Ruelle operator and transcendental entire maps

If $f$ is a transcendental entire function with only algebraic singularities we calculate the Ruelle operator of $f$. Moreover, we prove both (i) if $f$ has a summable critical point, then $f$ is not structurally stable under certain topological conditions and (ii) if all critical points of $f$ belonging to Julia set are summable, then there exists no invariant lines fields in the Julia set.


Introduction
If f is a transcendental entire map we denote by f n , n ∈ N , the n-th iterate of f and write the Fatou set as F (f ) = {z ∈ C; there is some open set U containing z in which {f n } is a normal family }. The complement of F (f ) is called the Julia set J(f ). We say that f belongs to the class S q if the set of singularities of f −1 contains at most q points. Two entire maps g and h are topologically equivalent if there exist homeomorphisms φ, ψ : C → C such that ψ • g = h • φ.
If we denote by M f , f ∈ S q the set of all entire maps topologically equivalent to f we can define on M f as in [1] a structure of (q + 2) dimensional complex manifold.
Fatou's conjecture states that the only structurally stable maps on M f are the hyperbolic ones. This conjecture is false in the case when there is an invariant line field in the Julia set of f , our result is a partial answer to this conjecture for transcendental entire maps with only finite number of algebraic singularities. P. Makienko [6,7] and G.M. Levin [5] have studied the Ruelle operator and the invariant line fields for rational maps, the idea of this work is to study an application of the proposed approach given in [7] for transcendental entire functions in class S q , where the singularities of f −1 are only algebraic.
Assumptions on maps. From now on we will assume that 1. f is transcendental entire and that the singularities of f −1 are algebraic and finite and all critical points are simple (that is f ′′ (c) = 0). Note that elements of generic subfamily of the family P 1 (z) + P 2 (sin(P 3 (z))) satisfy to assumptions above, here P i (z) are polynomials.
Let F n,m the space of forms of the kind φ(z)Dz m Dz n . Consider two formal actions of f on F n,m , say f * n,m and f * n,m , on a function φ at the point z by the formulas
where |z| < f and C(ǫ, f ) is constant does not depending on µ. Then .
Hence we can define the linear map β : We call β the Bers map as an analogy with Kleinian group (see for example [3]).
We can define an equivalence relation ∼ on H(f ) in the following way, Observe that (i) H 1 (f ) is linear complex space and (ii) there exists an injection Ψ such that Ψ : In structurally unstable cases β restricted on HD(f ) × J f is always injective.
If we show that β is onto, then we are done.
Proof of the claim. We will use here quasi-conformal theory (see for example the books of I. Kra [3] and S.L. Krushkal [4] and the papers of C. McMullen and D. Sullivan [8], [9]). Let ω be the Beltrami differential on S(f ) generated by ν (that is P * (ω) = ν). Let < ψ, φ > be the Petersen scalar product on S(f ), where φ, ψ ∈ A(S(f )) and where ρ is hyperbolic metric on disconnected surface S(f ). Then by (for example) Lemmas 8.1 and 8.2 of chapter III in [3] this scalar product defines a Hilbert space structure on A(S(f )). Then there exists an element α ′ ∈ HD(S(f )) such that equality Then the push forward operator P * : A(O) → A(S(f )) is dual to the pull back operator P * . Hence element P * (α ′ ) satisfies the next condition for any g ∈ A(O).

All above means that
, a ∈ J(f ). Hence the transcendental entire maps β(P * (α))(a) = β(ν)(a) on J(f ) and we have the desired result with α = P * (α ′ ). Thus the claim and the theorem are proved.

Calculation of the Ruelle operator
Let us recall that from above there exist a decomposition where p i are polynomials, h(z) is an entire function, {c i } are the critical points of f , and is absolutely convergent.
In order to use Bers' density theorem and the infinitesimal formula of quasi-conformal maps, see Remark 1. We will work with linear combinations of the following functions.
Proposition 3.1. Let γ a (z) as above. If f is any transcendental entire map with simple critical points, then The coefficients b i and c i comes from (1).
, with compact support, denoted by supp(ϕ) and ϕ(0) = ϕ(1) = 0. Now consider the following: the first equality is by the duality with the Beltrami operator, see Lemma 1 in Section 1.
Let us denote by ψ = ϕ(f ), so supp(ψ) = f −1 supp(ϕ) and is the union K i of compact sets if there is not asymptotic values on it. Hence applying the decomposition in (1) On the other hand making some calculations and applying Green's formula we have the following equalities. Since Applying again Green's formula we have: Applying (4) on (2) we obtain This is true for each ϕ, so the function inside the integral is by Weyl's lemma an holomorphic function on C \ {0, 1} which is integrable. In our case, this implies that

Formal Relations of Ruelle Poincare Series
In this section we want to study properties of series of the form where f * n denotes the n-th iteration of the Ruelle operator. Observe from Section 2 that in general we have for some coefficients c i j , determined by the Cauchy's product of two series A = a i and Define S(x, a, z) = x n f * n (γ a (z)) and A(x, a, z) = x n Proof.
Lemma 3. If a is summable with a ∈ C, then A(x, a, z) ∈ L 1 (C) for all |x| < 1. Proof.
now by the properties of potential function we have  Proof. Observe that we can choose N such that if sequence {f n (a)} is bounded, otherwise use the absolute convergence of , hence the corollary is proved.
where K and K 1 are constant depending only on ǫ and the points c i . As result for all |x| ≤ 1 we have This proves the Lemma. So we have that the following equality holds

Ruelle Operator and Line Fields
Let f be a transcendental entire map, we say that f admits an invariant line field if there is a measurable Beltrami differential µ on the complex plane C such that B f µ = µ a.e. |µ| = 1 on a set of positive measure and µ vanishes else were. If µ = 0 outside the Julia set J(f ), we say that µ is carried on the Julia set. See [8] for results of holomorphic line fields.
In Section 2 we consider the set F ix(B f ) = {µ ∈ L ∞ (C) : B f (µ) = µ}, with B f being the Beltrami operator. Consider now the following integrals The above equation is equal to the following expression, by the properties of the potential F µ By invariance of the Ruelle operator we have By Corollary 2 and Lemma 4 we can pass to the limit x → 1 in (5), as a result we have Definition. We say that (6) is a trivial relation if and only if Ψ i = 0, i = 2, 3 . . . q and Ψ 1 = 1.
For transcendental entire maps there are, in general, many critical points c k which are mapped to the critical value d 1 , even if the function is structurally stable.

Fixed Point Theory
In this section we want to prove Theorems A and B which were stated in the introduction. In order to prove the theorems we will give a series of results.

Proof. Let us remember that
, and so (6) is a non trivial relation.
Before we prove the above proposition we will prove a series of results which will help us to prove the proposition.
Consider the modulus of the Ruelle operator: |f * |α = α(ξ i )|ξ ′ i | 2 where ξ i are the inverse branches of z under the map f .
which is a contradiction, thus the claim is proved. Now by induction on the claim, we have that |α i | = | α i | and so f * |ϕ| = |ϕ|. This proves Lemma 5.

Remark 2. The measure σ(a) =
A |φ(z)| is a non negative invariant absolutely continue probability measure, where A ⊂ C is a measurable set. Proof. (i) Every non periodic point of the Fatou set has a back wandering neighborhood. By Remark 2 we have that ϕ = 0. Thus J(f ) = C and ϕ = 0 on every component of Y .
(iii) By using notations and the proof of Lemma 5 we have is an invariant line field. Thus the corollary is proved.

Lemma 6.
β j α j = k j ≥ 0 is a non negative function.
Proof. We have |1 + β j Hence γ j 2 = 0 and α j β j = γ j 1 is a real-valued function but α j β j is meromorphic function. So γ j 1 = k j is constant on every connected component of Y and the condition |1 + k j | = 1 + |k j | shows k j ≥ 0.
Proof of Proposition 6.4.
Proof. Let us show first that all postcritical values are in Z. Assume that there is some d i ∈ Y , then by the Lemma 6, ϕ(z) = (1 + k j )ϕ(ξ j (z))ξ ′ 2 j (z). Assume that the branch ξ j (z) is such that tends to c i when z tends to d i . Then ξ ′ 2 j tends to ∞ and so ϕ(c i ) = 0. Also for every k j we have that ϕ(c i ) = (1 + k j )ϕ(ξ j (c i ))ξ ′ 2 j (c i ), so ϕ(ξ j (c i )) = 0. This implies that if c is a preimage of a critical point, then ϕ(c) = 0, since J = C then ϕ = 0 in Y , which is a contradiction.
Let us show now that Z = f i (d 1 ). We will use a McMullen argument like in [8]. By Lemma 5 and Corollary 3, µ =φ |ϕ| is an invariant line field. That implies that ϕ is dual to µ and it is defined up to a constant. We will construct a meromorphic function ψ, dual to µ and such that ψ has finite number of poles on each disc D R of radius R centered at 0.
For that suppose that for z ∈ C there exists a branch g of a suitable f n , such that g(U z ) ∈ Y , where U z is a neighborhood of z. Then define ψ(ξ) = ϕ(g(ξ))(g ′ ) 2 (ξ), for all ξ ∈ U z . Note that ψ(ξ) is dual to µ and has no poles in U z . If there is no such branch g, then ξ is in the postcritical set, and there is a branch covering F from a neighborhood of ξ to U z , then define ψ(ξ) = F * (ϕ), with F * the Ruelle operator of F . The map ψ is a meromorphic function dual to µ in U z and has finite number of poles.
By the discussion above it is possible to construct a meromorphic function dual to µ in any compact disc D R . If we make R tends to ∞, we have a meromorphic function ψ defined in C dual to µ and with a discrete set of poles.
Observe now that such ψ is holomorphic in Y , then Z is discrete and so Z = f i (d 1 ) as we claim. Since every postcritical set is in Z, that implies that f is unstable and this proves the proposition.
The following propositions can be found in [7]. For completeness we prove them. Proof Assume that l(z) = 0 on Y. Let us calculate derivative ∂l in sense of distributions, then ω = ∂l = i b i δ a i and by standard arguments Such as a i = a j for i = j, then measure ω = 0 iff all coefficients b i = 0.
Let us check (1). Otherwise in this case we have that the function l is locally integrable and l = 0 almost everywhere and hence ω = ∂l = 0 in sense of distributions and hence ω = 0 as a functional on space of all continuous functions on Z which is a contradiction with the arguments above.
2) Assume that l = 0 identically out of Z. Let R(Z) ⊂ C(Z) denote the algebra of all uniform limits of rational functions with poles out of Z in the sup −topology, here C(Z) as usually denotes the space of all continuous functions on Z with the sup −norm. Then measure ω denote a lineal functional on R(Z). The items (2) and (3) are based on the generalized Mergelyan theorem (see [2]) which states If diameters of all components of C\Z are bounded uniformly below from 0, then every continuous function holomorphic on interior of Z belongs to R(Z).
Let us show that ω annihilates the space R(Z). Indeed let f (z) ∈ R(Z) be a transcendental entire map and γ enclosing Z close enough to Z such that f (z) does not have poles in interior of γ. Then such that l = 0 out of Z we only apply Fubini's theorem Then by generalized Mergelyan theorem we have R(Z) = C(Z) and ω = 0. Contradiction. Now let us check (3). We claim that l = 0 almost everywhere on ∪ i ∂O i .
Proof of the claim. Let E ⊂ ∪ i ∂O i be any measurable subset with positive Lebesgue measure. Then the function F E (z) = E dm(ξ) ξ−z is continuous on C\ ∪ i O i and is holomorphic onto interior of C\∪ i O i . Again by generalized Mergelyan theorem F E (z) can be approximated on C\ ∪ i O i by functions from R(C\ ∪ i O i ) and hence by arguments above and by assumption we have F E (z)dω(z) = 0. But again application of Fubini's theorem gives Hence for any measurable E ⊂ ∪ i ∂O i we have E l(z) = 0. The claim is proved. Now for any component O ∈ Y and any measurable E ⊂ ∂O we have E l(z) = 0. By assumption l = 0 almost everywhere on C. Contradiction thus the proposition is proved. Proposition 6.6. If f ∈ X, then A(d 1 , z) = ϕ(z) = 0 identically on Y in the following cases 2. if diameters of components of Y are uniformly bounded below from 0, Proof Let us prove (1). If f is structurally stable then relation (3) is trivial.
Assume now that the set X c 1 is bounded. Then by Proposition 6.5 we have that ϕ(z) = Other cases follows directly from Proposition 6.5 also.
Now let X c 1 be unbounded. Let y ∈ C be a point such that the point 1 − y ∈ Y, then the map g(z) = yz z+y−1 maps X c 1 into C. Let us consider the function G(z) = 1 z i , then by proposition 6.5 G(z) = 0 identically on g(Y ). Now we Claim that Under condition of theorem A G(g(z))g ′ (z) = φ(z).

Proof of Theorem A
Theorem A. Let f ∈ X. If f has a summable critical point, then f is not structurally stable map.
Proof. It follows from Proposition 6.6 that ϕ = 0 on Y , then (6) is non a trivial relation by Proposition 6.4, then applying Proposition 5.2 the map f is not stable. Therefore Theorem A is proved.
Corollary A. Let f transcendental entire map with summable critical point c ∈ J(f ). If ϕ = 0 onto C\X c , then f is an unstable map.

Proof of Theorem B
Theorem B. If f ∈ W is summable, then there exists no invariant line fields on J(f ).
Proof. As we observe in equation (6) in Section 4, each summable critical point restricts the image of the β operator. The image of the β operator , belongs to the common solutions of the equations for all c i ∈ J(f ), hence if this system is linearly independent, then the dimension of the image of β will be 0 and so we will have J f = ∅. So we have to assume that the above system is linearly dependent.
That means in this case, that there are constants B i such that the function ϕ(z) = B i A(z, f (c i )) is a fixed point of the Ruelle operator f * . As in the Lemmas above, the measures ∂ϕ ∂z = i B i n δ f n (f (c i )) (f n ) ′ (f (c i )) = 0.
Then B i = 0, this proves Theorem B.