, in proving the Feigenbaum-Coullet-Tresser universality conjecture [28, 20, 15]. Perhaps, more striking for us are Douady-Hubbard’s examples, or alike, of infinitely-renormalizable quadratic polynomials with non-locally connected Julia sets [17, 27, 10, 11, 12, 4, 3]. As for the Feigenbaum polynomial , all the renormalizations of such maps are satellite, although, contrary to , combinatorics is unbounded (which, in turn, implies that those maps cannot have complex bounds [1]).
Dynamics of every holomorphic endomorphism of the Riemann sphere classically splits into two subsets: the Fatou set and its complement the Julia set , where is the maximal (possibly, empty) open set where the sequence of iterates , forms a normal (i.e., a precompact) family. See e.g. [2], [16] for the Fatou-Julia theory and [26] for a recent survey.
If is a polynomial, then the Julia set coincides with the boundary of the basin of infinity of . The complement is called the filled Julia set of the polynomial . The compact is connected if and only if it contains all critical points of in the complex plane.
A quadratic polynomial with connected filled Julia set is renormalizable if, for some topological disks around the critical point of , and some (period of the renormalization), the restriction is a proper branched covering map (called polynomial-like map) of degree and the non-escaping set (called the filled Julia set of the polynomial-like map ) is connected. The map is then a renormalization of and the set is a "small" (filled) Julia set of . By the theory of polynomial-like mappings [5], there is a quasiconformal homeomorphism of , which is conformal on , that conjugates on a neighborhood of to a uniquely defined another quadratic polynomial with connected filled Julia set. If is renormalizable by itself, then is called twice renormalizable, etc. If admits infinitely many renormalizations, it is called infinitely-renormalizable. The renormalization is simple if any two sets , , , are either disjoint or intersect each other at a unique point which does not separate either of them. A simple renormalization is called primitive if all sets , , are disjoints and satellite otherwise.
To state our main results, Theorems 1.1, let be infinitely renormalizable. Then its Julia set coincides with the filled Julia set and is a nowhere dense compact full connected subset of . Let be the sequence of consecutive periods of simple renormalizations of and denote the "small" Julia set of the -renormalization (where ). Then is an integer, , for any , and -orbits of ,
, form a strictly decreasing sequence of compact subsets of . Let
be the intersection of the orbits of the "small" Julia sets . For every , repelling periodic orbits of are dense in , while each component of is wandering. In particular, contains no periodic points of .
Let
be the postcritical set of . Clearly,
Moreover, the critical point is recurrent, hence,
where is the omega-limit set of a point .
We prove in [9] that cannot contain any hyperbolic set. On the other hand, a hyperbolic set of a rational map always carries an invariant measure with a positive Lyapunov exponent. So a generalization of [9] would be that never carries such a measure. Here we prove this generalization for a class of "satellite" infinitely-renormalizable quadratic polynomials:
Theorem 1.1.
Suppose that admits infinitely many satellite renormalizations. Then has no invariant probability measure with positive Lyapunov exponent.
Remark 1.1.
Conjecturally, the same conclusion should hold for any infinitely-renormalizable . One can show this assuming that the Julia set of is locally-connected (e.g., this is the case if admits complex bounds). Indeed, if had an invariant probability measure with positive Lyapunov exponent, then, taking a typical point of this measure and repeating the proof of [14], Corollary 5.5, we would conclude that the Julia set of is not locally-connected (in fact, contains a non-trivial continuum). Thus the only open case remains when has only finitely many satellite renormalizations and contains a non-trivial continuum.
Remark 1.2.
For every rational map (in particular, quadratic polynomial) and every invariant probability measure supported on Julia set Lyapunov exponents are non-negative, see [22] (compare a remark preceding Corollary 1.3). On the other hand, if is hyperbolic or non-uniformly hyperbolic (topologically Collet-Eckmann) Lyapunov exponents for all invariant probability measures supported on Julia set are positive and bounded away from , see [24].
Let us comment on the behavior of the restriction map where as in Theorem 1.1. First, by [9], the postcritical set must intersect the omega-limit set of each . At the same time, dynamics and topology of the further restriction can vary. Indeed, there are infinitely renormalizable quadratic polynomials with all renormalizations being of satellite type such that at least one of the following holds111A more complete description of should follow from the methods developed in [3].:
(1) is not minimal. This case happens in Douady-Hubbard’s type examples. Indeed, by the basic construction [17], then contains a closed invariant set (which is the limit set for the collection of -fixed points of renormalizations) such that . By [9], is non-empty. Thus is an invariant non-empty proper compact subset of .
(2) is a so-called "hairy" Cantor set, in particular, contains uncountably many non-trivial continua. This case takes place following [3].
(3) is a Cantor set and is minimal; this happens whenever either admits complex bounds (which then imply ) or is robust [19]222The ”robustness” can happen without ”complex bounds” as it follows from [3] combined with [1].. Under either of the two conditions, is a minimal homeomorphism, which is topologically conjugate to acting on the projective limit of the sequence of groups ; in particular, (hence, also , as it follows from the next Corollary 1.3) is uniquely ergodic in this case.
Theorem 1.1 yields the following dichotomy about the measurable dynamics of on the Julia set of . Recall that, by [22], any invariant probability measure on the Julia set of a rational function has non-negative exponents.
Corollary 1.3.
Let be an invariant probability ergodic measure of . Then either
-
(i)
and its Lyapunov exponent ,
or
-
(ii)
and .
In particular, the set is "measure invisible", see also Proposition 6.1 which is a somewhat stronger version of Corollary 1.3.
Corollary 1.4.
If admits infinitely many satellite renormalizations, then
(1.1) |
and
(1.2) |
As in [9], we use heavily a general result of [23] on the accessibility although the main idea of the proof is different. Indeed, in [9] we utilize the fact that the map cannot be one-to-one on an infinite hyperbolic set. At the present paper, to prove Theorem 1.1 we assign, loosely speaking, an external ray to a typical point of a hypothetical measure with positive exponent such that the family of such rays is invariant and has a controlled geometry. Given a satellite renormalization we use the measure and the above family of rays to choose a point and build a special domain that covers a "small" Julia set such that there is a univalent pullback of the domain by along the renormalization that enters into itself, leading to a contradiction. The choice of is ’probabilistic’, i.e., made from sets of positive measure, and the construction of the domain differs substantially depending on whether all satellite renormalizations of are doubling or not.
Acknowledgment. The conclusion (1.2) of Corollary 1.4 that the Lyapunov exponent at the critical value equals zero answers partly a question by Weixiao Shen333 Shen asked the following question, in relation with Corollary 5.5 of [14]: Is it possible that the upper Lyapunov exponent at a critical value of a polynomial is positive assuming that is infinitely-renormalizable around ? See also Remark 1.1 which inspired the present work as well as the prior one [9]. The authors thank the referee for careful reading the paper and many helpful comments.
2. Preliminaries
Here we collect, for further references and use throughout the paper, necessary notations and general facts. (A)-(D) are slightly adapted versions of (A)-(D) in Sect. 2, [9] which are either well-known [19], [18], or are proved here.
Let be infinitely renormalizable. We keep the notations of the Introduction.
(A). Let be the Green function of the basin of infinity of with the standard normalization at infinity . The external ray of argument is a gradient line to the level sets of that has the (asymptotic) argument at . is called the (Green) level of and the unique such that is called the (external) argument (or angle) of . A point is accessible if there is an external ray which lands at (i.e., converges to) . Then is called an (external) argument (angle) of .
Let be the doubling map . Then .
Every point of a repelling cycle of period is the landing point of an equal number , , of external rays where coincides with the number of connected components of . Their arguments are permuted by according to a rational rotation number (written in the lowest term); is the number of cycles of rays landing at . If , there is an alternative [18]:
, then so that each of two external ray landing at is fixed by ,
, i.e., , then , i.e., the arguments of rays landing at form a single cycle of .
(B). All periodic points of are repelling. Given a small Julia set containing , sets , , are called small Julia sets of level . Each contains small Julia sets of level . We have . Since all renormalizations are simple, for , the symmetric companion of can intersect the orbit of only at a single point which is periodic. On the other hand, since only finitely many external rays converge to each periodic point of , the set contains no periodic points. In particular, each component of is wandering, i.e., for all . All this implies that if and only if .
Given , for every , let be the unique such that . Let be a small Julia set of level containing and , a component of containing .
In particular, is the component of containing and , the component containing .
Note that either as or for some and all , that is, . This is so since the sequence of the sets is non-increasing, hence non-increasing, hence (the time to reach ) non-decreasing.
The map is two-to-one if and one-to-one otherwise. Moreover, for every , consists of two points if and consists of a single point otherwise. Denote
We conclude that:
is a homeomorphism. Given and , denote and
that is, the only point .
(C). Given , the map has two fixed points: the separating fixed point (that is, has at least two components) and the non-separating (so that has a single component).
For every , there are such that two rays and land at the non-separating fixed point of and the component of which does not contain has two characteristic propertiers [18]:
(i) contains and is disjoint with the forward orbit of ,
(ii) for every , consider arguments (angles) of external rays which land at . The angles split into finitely many arcs. Then the length of any such arc is bigger than the length of the arc
Denote
The rays , land at a common point . Introduce an (unbounded) domain with the boundary to be two curves and . Then and is a two-to-one branched covering. Also,
Let
so that and argument of any ray to lies in .
Let us iterate this construction. Given , let be one of the two arcs of with end points
such that arguments of any ray to lies in . Let
where . Then
and argument of any ray to lies in fact in . Note that
(2.1) |
So is a homeomorphism and has two components (’windows’) and of equal length.
Let and . The domain is bounded by two rays converging to and completed by along with two rays completed by their common limit point where .
By (i)-(ii), for a fixed , domains , , are pairwise disjoint.
Let be a component of which is contained in . Then
(2.2) |
is a two-to-one branched covering and
Let be the set of arguments of rays entering . Then consists of components so that map homeomorphically each of these components onto one of the ’windows’ of .
Furthermore, let
Unlike the map (2.2), the map
(2.3) |
is a two-to-one branched covering only assuming is a homeomorphism, which holds if and only if is a homeomorphism. In the latter case,
Primitive vs satellite renormalizations. Let and be the rotation number of . The next claim is well-known, we include the proof for reader’s convenience.
Lemma 2.1.
-
(1)
the renormalization is primitive if and only if , the period of is and is the landing point of exactly two rays and they are fixed by ,
-
(2)
points , are all different,
-
(3)
is satellite if and only if the -fixed point of coincides with the -fixed point of . In particular, and . Moreover, each of points of the orbit of is the landing points of precisely rays which are permuted by according to the rotation number . Completed by the landing point they split into "sectors" such that the closure of each of them contains a unique "small" Julia set of level sharing a common point with the boundary of the "sector".
Proof.
(1). is satellite if and only if meets at some other iterate of , hence, , and vice versa. (2). assume for some . As , the period of is smaller than . It follows that contains two small Julia sets of level that meet at , hence, separates , a contradiction as does not. (3). By (1), is satellite if and only if . Let . Then is an integer and is equal to the period of . It follows that sets are split into connected closed subsets , where and , . Moreover, and , , . By [19, Theorem 8.5], is a simple renormalization and , are subsets of its small Julia sets. Since are all consecutive periods of simple renormalizations, then for some . Therefore, -fixed point of is -fixed point of . As all renormalizations are simple, if that would imply that , a contradiction with (2). The claim about "sectors" follows since each map is one-to-one in a neighborhood of and the closure of contains a single "small" Julia set of level sharing a common point with . ∎
We need a more refined estimate provided the renormalization is not doubling. Assume is satellite so that with and the rotation number of is .
Lemma 2.2.
Assume is satellite and , i.e., is not doubling. Then
(2.4) |
In particular, given , the length of tends to zero as uniformly in (where is the integer part of ).
Moreover, for every , and the map is a two-to-one branched covering such that
Proof.
Let . Then is a two-to-one covering of degree and the critical value .
(1) Recall that consists of two ’windows’ so that is orientation preserving homeomorphism of either ’window’ onto .
(2) Consider rays to . The map is a local homeomorphism near which permutes the rays to according to the rotation number . In particular, maps any pair of adjacent rays to onto another pair of adjacent rays to .

(3) Not all arguments of these rays lie in a single ’window’ of because otherwise, by (1), the set of those arguments would lie in the non-escaping set of an orientation preserving homeomorphism , which consists of a fixed point of this map, a contradiction with the fact that .
(4) The rays split into disjoint domains , . By the "ideal boundary" of we will mean the usual (topological) boundary (in our case, the set of boundary rays completed by their landing points) along with the "boundary at infinity" which is the set of arguments of rays entering . Then define on to be on and on the "boundary at infinity" of .
(5) By (3), one of , called , has in its boundary, and another one, called , has in the boundary. In particular, the boundary of any other , , consists of a pair of adjacent rays to whose arguments belong to a single ’window’ of . Therefore, by (1), the rest of indices can be ordered in such a way that is a one-to-one map for (note that the "boundary at infinity" of each , , consists of a single "arc at infinity"). Therefore, is a homeomorphism for . The map on is also a one-to-one map on its image where is bounded by two adjacent rays to . cannot contain because otherwise would contain , a contradiction. Thus must contain . That is, covers .
Thus, for , is a homeomorphism, and is also a homeomorphism where the image covers and has two common rays with the boundary of .
(6) The critical value of has a unique preimage by g (the critical point of ). As and is bounded by two adjacent rays to , for some . If , then while would not be a homeomorphism of on its image. This shows that .
∎
(D). Given a compact set denote by (or simply , if the map is fixed) the set of arguments of the external rays which have their limit sets contained in . It follows from (C) that , i.e., it is either a single-point set or a two-point set.
Since contains at most two angles, contains at most two different accessible points. More generally, given let
Then so that
is not empty and consists of either one or two components. Since for we conclude using (2.1):
consists of either a single point or two different points. In particular, for any component of which is not one of , , there is either one or two rays tending to .
From now on, is an -invariant probability ergodic measures supported in : , and having a positive Lyapunov exponent
(E). We start with the following basic statement. Parts (i)-(ii) are easy consequences of the invariance of and (B) while (iii) is a part of Pesin’s theory as in [25, Theorem 11.2.3] coupled with the structure of , see (B). Recall that .
Proposition 2.3.
(i) For every and , .
(ii) has no atoms and for every component of .
(iii) and is a -measure preserving homeomorphism. There exists a measurable positive function on such that for -almost every , and all , if is the unique point of with , then a (univalent) branch of is well-defined such that ,
Remark 2.4.
The branch of depends on and but it should be clear from the context which points and are meant.
Using the Birkhoff Ergodic Theorem and Egorov’s theorem, Proposition 2.3 implies immediately (e1)-(e3) of the next corollary. The proof of (e4)-(e5) is given right after it.
Corollary 2.5.
For every , there exists a closed set and constants , such that:
() ,
() there exists another closed set such that as follows. For every and every there exists a (univalent) branch of such that and , for every . Moreover, as uniformly in ,
() for every there exists a sequence of positive integers , , such that and for all ,
() given and , let be the unique such that . Then as uniformly in ,
() for , we have: and
as uniformly in .
Proof of ()-(): assuming the contrary in (), we find some and sequences and , , such that (see (B)) hence, , for all . Since is closed, one can assume . Hence, , a contradiction. Now, for () using (), uniformly in .
3. External rays to typical points
We define a telescope following essentially [23]. Given , , , and , an -telescope at is collections of times and disks , such that, for every : (i) , (ii) there is a univalent branch of so that and, for , (clearly, here is a branch of that maps to ). The trace of the telescope is a collection of sets , . We have: .
By the first point of intersection of a ray , or an arc of , with a set we mean a point of with the minimal level (if it exists).
Theorem 3.1.
[23] Given , , and there exist , and such that for every -telescope the following hold. Let . Let be any point at the boundary of such that . Then there are indexes such that , , as follows. Let and let be an infinite arc of an external ray through between the point and . Let and, for , let be the first point of intersection of with . Then, for ,
Next corollary of Theorem 3.1 is a key one.
Proposition 3.1.
Given there exists a closed set as follows. First, and where is the set defined in (E) and satisfies ()-(). There exists and, for each there is as follows.
(1) Let . Then is the landing point of an external ray of argument . Moreover, the first intersection of with has the level at least .
(2) for each , a branch of is well-defined such that , , for every and as uniformly in ,
(3) if , then .
Proof.
(1)-(2) will hold already for the set which follows from Theorem 3.1 as in [23] and uses only that has a positive exponent; (3) will follow in our case as we shrink a bit the set since each point admits at most two external arguments.
Here are details. Let and as in the properties ()-() of the set . Then, by ()-(), there is such that, for each , every admits -telescope with the times that appear in the property () of . On the other hand, there exists such that for every there is a point with the level . This is so due to .
Given this , let , and be as in Theorem 3.1.
Let and as in (). Fix . Let be the corresponding trace. By Theorem 3.1, there are such that , , . Let be an arc of an external ray between the point and . Let be the first intersection of with . Then, for ,
(3.1) |
For all ,
(3.2) |
Denote by the argument of an external ray that contains the arc .
Now, given a sequence
(3.3) |
such that , we get a sequence of arguments and a sequence of arcs of external rays of the corresponding arguments . Passing to a subsequence in the sequence , if necessary, one can assume that , for some argument .
Fix any and choose such that,
Then, by Theorem 3.1, for each , the first intersection of the ray with the boundary of has the level at least . It follows, for any , the sequence of arcs of the rays between the levels and do not exit for all . As , it follows that the arc of the ray between levels and stays in too. As and can be chosen arbitrary small, must land at and satisfy (1) with replaced by .
Let us call the above procedure of getting from the constants , , the point and the sequence (3.3) the -procedure.
Note that (2) is property () of the set .
In order to satisfy property (3), we shrink the set and correct changing it to some (if necessary) as follows. Using the Birkhoff Ergodic Theorem and Egorov’s theorem, choose a closed subset of such that and, for each , the set is infinite. Note that . We have proved that, for each , (1) holds for the point instead of , in particular, is an argument of . On the other hand, by (D1), each admits at most two external arguments, hence, all possible external arguments of the forward orbit , , belong to at most two different orbits of . Hence, there is one of those orbits, for some , such that the intersection is an infinite set, so that for an infinite sequence .
Let’s start over with the -procedure for the point and the sequence . Then, by the construction, for all , hence, (1) holds with instead of the previous . If is any other point of the grand orbit (remember that is invertible), the -procedure works for with the same (perhaps, truncated) sequence , which ensures that (3) holds (for the corrected arguments) too. ∎
Remark 3.2.
Given , we cannot just set to satisfy property (3) because this would change in the definition of telescope, so we might loose property (1). Notice that correcting (flipping) to does not change The same for flipping any in the grand orbit of . But the flipping can make for where with , thus yielding (3).
4. Lemmas
Recall that for any we define for any . This makes sense since is invertible on , see (E).
Lemma 4.1.
Let where .
(a) If then .
(b) yields . If, additionally to (a), for all and where , where is always either or then and are in the same component of .
(c) If for all and (where , are defined in Proposition 3.1), then the ray lands at the limit point . In particular, given there is such that for some whenever .
Proof.
(a) Assume the contrary. Then there is such that . As, for any , where the latter union is a subset of , the distance between and is at least , a contradiction.
(b) by combinatorics and definitions of points . In particular, for every , and are in the same component of . By (a), any limit set of the sequence of compacts in the Hausdorff metric is a subset of . On the other hand, is connected as each set is connected. This proves (b).
(c) We prove only the first claim as the second one directly follows from it. Fix any and choose such that for any , . Then, by Proposition 3.1, part (1), for each , the first intersection of the ray with the boundary of has the level at least . It follows, for any , the sequence of arcs of the rays between the levels and do not exit for all . As and can be chosen arbitrary small, must land at . ∎
By lemma 4.1(c), if arguments of are close then are close as well.
Definition 4.2.
Given and we define as follows. First, for and , we define . Namely, let be so that the distance between the equipotential of level and is bigger than . Then is such that for any , if lie on the same equipotential of level and the difference between external arguments of is less than then the length of the shortest arc of the equipotential between and is less than . Apply Lemma 4.1(c) with and find the corresponding . Let
where is defined in Proposition 3.1.
In the next two lemmas we construct curves with special properties. The idea is as follows. Let . Then . It is easy to get a curve in as follows: begin with an arc from a point to and then iterate this arc by . In this way we get a curve such that , hence, lands at a fixed point of . We show in the next lemma (in a more general setting) that if both points are either in the range of the covering (2.2) (condition (I)) or in the range of the covering (2.3) (condition (II)) then . This implies that has to be the -fixed point of . In Lemma 4.5 assuming additionally that is satellite, we ’rotate’ the curve by to put the set in a ’sector’ bounded by and by its ’rotation’. In Lemma 4.7-4.8 we consider the case of doubling for which the condition (II) usually does not hold.
In what follows, we use the following notation: given , let
It is a closed subset of of points such that for . As is a -automorphism, . Notice that this bound is independent of .
Recall that for a semi-open curve , we say that lands at, or tends to, or converges to a point if there exists . Then , are endpoints of the curve and is called also the starting point of .
Lemma 4.3.
Fix and consider the set with the corresponding constant . Fix . Let from Definition 4.2. For every there exist , as follows.
For every consider the closed set . Fix . Assume for an arbitrary as above, that either (I) and belong to a single component of , or (II) the map is a homeomorphism and the length of the arc is less than .
Then:
(a) the map has a unique fixed point and ,
(b) there is a semi-open simple curve
such that:
-
(1)
it lands at and . Another end point of lies in and ,
-
(2)
where the ’fundamental arc’ consists of an arc of an equipotential of the level at least that joins a point with a point , being extended by an arc of the ray between points and ; in particular, the Green function at a point is not increasing as moves from to along ,
-
(3)
the point is the landing point of a ray which is fixed by and which is homotopic to through a family of curves in with the fixed end point .
-
(4)
arguments of all points of the curve lie in a single component of in the case (I) and in a single component of in the case (II) (recall that has components and has components, see Sect 2, (C)).
Besides,
(4.1) |
as , uniformly in and .
(c) if then where , the non-separating fixed point of . Moreover,
as .
Remark 4.4.
Note that while .
Proof.
Fix such that, for every and , the length of each ’window’ of is less than . Therefore, for , in either case (I), (II),
(4.2) |
which implies, in particular, that .
Denote which is a holomorphic univalent function in . Since are uniform contractions there is such that whenever . Let .
Let also , where is defined in Proposition 3.1.
Let be the unique fixed point of the latter map . We construct the curve to the point as follows. First, joint a point , , to a point by an arc of the equipotential . By the choice of , . Secondly, connect to the point by an arc . Let now . Then properties (1), (2) in (b) are immediate and (3) follows from general properties of conformal maps. Now, by Proposition 3.1(2) and (4.2), for all big enough, , moreover, the modulus of the annulus tends to as . Therefore, (4.1) follows from Koebe and Proposition 3.1(2).
It remains to show the property (3) and that . Consider the case (II), which is equivalent to say that the map is a homeomorphism on each of two components of . Let be the set of arguments of points of the curve . Let be a component that contains . Assume, by a contradiction, that contains which is in the boundary of . Then is the argument of a point of , for some , hence, is simultaneously the argument of a point of and in the boundary of , a contradiction. The case (I) is similar. Property (3) is verified. In fact, we proved more: for , the set is a subset of a single (depending on ) component of in the case (II) and a single component of in the case (I). This implies that all point , , of the cycle of containing belong to the closure of in the case (II) and to the closure of in the case (I). Therefore, this cycle lies in , in particular, .
Proof of (c): if then is a fixed point of and, moreover, the ray lands at and is fixed by . Hence, the rotation number of w.r.t. the map is zero. On the other hand, is the only such a fixed point, i.e., as claimed. Then (4.1) implies that . ∎
For the rest of the paper, let us fix , , , , , and as follows:
, , is such that
This choice is motivated by the following
Fact ([21], [13], [8]): if a repelling fixed point of is the landing point of rays, then . Hence, if , then .
Furthermore, fix such that , apply Proposition 3.1 and Lemma 4.3 and find, first, , then fix and find the corresponding , and .
Let
Let us analyze several possibilities.
Lemma 4.5.
There is as follows. Let and . Consider so that .
Let and . Assume that either (I) , belong to a single component of , or (II) the map is a homeomorphism and the length of the arc is less than .
Then:
(i) as and for .
(ii) assume that is satellite, i.e., (by Lemma 2.1) has period , with rotation number of , and is the (i.e., separating) fixed point of . Then and
(4.3) |
There exist two simple semi-open curves and that satisfy the following properties:
-
(1)
and tend to and ,
-
(2)
consist of arcs of equipotentials and external rays; the starting point of lies in an arc of and the starting point of lies in an arc of where for some , such that levels of and are equal and at least ,
-
(3)
one of the two curves (say, ) is homotopic, through curves in tending to , to the ray , and another one - to the ray ;
-
(4)
, ,
-
(5)
, , in particular, are disjoint; being completed by their common limit point and two other arcs: an arc of the ray from to and an arc of the ray from to , they split the plane into two domains such that one of them contains and another one contains all other different iterates , . The intersection of closures of all those sets consists of the fixed point of .
Remark 4.6.
Beware that the point that determines both curves , does not belong to either of these curves.
Proof.
(i) follows from Lemma 4.3 where we take . Fix such that for all .
Let us prove (ii). Here we build a "flower" of arcs at the fixed of the satellite starting with an arc which is fixed by and then "rotate" this arc by a branch of (for which the same point is also a fixed point, see (C)). Let where the latter curve is defined in Lemma 4.3. Then properties (1)-(3) of the curve are satisfied also for . In particular, is homotopic to .
As both are external arguments of which is a -periodic point of , there is such that . Now we use that and that to prove (4.3). Indeed, for each , since is a homeomorphism and , we have: . Hence, if , then implying that . Then which along with Proposition 3.1, part (2) imply (4.3).
In turn, (4.3) implies that, provided is big, uniformly in where is either or .
Now we consider a curve that starts at and tends to . By Proposition 3.1 coupled with (4.3), one can join by an arc of the ray inside of up to a point of level . This will be the required curve . To get the curve we modify by cutting off the arc of an equipotential: (see Lemma 4.3 for details about ). Properties (1)-(5) follow.
∎
Given a point and such that , where , let and , the arguments of , as in Proposition 3.1. We call -friendly if and lie in the same component of and -unfriendly otherwise (or simply friendly and unfriendly if is clear from the context). The name reflects the fact that for an -friendly point the condition (I) of Lemma 4.5 always holds for and , so Lemma 4.5 always applies.
When the rotation number of is equal to we have:
Lemma 4.7.
There is (depending only on fixed and ) as follows. Suppose that, for some , the rotation number of the separating fixed point is equal to . Let and , . Assume that all three points are -unfriendly.
Then there exist two (semi-open) curves and consisting of arcs of rays and equipotentials with the following properties:
(i) , , moreover, arguments of points of lie in one ’window’ of while arguments of points of lie in another ’window’ of ,
(ii) and converge to a common point which is a fixed point of (i.e., is either the non-separating fixed point or the separating fixed point ,
(iii) starting points of have equal Green level which is bigger than ,
(iv) , , as .
Proof.
As , lengths of ’windows’ of tend uniformly to zero as . It follows from the definition of friendly-unfriendly points that are in one ’window’ of and are in another ’window’ of . Therefore, condition (I) of Lemma 4.3 holds for each pair and . Now, apply Lemma 4.3 to , first, with , , and then with , . Let and . Then (i),(iii) hold. To check (ii), note that these curves converge to some points , which are fixed by On the other hand, since the rotation number of is , has no -cycle. Therefore, one must have either or , i.e., (ii) holds too. As and as , , too, by Lemma 4.1. Besides, by (4.1), as . As , (iv) also follows. ∎
Lemma 4.8.
Let . Assume that is satellite and doubling, i.e., and the rotation number of is equal to (in particular, ). For some , denote . Let , be the two small Julia sets of the next level which are contained in (note that contains the critical point and contains the critical value of the map ). Let be such that all its forward iterates , . Then there exist two simple semi-open curves , consisting of arcs of rays and equipotentials that satisfy essentially conclusions of the previous lemma where is replaced by , i.e.:
(i) , moreover, arguments of points of lie in one ’window’ of while arguments of points of lie in another ’window’ of ,
(ii) and converge to a common point which is a fixed point of (i.e., is either the non-separating fixed point or the separating fixed point ,
(iii) starting points of have equal Green level which is bigger than ,
(iv) , as uniformly in .
Remark 4.9.
Condition , , is equivalent to the following: .
Proof.
To fix the idea let’s replace , using a conjugacy to a quadratic polynomial, by a quadratic polynomial (denoted also by ) so that now where and is satellite with two small Julia sets , that meet at the -fixed point of and rays of arguments , land at . Here , , is a homeomorphism while is a two-to-one map. If a ray of has its accumulation set in then and if accumulates in then . This implies that if lands at and lies in one of the two ’windows’ , then lands at where must be in a different ’window’ (in other words, points of are ’unfriendly’). Coming back to this means that, for , are always in different components (where by ’component’ we mean a component of ). Besides, for , and are always in different , . This leaves us with the only possibilities:
(i) are in different components; this implies that are in different components and are in different components, that is, points are all unfriendly;
(ii) are in the same components; there are two subcases:
(ii’) are in different components, i.e., (i) holds with replaced by which implies that are all unfriendly;
(ii”) are in the same component which then means that and are both friendly.
5. Proof of Theorem 1.1
Every invariant probability measure with positive Lyapunov exponent has an ergodic component with positive exponent. So let be such an ergodic -invariant measure component supported in . First, we have the following general
Remark 5.1.
Given such that as in Proposition 2.3, and given , the set cannot be covered by because otherwise the branch of , which sends to meets the critical value along the way so cannot be well-defined. Thus , for each , and . In particular, for all and .
We need to prove that has finitely many satellite renormalizations. Assuming the contrary, let be an infinite subsequence such that is a satellite renormalization of for each .
We arrive at a contradiction by considering, roughly speaking, two alternative situations. In the first one, we find a point , , and two curves in where that tend to the -fixed points of such that another ends of the curves can be joined by an arc of equipotential in thus ’surrounding’ by a ’triangle’ in which would be a contradiction as in Remark 5.1. The second situation is when the first one does not happen. Then we use several curves to ’surround’ by a ’quadrilateral’ in , ending by the same conclusion. The curves we use have been constructed in Lemmas 4.5, 4.8.
The first situation happens in cases A and B1, and the second one in B2.
Case A: contains an infinite sequence of indices of non-doubling renormalizations. Passing to a subsequence one can assume that is satellite not doubling for every .
Fix . By Lemma 2.2, for each and each , the map is a homeomorphism and the length as uniformly in . Fix such that for each , . For , let
Let , . Denote , . The intersection contains all with , . Hence,
as . Therefore, fixing =1/8, there are such that for each , , ,
Fix such , assume also that where is defined in Lemma 4.5 and recall the set
Since , there is and, by the choice of , the assumption (II) of Lemma 4.5 holds for . Therefore, there exist two simple semi-open curves and that satisfy the following properties: and tend to , and consist of arcs of equipotentials and external rays; the starting point of and the starting point of have equal levels which is at least ; , ; finally, being completed by their common limit point and arcs of rays from to and from to , they split the plane into two domains such that one of them contains and another one contains all other iterates , . Now, since and by the choice of , the distance between arguments of the points and inside of is less than . By the definition of , and can be joined by an arc of equipotential inside of . Consider a Jordan domain with the boundary to be the arc and semi-open curves , completed by their common limit point . Then . By the properties of the curves , contains either or its iterate , for some , in a contradiction with Remark 5.1.
Complementary to A is
Case B: for all big , every satellite renormalization is doubling, i.e., and for every .
Let and . Note that .
For every , let
As , it follows,
Since we are in case B, each contains precisely two small Julia sets of the next level each of them of measure . Hence, the measure of intersection of each of these small Julia sets with is bigger than . By Lemma 4.8, choosing for every a point we get a pair of curves consisting of arcs of rays and equipotentials as follows: (i) , moreover, arguments of points of lie in one ’window’ of while arguments of points of lie in another ’window’ of , (ii) and converge to a common point which is a fixed point of (i.e., is either the non-separating fixed point or the separating fixed point , (iii) start points of have equal Green level which is bigger than , (iv) as uniformly in and . We add one more property as follows. Let
Then: (v) is a simple curve; the level of is positive and decreases (not strickly) from to zero along and then increases from zero to along ; moreover, if , , then , are either disjoint or meet at the unique common point and then disjoint with all others , . This is because, by property (i), where (by (C), Sect 2) any two , , , are either disjoint or meet at in which case is satellite. In the considered case, any satellite is doubling so for all different from .
We assign, for the use below, a ’small’ Julia set to each as follows: by the construction, is either the -fixed point of , or the -fixed point of . In the former case, let , and in the latter case, (one of the two small Julai sets of the next level that are contained in . Observe that and is disjoint with any other provided , are disjoint.
There are two subcases B1-B2 to distinguish depending on whether arguments of end points of become close or not. If yes, then one can join the end points of some by an arc of equipotential inside of to surround a small Julia set as in case A, which would lead to a contradiction. If no, the construction is more subtle: we build a domain (’quadrilateral’) in bounded by two disjoint curves as above completed by two arcs of equipotential that join ends of different curve, so that the obtained quadrilateral again contains a small Julia set.
B1: .
By property (i) listed above and the definition of , there are a sequence , and as above, such that two ends of each curve can be joined inside of by an arc of equipotential of fixed level such that all arguments of points in belong to . Then we arrive at a contradiction as in case A.

B2: for all big and all .
Fix , . Define a subset of as follows:
As ,
For each we define further
Then
as otherwise and, therefore, , i.e., , a contradiction because .
Fix . Thus contains pairwise different indices , . As , we find curves . By property (v), if two of them meet, they are disjoint with all others. Therefore, there are at least of them denoted by , , which are pairwise disjoint. Let be two ends of .
For each , arguments of points of lie in different ’windows’ of . On the other hand, by the choice of , . As is big enough, lengths of ’windows’ of are less than . But since we are in case B2, the length of is bigger than . One can assume, therefore, that, for , arguments of lie in one window of while arguments of are in another window. Therefore, differences of arguments of all tend to zero as , and the same for . As all , this implies by Lemma 4.1 that . This along with property (iv) implies that , , for all big . Since, for big , differences of arguments of all are less than , and the same for , one can joint all by an arc of equipotential of level and all by an arc of equipotential of the same level such that . Let the end points of be, say, and , so that is in between. Since all curves , , are pairwise disjoint, the end points of have to be then and , so that is in between. Therefore, we get a ’big’ quadrilateral bounded by where , . The curve splits into two ’small’ quadrilaterals with a common curve in their boundaries. Recall now that the curve comes with a small Julia set of level either or , such that is a single point while is disjoint with , . Therefore, , a contradiction with Remark 5.1.
6. Proof of Corollaries 1.3-1.4
Corollary 1.3 follows directly from the following
Proposition 6.1.
Let be an infinitely renormalizable quadratic polynomial. Then conditions (1)-(4) are equivalent:
-
(1)
has no invariant probability measure with positive exponent,
-
(2)
for every neighborhood of and every there exist and such that, for each and with ,
additionally, has no invariant probability measure with positive exponent,
-
(3)
every invariant probability measure of is, in fact, supported on and has zero exponent,
-
(4)
for every invariant probability ergodic measure of on the Julia set of , either and its Lyapunov exponent , or and .
Proof.
(1)(2). Assume the contrary. Let . Since is a neighborhood of a compact set , the Euclidean distance . By a standard normality argument, as all periodic points of are repelling, there are and such that whenever and . As (2) does not hold, find , a sequence , points and a sequence such that, for each ,
Fix a big such that and consider the times where such that . Let so that . Therefore, by the choice of and , where . In this way we get a sequence of measures such that the Lyapunov exponent of is at least . Passing to a subsequence one can assume that converges weak-* to a measure . Then is an -invariant probability measure on with the exponent at least , a contradiction with (1).
(2)(3), by the Birkhoff Ergodic Theorem along with [22].
(3)(4): let be as in (4) and for some open set with . Let be the first return map equipped with the induced invariant measure . By the Birkhoff Ergodic Theorem and by an argument as in (1)(2), the exponent of w.r.t. is strictly positive. Hence, is positive too. This proves the implication.
And (4) obviously implies (1). ∎
Proof of Corollary 1.4.
If were strictly positive, for some , that would imply, by a standard argument (see the proof of Corollary 1.3), the existence of an -invariant measure with positive exponent supported in , with a contradiction to Theorem 1.1. This proves (1.1). By [14], . On the other hand, by (1.1), , which proves (1.2). ∎
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