No entire function with real multipliers in class S

We prove that there is no entire transcendental function in class S with real multipliers of all repelling periodic orbits.


Introduction
Already Fatou in [4, section 46] pointed out that if the Julia set of a rational function is a smooth curve, then all periodic points in the Julia set have real multipliers. Recently Eremenko and van Strien proved the converse statement [3, Theorem 1], i.e. if all repelling periodic orbits of a rational function f : C → C (of degree at least 2) have real multipliers, then either the Julia set is contained in a circle or f is a Lattés map. Moreover, they gave a detailed description of these rational maps whose Julia set is contained in a circle.
Their result (as well as our previous considerations in [1,Lemma 3.7]) was a motivation for us to study transcendental functions with real multipliers of all repelling periodic orbits. We prove the following claim: Theorem 1.1. There exists no entire transcendental function in class S with real multipliers of all repelling periodic orbits.
In order to prove it, we will construct on C an invariant line field, univalent in a neighbourhood of the Julia set. This will contradict Theorem 6.1 from [6] saying that such a line field cannot exist in the transcendental case. The construction of a line field uses a global linearization function associated with a repelling periodic point that posses only regular preimages. We will show that this linearization function is automorphic with respect to a discrete group of affine maps. The proof of this part strongly relies on methods introduced by Eremenko and van Strien in [3,.
Note that to obtain the contradiction it is enough to assume that f has real multipliers of repelling cycles in some relatively open set in the Julia set. Let us also remark that the proof remains valid if f is an entire transcendental function from class B (i.e. its finite singular values form a bounded set) for which there exists in the Julia set a relatively open set disjoint from orbits of singular values.

Preliminaries
Throughout the entire article f : C → C denotes an entire transcendental function from class S, i.e. with finitely many singular values. By singular value we mean critical value or finite asymptotic value. Note that in this case, similarly as for rational functions, only finitely many periodic points may belong to the forward orbits of singular values. It means that from among an infinite number of repelling cycles we can always choose such that have only regular preimages.
Recall that the Fatou set F (f ) consists of points in whose neighbourhood the iterates {f n } n≥1 form a normal family while the Julia set is its complement J(f ) = C \ F (f ). The Julia set may be equivalently defined as the closure of all repelling periodic orbits of f . Moreover, if z ∈ J(f ) is not an exceptional value (i.e. its backward orbit is infinite), then For these and more properties of Fatou and Julia sets see [2]. We will also use the fact that the Hausdorff dimension of the Julia set of an entire transcendental function with finitely many singularities is greater than one (see [7]). Our proof uses the notion of invariant curves and unstable manifolds introduced in [3, Section 1.1].
Definition 2.1. A simple curve γ : (0, 1) → C passing through a repelling periodic point p of period n is called an unstable manifold for p if there exists a subarc γ * ⊂ γ containing p such that f n maps γ * diffeomorphically onto γ.
We say that γ is an invariant curve for q ∈ f −m (p) if γ is contained in an unstable manifold for p, q ∈ γ and f m (γ ∩ V ) ⊂ γ for some neighbourhood V ∋ q.
Note that this is not what we usually mean by an unstable manifold for a hyperbolic periodic point of a smooth dynamical system. We decided, however, to apply the notion from [3] for the convenience of the reader, since our proof closely follows their.
As we mentioned in the introduction our first aim will be to show that if f has only real multipliers of repelling periodic orbits, then a linearization function Ψ associated with a repelling point (with regular preimages) is automorphic with respect to a discrete group of isometries of the plane C (see [5,Definition 2.4]).
It was shown in [5, Lemma 2.5] that the transitivity condition (2) in the above definition can be replaced by the following weaker assertion: (2') Γ acts transitively on one regular fibre, i.e. there is w ∈ Ψ(C) which is not a critical value of Ψ such that for every pair z 1 , z 2 ∈ Ψ −1 (w) we can find some γ ∈ Γ with γ(z 1 ) = z 2 .
We will use the linearization function to construct an invariant line field on C univalent in an open set intersecting the Julia set (cf. [6, Definition 4.1]). Such line field cannot exist for transcendental functions by the result of Rempe and van Strien [6, Theorem 6.1]. Recall that an invariant line field on C is a measurable choice of directions for any z ∈ C invariant under f . Since any direction (straight line through the origin) is determined by two angles which differ in π, it is convenient to associate an invariant line field with a measurable function u : C → S 1 satisfying the invariance condition for almost all z ∈ C.
3 Proof of Theorem 1.1 Suppose f : C → C is an entire transcendental function from class S such that all repelling cycles have real multipliers. Choose a repelling periodic point p ∈ J(f ) which does not belong to the forward orbit of singular values. Replacing if necessary f by its iterate we may assume that p is a fixed point.
Denote by Ψ a local linearization of f at p, i.e. a univalent function defined on a small disc O 0 centered at the origin such that for z close to 0, which can be chosen arbitrarily. We can extend Ψ to a global linearization Ψ : C → C satisfying the same relation (3.1) by the formula Note that, since Ψ is univalent on O 0 , it immediately implies the following fact.
Lemma 3.1. If p is any repelling periodic point which is not an iterate of a singular value and Ψ is a global linearization at p, then for any Additionally, since O 0 was chosen to be a round disc, We are going to prove that T is actually of the form ±z + Q. To do this we will show that T ′ is real on an infinite family of straight lines passing through the origin. This part of the proof closely follows [6, sections 1.1-1.2] however we needed to adapt it to the case of entire transcendental function. For the convenience of the reader we present it with all details and keep similar notation.
hence there exists a point z n ∈ V n such that Ψ(z n ) is a repelling fixed point for f n . We obtain this way a whole sequence z n → 0, for n ≥ N, as stated in the lemma. 2. For n large enough there exists a sequence z n ∈ L such that z n → 0, λ n z n → Q, Ψ(z n ) is a repelling point of period n and γ is an unstable manifold for Ψ(z n ); 3. For n large enough γ is an invariant curve for Ψ(λ −n Q) ∈ f −n (p); Proof. Let z n be the sequence defined in the proof of Lemma 3.2 and let x n = Ψ(z n ) be the corresponding periodic point. We are going to show that Recall that multipliers of all repelling cycles are real, thus Note also that Ψ(λ n z n ) = f n (Ψ(z n )) = f n (x n ) = x n = Ψ(z n ). But for n large enough we have: z n ∈ O 0 , λ n z n ∈ O 1 , and since T : If T ′ is constant then T is an affine map, T (z) = az + Q, where a = T ′ (0) ∈ R and the identity Ψ•T = Ψ extends to the whole plane C. But this is possible only if |a| = 1, so we conclude that T (z) = ±z + Q. Then obviously T (L) = L, where L is the line through 0 and Q, in particular T (L ∩ O 0 ) = L ∩ O 1 . Moreover, (3.3) implies that z n = Q λ n ±1 ∈ L. Suppose now that T ′ is not constant, then the set X = {z ∈ O 0 : T ′ (z) ∈ R} is a finite union of real analytic curves. We will show that one of these curves is L ∩ O 0 . For simplicity we may assume without loss of generality that L is the real line, thus Q ∈ R.
Let β be a curve in X containing infinitely many points z n . Since λ n z n → Q ∈ R, thus arg(z n ) → 0 and we conclude that β is tangent to L = R at 0. Note that if T | R is real, then we are done. Suppose that T | R is not real, then where m is chosen so that a 1 , . . . , a m ∈ R and a m+1 / ∈ R. Since a 1 = T ′ (0) ∈ R \ {0}, we have that m ≥ 1.
We are going to show that β ⊂ R. Suppose on the contrary that this is not the case and recall that β is tangent to L = R at 0 so it is of the form Let k ≥ 2 be the smallest subscript for which a k = 0. Since a m+1 / ∈ R, therefore k ≤ m + 1. If k = m+ 1, then the condition T ′ | β ∈ R implies ℑT ′ (β(x)) = (m+ 1)ℑa m+1 x m + . . . ≡ 0, what is impossible since ℑa m+1 = 0. It follows that k ≤ m.
Note that, since a k = 0 and ℑa m+1 = 0, this is possible only if k − 2 + K = m. Therefore, Since z n ∈ β, it is of the form z n = t n +ibt K n +o(t K n ), where t n → 0 and so arg z n ∼ t K−1 n . On the other hand, we have that ℜT (z n ) → Q = 0 and ℑT (z n ) = O(t K n ) (in view of (3.4) and the form of T ), therefore arg T (z n ) = O(t K n ). But this clearly contradicts (3.3) what proves the second part of point 5. And since any curve from X containing infinitely many points z n is actually the line L, therefore for all n bigger than some N, we have z n ∈ L. Now, T : O 0 → O 1 is biholomorphic with T (0) = Q and T ′ | L∩O 0 is real, thus for any z ∈ L ∩ O 0 , we have T (z) ∈ L, giving 4.
If we take γ * = Ψ(λ −1 (L ∩ O 0 )), then it is clear that the curve γ = Ψ(L ∩ O 0 ) is an unstable manifold for p. In order to prove 2., recall that f n : The same argument proves also 3., since λ −n Q ∈ L ∩ V n .
Recall that Q was an arbitrary Ψ-preimage of p different from 0 and T : O 0 → O 1 was a biholomorphic map such that Ψ • T = Ψ on O 0 . So far we showed that T ′ is real on the line L, in what follows we are going to prove that T ′ is actually constant. Proof. Notice that Ψ −1 (J(f ))) cannot be contained in a finite family of lines since the Hausdorff dimension of J(f ) is greater than one (see [7]). In particular Ψ −1 (J(f )) L. Recall that p is not an asymptotic value so it is not exceptional, hence its backward orbit is dense in J(f ). There exists thus a point and denote by L ′ the line through 0 and Q ′ . Note that we can construct infinitely many such lines. We will show that T ′ is real on L ′ . By Lemma 3.1 we have Ψ ′ (Q ′ ) = 0 so we can choose O ′ 0 and O ′ 1 , neighbourhoods of 0 and Q ′ respectively, on which Ψ is univalent and such that Ψ(O ′ 0 ) = Ψ(O ′ 1 ). As before we want O ′ 0 to be a round disc. Applying Lemma 3.2 and 3.3 to the point Q ′ we obtain a sequence z ′ k ∈ L ′ such that z ′ k → 0, λ k z ′ k → Q ′ and x ′ k = Ψ(z ′ k ) is a repelling point for f of period k. Fix a point z = z ′ k so that λ k z ∈ O ′ 1 and x = Ψ(z), the corresponding periodic point, does not belong to the forward orbit of singular values. Note that there are infinitely many points z satisfying these conditions and we will show that for any such point, is an unstable manifold for p and for x. Note that it is connected, smooth and has no intersections (Ψ| O ′ 0 is univalent). Since γ ′ is an unstable manifold for x, there exists a nested sequence of curves Take i so large that the curve γ ′ i, * is short enough and there exists a curveL ′ i ∋ 0 which is mapped byΨ differomorphically onto γ ′ i, * . Note that is a diffeomorphism onto γ ′ and thereforeΨ restricted to the curveL ′ = µ iL′ i is also a diffeomorphism andΨ(L ′ ) = γ ′ . As p ∈ γ ′ , there existsŵ ∈L ′ such thatΨ(ŵ) = p.
Since γ ′ is an invariant manifold for x andΨ is the linearization associated with x, the curveL ′ must be invariant under z → µz. But the only smooth curve through 0 invariant under multiplication by a real number is a line segment, thusL ′ must be contained in a lineM through 0.
Let z ′ = T (z) ∈ O 1 , where z ∈ L ′ was the point chosen from the sequence z ′ k → 0. For all large j we have that w j = λ −jk z ′ ∈ O 0 . Note that Ψ(w j ) → p = Ψ(ŵ) as j → ∞. SinceΨ|L ′ is a diffeomorphism andŵ ∈L ′ , therefore for every j large enough there exists nearŵ unique pointŵ j such thatΨ(ŵ j ) = Ψ(w j ).
Note thatΨ  Denote by T q (l) a line tangent to a curve l at a point q. Since λ jk w j = z ′ and the curves λ jk M j and T (M j ) both pass through z ′ , by (3.7) these curves actually agree near z ′ . In particular Recall thatM ′ j is the line containing [ŵ j , 0],ŵ j →ŵ as j → ∞ andM is the line through 0 andŵ, thereforeM ′ j converges toM. We also havê Sincê Ψ is a diffeomorphism on a neighbourhood of [0,ŵ], we get therefore that M j converges to a line segment in L ′ in C 1 -topology.
We conclude that T z ′ (λ jk M j ) = T w j (M j ) → T 0 (L ′ ) = T z (L ′ ) as j → ∞ (since z ∈ L ′ ) and also T z ′ (T (M j )) → T z ′ (T (L ′ )) . This and (3.8) imply that hence T ′ (z) ∈ R as we claimed. Since it holds for a sequence of points z = z ′ k ∈ L ′ , z ′ k → 0, we obtain that T ′ (z) ∈ R for every z ∈ L ′ .
Since there are infinitely many such lines passing through 0 on which T ′ is real, we conclude that T ′ must be constant. Thus T is an affine map and the identity Ψ • T = Ψ holds on the whole C. But, as we mentioned in the proof of Lemma 3.3, it may happen only if it is of the form T (z) = ±z + Q.
Recall that, by Lemma 3.1, Ψ −1 (p) is a regular fibre. As a consequence of Lemma 3.4, for every pair Q 1 , Q 2 ∈ Ψ −1 (p), there exists an affine map T (z) = ±z + c such that Ψ • T = Ψ on C and T (Q 1 ) = Q 2 . It follows from (2') of Definition 2.2 that Ψ is automorphic with respect to a discrete group of isometries Γ ⊂ {T : T (z) = ±z + c}.
This allows us to define a measurable line field almost everywhere on C by the formula for any z ∈ ψ −1 (w).
Note that, since Ψ is automorphic with respect to Γ, the line field u is well defined almost everywhere: if z 1 , z 2 ∈ ψ −1 (w), then ψ ′ (z 1 ) = ±ψ ′ (z 2 ). Moreover, since λ ∈ R, (3.1) immediately implies that where z ∈ Ψ −1 (w), hence u is f -invariant. Thus, we have defined a measurable f -invariant line field u on C which is univalent in a neighbourhood of the Julia set J(f ) (cf.