Lyapunov spectrum
for exceptional rational maps
Abstract.
We study the dimension spectrum for Lyapunov exponents for rational maps acting on the Riemann sphere and characterize it by means of the LegendreFenchel transform of the hidden variational pressure. This pressure is defined by means of the variational principle with respect to nonatomic invariant probability measures and is associated to certain finite conformal measures. This allows to extend previous results to exceptional rational maps.
Key words and phrases:
2000 Mathematics Subject Classification:
Primary: 37D25, 37C45, 28D99, 37F101. Introduction and main results
We are going to study the Lyapunov exponents of a rational function acting on the Riemann sphere, of degree at least . In particular, continuing the investigations in [5], we are interested in the case that the map is exceptional. Slightly modifying [8, Section 1.3], we call exceptional if there exists a finite, nonempty, and forward invariant set such that
(1) 
Here is the Julia set of and is the set of critical points of . Every such set has at most points (see Lemma 1), hence there is a maximal set with this property, which we denote by . If is nonexceptional we put . When is clear from the context we denote simply by .
1.1. Main results
Given , denote by and the lower and upper Lyapunov exponent at , respectively. If both values coincide then we call the common value the Lyapunov exponent at and denote it by . Similarly, for a invariant probability measure we denote by its Lyapunov exponent. Let be the set of all invariant Borel probability measures supported on and be the one of all nonatomic ones. Let and be the sets of ergodic measures contained in and , respectively. Let
(see Corollary 1 for equivalent definitions of ).
For given numbers we consider the level sets
We denote by the set of Lyapunov regular points with exponent . We will describe the complexity of such level sets in terms of their Hausdorff dimension . To do so, given a parameter let us consider the potential and the pressure function
(2) 
Notice that if is nonempty the potential is unbounded and does not coincide with the the classical topological pressure for (see [17]). We define the hidden variational pressure
(3) 
(following the terminology in [15]). After Makarov and Smirnov [8, Theorem B], the pressure function fails to be real analytic on the interval if and only if is exceptional and
Moreover, by [8, Theorem A] the function is real analytic on the interval and
(4) 
For any let
(5) 
Our main result is the following theorem.
Theorem 1.
Let be a rational function of degree at least . For any numbers , with we have
In particular, for any we have
For we have
Moreover,
and
The result of the above theorem has been shown in [5] in the particular case that is nonexceptional.
To prove our main result, in this paper we will create new technical tools in order to deal with exceptional rational maps and then show how these tools can be applied to adapt the original proofs in [5]. The paper is organized as follows. In Section 2 we collect some known results about exceptional maps that will be used in the rest of the paper. In Section 3 we will introduce the concept of hidden pressure using backward branches of , analogously to the tree pressure from [15]. In the case of exceptional rational maps we not always have at hand a finite conformal measure with dense support, see Proposition 1. For that reason, in Section 4 we introduce finite conformal measures that are associated to the hidden pressure. Finally, in Section 5 we apply these tools to prove Theorem 1. In Section 5.1 we provide a lower bound for dimension using the fact that for any rational map we can find an increasing family of uniformly expanding Cantor repellers contained in using a construction of bridges that has been established in [5] and applies to the setting of this paper without changes. In Section 5.2 we provide an upper bound for dimension applying Frostman’s Lemma to an appropriate conformal measure at a conical point. Finally, in Section 5.3, we show the existence of periodic orbits in with exponent as large as possible.
2. Exceptional maps and phase transitions
For a critical point we will denote by the local degree of at . The following result has been proved by the same computation first in [4, Lemma 2].
Lemma 1.
If is a finite subset of such that , then . If is a polynomial then .
Proof.
Using that has critical points counted with multiplicity, by (1) we have
so . If is a polynomial, then it has at most finite critical points counted with multiplicity, so in this case . ∎
The following is an example of a one parameter family of exceptional rational maps such that for some parameters the exceptional set contains a critical point: for put
The point is critical of multiplicity , the point is fixed of multiplier , and the point is critical of maximal multiplicity and the only preimage of . Thus, when belongs to the Julia set we have . By choosing suitable , the fixed point could be repelling, Cremer, etc.
If is exceptional, then the set contains at least one periodic point. Observe that it hence must consist of a finite number of periodic points plus possibly some of their preimages. We write , where denotes the subset of all neutral periodic points in plus its preimages and where denotes the subset of all expanding periodic points in plus its preimages. We refer to [7] for further details on exceptional maps and numerous examples.
We will say that has a phase transition in the negative spectrum if the function fails to be real analytic on . In this case we put
We have and, since the function is convex, for each we have .
In the following proposition we gather several results in [8, 15]. A measurable subset of is said to be special if is injective. Given a function , a Borel probability measure on is said to be conformal outside if for every special set we have
If we simply say that is conformal.
Proposition 1.
Let be a rational map of degree at least and let . Then we have the following properties.

Suppose that does not have a phase transition in the negative spectrum, or that has a phase transition in the negative spectrum and . Then and there is a finite conformal measure whose support is equal to .

Suppose that has a phase transition in the negative spectrum and that . Then is exceptional, there is an expanding periodic point such that and for every neighborhood of and every measure that is conformal measure outside we have
Proof.
The equality in part follows from the definition of . The existence of the conformal measure in part follows from [8, Lemma ] if and from [15, Theorem A] if .
The fact that is exceptional and that there is an expanding periodic point such that in part is given by [8, Theorem B]. To complete the proof of part , let be a conformal measure . Since is topologically exact on , it follows that the support of is equal to . Let be the period of and let be sufficiently small so that and so that the inverse branch of fixing is defined on and satisfies . Then there is a distortion constant such that, if we put , then for each integer we have by the conformality of
Thus
proving the proposition. ∎
3. Hidden tree pressure
The goal of this section is to prove equivalence of three pressure functions: the hidden variational pressure defined in (3) as well as the hidden hyperbolic pressure and the hidden tree pressure defined in (6) and (8) below.
Given , the hidden hyperbolic pressure is defined as
(6) 
where the supremum is taken over all compact invariant (i.e. ) isolated expanding subsets of . We call such a set uniformly expanding repeller. Here isolated means that there exists a neighborhood of such that for all implies .
Proposition 2.
for every .
Proof.
The inequality follows from the variational principle. On the other hand [16, Theorem 11.6.1] implies that for any we have and hence . ∎
Before defining the hidden tree pressure, let us recall some concepts from [13], [15], and [16, Chapter 12.5]. Given and , we consider the tree pressure of at defined by
A point is said to be safe if
where denotes the spherical distance. A point is said to be expanding if there exist numbers and such that for all sufficiently large the map is univalent on and satisfies . Here, for a subset of and we denote by the connected component of containing .
We point out that every point in outside a set of Hausdorff dimension zero is safe and that for each safe point we have , see [13, 15] and compare with [13, Theorem 3.4].
Lemma 2.
There exists an expanding safe point in .
Proof.
Notice that
Since is finite and for any and , this inclusion implies that the set of points that fail to be safe has zero Hausdorff dimension. Thus, the existence of an expanding safe point outside follows from the existence of uniformly expanding Cantor repellers outside , for example as derived in [5, Lemma 4]. Note that such repellers always have a positive Hausdorff dimension by Bowen’s formula (see, for example, [16, Chapter 9.1]). ∎
Let us now define the hidden tree pressure that is an analogue of the tree pressure, obtained by considering a restricted tree of preimages. Given a subset of and which is not in the forward orbit of a critical point, we define
(7) 
and we consider the hidden tree pressure of at defined by
(8) 
Usually the point will be expanding safe and the set will be a neighborhood of . Note that after Lemma 2 there are such and .
Lemma 3.
If , is a sufficiently small neighborhood of , and is expanding safe, then the pressure does not depend on .
To prove the above lemma we need the following technical lemma.
Lemma 4.
For an arbitrary neighborhood of and an arbitrary number there exists a number and positive integers such that for every point there exist numbers , and a point such that the set is dense in and satisfies
Proof.
By the locally eventually onto property of on there is an integer such that for each the set is dense in . We put
For each integer let be defined by
We will show that for each there is an integer such . Since for each the function is continuous and for each the sequence is nondecreasing, it follows that there is a number so that is strictly positive on . This will imply the desired assertion with
We distinguish three cases:
1) If is not in the forward orbit of a critical point then .
2) If is in the forward orbit of a citical point that is not preperiodic then there exists a number such that is disjoint from . Hence, we obtain that .
3) If is in the forward orbit of a preperiodic critical point then, there is and an infinite backward trajectory starting at that is disjoint from and in particular this backward trajectory is longer than . Hence, we can choose numbers , and a point such that is not in the forward orbit of a critical point and such that for each we have . In particular, we have . Thus, if we put
then . ∎
Proof of Lemma 3.
Let , be two neighborhoods of . Without loss of generality we can assume that . By Lemma 4, every backward branch of starting at and ending at some point can be modified to end at some . The modification involves only removing at most last steps, that decreases at most by a constant factor because , and replacing them by at most steps, which stay in a uniformly bounded from below distance from critical points. Hence we conclude that and differ at most by . This proves the lemma. ∎
We denote by the maximal distortion of a map on a set . We establish one preliminary approximation result.
Proposition 3.
Given , a sufficiently small neighborhood of and an expanding safe point , for every there exists a uniformly expanding repeller such that
Proof.
We start by recalling the construction used in [15, Proposition 2.1] to prove an analogous statement for and then we modify it using Lemma 4 to prove the proposition.
As is expanding safe, there exist , and so that for all the map is univalent on and . Hence, in particular, the distortion is bounded from above uniformly in by some number . Given , let be the smallest integer satisfying . Hence, with the above, we have and , where and . Let be such that for any .
Let us choose positive constants , and large enough so that and that for every , , for every point on the component the map is univalent and satisfies
(9) 
Note that with this choice we have for large ,
As and covers , we can conclude that for every preimage there exists a component of contained in . The map , and hence , is univalent on . Thus, the map
(10) 
has no critical points, and is a uniformly expanding repeller with respect to .
Let us now slightly modify the construction of by (10) and ignore all those backward branches that correspond to a point . Given , let us consider the positive integers and the number provided by Lemma 4. Then, by Lemma 4, for each point there exist numbers , , a point in . Any such branch stays far from . Note that the distortion of on
is bounded by a constant independent of and . Given an integer , put
Note that for distinct and in the sets and are disjoint. Setting
the sets
are uniformly expanding repellers for and , respectively. Both of these sets are disjoint from by construction. On the other hand there, letting
we have
Since , there is such that
Hence, if we put , then
Since , , are independent of and , we obtain the desired assertion by taking a sufficiently large . ∎
We are now ready to prove one further equivalence.
Proposition 4.
Given a sufficiently small neighborhood of and an expanding safe point , for every we have
Proof.
By Proposition 3, we have .
In view of Lemma 3, to prove the inequality it is enough to show that for each expanding repeller that does not intersect , there is a neighborhood of such that . Notice that for every and every neighborhood of disjoint from we have,
This follows easily considering the contribution of the backward branches of contained in in the sum in (7).
Let be a neighborhood of on which is uniformly expanding. Thus, there is a constant and for every there is such that for every integer and every there is shadowing and so that
It follows that for each neighborhood disjoint from we have .
By the eventually onto property of on , we have for some . Fix and let neighborhood of disjoint from . Then we have
This shows and completes the proof of the inequality . ∎
4. finite conformal measures
Recall that is a rational map of degree at least . If is exceptional, then is the maximal finite and forward invariant subset of satisfying . Otherwise .
In the following proposition we adapt the classical method by Patterson and Sullivan to construct a conformal measure on for each . For a map without a phase transition in the negative spectrum or for a map with a phase transition in the negative spectrum at some parameter , we obtain a finite conformal measure supported on , as in part of Proposition 1. For a map having a phase transition in the negative spectrum at some parameter this construction gives us a conformal measure outside , which is finite outside each neighborhood of . Recall that by part of Proposition 1, existence of phase transition implies that there does not exist a finite conformal measure for .
Proposition 5.
Let be a rational function of degree at least . For each there exists a Borel measure on that is conformal outside , finite outside any neighborhood of , gives zero measure to and whose support is equal to .
Proof.
As a first step we will apply the PattersonSullivan method while considering only those inverse branches outside a given neighborhood of . We obtain in this way a measure that is conformal outside the set . We will obtain a measure conformal outside by taking the limit of the measures obtained by repeating this construction with replaced by smaller and smaller neighborhoods.
We start with the following lemma. Recall that denotes the set of neutral periodic points in plus its preimages.
Lemma 5.
Given , for every there exist positive numbers and such that for every and every integer we have
where the sum is taken over all satisfying for every .
Proof.
Let be sufficiently small so that for each periodic point of minimal period we have
Hence, there is some constant such for every integer and every point satisfying for every we have
Reducing if necessary, we may assume that for every the map is injective on and the set is disjoint from . So for each and there is at most one point such that . By induction we can conclude that for each , , and , there is at most one point