Dynamics of transcendental Hénon maps

Mathematische Annalen, Jan 2018

The dynamics of transcendental functions in the complex plane has received a significant amount of attention. In particular much is known about the description of Fatou components. Besides the types of periodic Fatou components that can occur for polynomials, there also exist so-called Baker domains, periodic components where all orbits converge to infinity, as well as wandering domains. In trying to find analogues of these one dimensional results, it is not clear which higher dimensional transcendental maps to consider. In this paper we find inspiration from the extensive work on the dynamics of complex Hénon maps. We introduce the family of transcendental Hénon maps, and study their dynamics, emphasizing the description of Fatou components. We prove that the classification of the recurrent invariant Fatou components is similar to that of polynomial Hénon maps, and we give examples of Baker domains and wandering domains.

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Dynamics of transcendental Hénon maps

Dynamics of transcendental Hénon maps Leandro Arosio 0 1 2 Anna Miriam Benini 0 1 2 John Erik Fornaess 0 1 2 Han Peters 0 1 2 Leandro Arosio arosio@mat.uniroma 0 1 2 0 Korteweg de Vries Institute for Mathematics, University of Amsterdam , Amsterdam , The Netherlands 1 Department of Mathematical Sciences, NTNU , Trondheim , Norway 2 Dipartimento Di Matematica, Università di Roma “Tor Vergata” , Rome , Italy The dynamics of transcendental functions in the complex plane has received a significant amount of attention. In particular much is known about the description of Fatou components. Besides the types of periodic Fatou components that can occur for polynomials, there also exist so-called Baker domains, periodic components where all orbits converge to infinity, as well as wandering domains. In trying to find analogues of these one dimensional results, it is not clear which higher dimensional transcendental Leandro Arosio and Anna Miriam Benini were supported by the SIR Grant “NEWHOLITE-New methods in holomorphic iteration” No. RBSI14CFME. Han Peters was supported by the NFR Grant No. 10445200. Part of this work was done during the international research program “Several Complex Variables and Complex Dynamics” at the Center for Advanced Study at the Academy of Science and Letters in Oslo during the academic year 2016/2017. - maps to consider. In this paper we find inspiration from the extensive work on the dynamics of complex Hénon maps. We introduce the family of transcendental Hénon maps, and study their dynamics, emphasizing the description of Fatou components. We prove that the classification of the recurrent invariant Fatou components is similar to that of polynomial Hénon maps, and we give examples of Baker domains and wandering domains. Mathematics Subject Classification 32H50 · 37F50 · 37F10 Contents 1 Introduction Our goal is to combine ideas from two separate areas of holomorphic dynamics: the study of transcendental dynamics on the complex plane, and the study of polynomial Hénon maps in C2. Recall that a polynomial Hénon map is a map of the form F : (z, w) → ( f (z) − δw, z), where f is a polynomial of degree at least 2, and δ is a non-zero constant. Here we consider maps of the same form, but where f is a transcendental entire function. We call such F a transcendental Hénon map, and it is easy to see that F is a holomorphic automorphism of C2 with constant Jacobian determinant δ. Special cases of transcendental Hénon maps, namely transcendental perturbations of polynomial Hénon maps, have been first considered in [13]. The main reason for considering transcendental Hénon maps and not arbitrary entire maps in C2 is that the space of entire maps is too large. Even the class of polynomials maps in two complex variables is often considered too diverse to study the dynamics of these maps all at the same time. On the other side, the family of polynomial automorphisms of C2 has received a large amount of attention. It portrays a wide variety of dynamical behavior, yet it turns out that this class of maps is homogeneous enough to describe its dynamical behavior in detail. A result of Friedland and Milnor [22] implies that any polynomial automorphism with non-trivial dynamical behavior is conjugate to a finite composition of polynomial Hénon maps. It turns out that finite compositions of polynomial Hénon maps behave in many regards similarly to single Hénon maps, and the family of Hénon maps is sufficiently rigid to allow a thorough study of its dynamical behavior. Very little is known about the dynamics of holomorphic automorphisms of C2, although there have been results showing holomorphic automorphisms of C2 with interesting dynamical behavior, such as the construction of oscillating wandering domains by Sibony and the third named author [20], and a result of Vivas, Wold and the last author [28] showing that a generic volume preserving automorphisms of C2 has a hyperbolic fixed point with a stable manifold which is dense in C2. Transcendental Hénon maps seems to be a natural class of holomorphic automorphisms of C2 with non-trivial dynamics, restrictive enough to allow for a clear description of its dynamics, but large enough to display interesting dynamical behaviour which does not appear in the polynomial Hénon case. We classify in Sect. 4 the invariant recurrent components of the Fatou set of a transcendental Hénon map, that is, components which admits an orbit accumulating to an interior point. Invariant recurrent components have been described for polynomial Hénon maps in [5]; our classification holds not only for transcendental Hénon maps but also for the larger class of holomorphic automorphisms with constant Jacobian. Moreover, using the fact that f is a transcendental holomorphic function, we obtain in Sect. 3 results about periodic points and invariant algebraic curves. We show that the set Fix(F 2) is discrete, and (if δ = −1) that F admits infinitely many saddle points of period 1 or 2, which implies that the Julia set is not empty. We also show that there is no irreducible invariant algebraic curve (the same was proved by Bedford–Smillie for polynomial Hénon maps in [4]). The dynamical behavior can be restricted even further by considering transcendental Hénon maps whose map f has a given order of growth. For example, if the order of growth is smaller than 21 , then Fix(F k ) is discrete for all k ≥ 1. We then give examples of Baker domains, escaping wandering domains, and oscillating wandering domains. Such Fatou components appear in transcendental dynamics in C, and for trivial reasons they cannot occur for polynomials. The existence of the filtration gives a similar obstruction for polynomial Hénon maps, but this filtration is lost when considering transcendental Hénon maps. For a transcendental function a Baker domain is a periodic Fatou component on which the orbits converge locally uniformly to the point ∞, which is an essential singularity [7]. We give an example in Sect. 5 of a transcendental Hénon map with a two-dimensional analogue: a Fatou component on which the orbits converge to a point at the line at infinity ∞, which is (in an appropriate sense) an essential singularity. In one complex variable for any Baker domain there exists an absorbing domain, equivalent to a half plane H, on which the dynamics is conjugate to an affine function, and the conjugacy extends as a semi-conjugacy to the entire Baker domain. In our example the domain is equivalent to H × C, and the dynamics is conjugate to an affine map. The final part of the paper is devoted to wandering domains. Recall that wandering domains are known not to exist for one-dimensional polynomials and rational maps [30], but they do arise for transcendental maps (see for example [7]). In higher dimensions it is known that wandering domains can occur for holomorphic automorphisms of C2 [20] and for polynomial maps [1], but whether polynomial Hénon maps can have wandering domains remains an open question. We will consider two types of wandering Fatou components, each with known analogues in the one-dimensional setting. We construct in Sect. 6 a wandering domain, biholomorphic to C2, which is escaping: all orbits converge to the point [1 : 1 : 0] at infinity. The construction is again very similar to that in one dimension. However, the proof that the domain and its forward images are actually different Fatou components is not the proof usually given in one dimension. Instead of finding explicit sets separating one component from another, we give an argument that uses exponential expansion near the boundary of each of the domains. Finally, we construct in Sect. 7 a transcendental Hénon map F with a wandering domain , biholomorphic to C2, which is oscillating, that is it contains points whose orbits have both bounded subsequences and subsequences which converge to infinity. Up to a linear change of variable, the map F is the limit as k → ∞ of automorphisms of C2 of the form Fk (z, w) := ( fk (z) + 21 w, 21 z), all having a hyperbolic fixed point at the origin. The family (Fk ) is constructed inductively using Runge approximation in one variable to obtain an entire function fk+1 which is sufficiently close to fk on larger and larger disks, in such a way that the orbit of an open set U0 ⊂ C2 approaches the origin coming in along the stable manifold of Fk and then goes outwards along the unstable manifold of Fk , over and over for all k ∈ N. Regarding the complex structure of those Fatou components, in both the Baker domain and the oscillating wandering domain case one encounters the same difficulty. Namely, in both cases one finds a suitable invariant domain A of the Fatou component on which it is possible to construct, using the dynamics of F , a biholomorphism to a model space (H × C and C2 respectively, where H denotes the right half-plane). One then needs to prove that the domain A is in fact the whole Fatou component, and this is done by using the following plurisubharmonic method: If A is strictly smaller than then we can construct a plurisubharmonic function u : → R ∪ {−∞} for which the submean value property is violated at points in ∂ A ∩ . We note that a somewhat similar argument was given by the third author in [18], and we believe that this method can be applied in a variety of similar circumstances. It is important to point out that for an entire map F : C2 → C2 there are two natural definitions of the Fatou set, which correspond to compactifying C2 either with the onepoint compactification C2, or with P2. In one dimension the two Fatou sets coincide, and the same is true for polynomial Hénon maps, since by the existence of the filtration all forward orbits that converge to infinity converge to the same point on the line at infinity ∞ = P2 \ C2. For a general entire self-map of C2 these two definitions can give two different Fatou sets (see Example 2.6). Notice that, if we compactify with C2, any open subset of C2 on which the sequence of iterates F n diverges uniformly on compact subsets would be in the Fatou set regardless of how the orbits go to infinity. This seems to be too weak a definition in two complex variables. We thus define the Fatou set compactifying C2 with P2 (which has the additional advantage of being a complex manifold). Section 2 is devoted to this argument. 2 The definition of the Fatou set Let n ∈ N, n ≥ 1 and let X be a complex manifold. There are (at least) two natural definitions of what it means for a family F ⊂ Hol(X, Cn ) to be normal. We denote by Cn the one-point compactification of Cn, and with the symbol ∞ we denote both the point at infinity and the constant map z → ∞. Definition 2.1 A family F ⊂ Hol(X, Cn ) is Pn-normal if for every sequence ( fn) ∈ F there exists a subsequence ( fnk ) converging uniformly on compact subsets to f ∈ Hol(X, Pn ). In other words, F is relatively compact in Hol(X, Pn ). A family F ⊂ Hol(X, Cn ) is Cn-normal if for every sequence ( fn ) ∈ F which is not divergent on compact subsets there exists a subsequence ( fnk ) converging uniformly on compact subsets to f ∈ Hol(X, Cn ). This is equivalent to F being relatively compact in Hol(X, Cn ) ∪ ∞ ⊂ C 0(X, Cn ). Remark 2.2 When n = 1 the two definitions are equivalent. A family F ⊂ Hol(X, Cn ) is Pn-normal if and only if it is equicontinuous with respect to the Fubini-Study distance on Pn. This follows from the Ascoli–Arzelà theorem and from the fact that Hol(X, Pn ) is closed in C 0(X, Pn ). One may think that, similarly, a family F ⊂ Hol(X, Cn ) is Cn-normal if and only if it is equicontinuous with respect to the spherical distance dCn on Cn, but this is not the case, as the following example shows. Example 2.3 For n ≥ 2, the family Hol(D, Cn) ∪ ∞ is not closed in C 0(D, Cn). As a consequence, for a family F ⊂ Hol(D, Cn), being relatively compact in C 0(D, Cn) is not equivalent to being Cn-normal. Proof Let n = 2. Let sn ≥ 0 be an increasing sequence of real numbers converging to 21 . Let αn ≤ π2 be a decreasing sequence converging to 0. We define the Pac-man Pn := D \ {sn + ρeiθ : ρ > 0, |θ | < αn}. Let rn ≥ 0 be an increasing sequence converging to 21 . Let βn be a sequence decreasing to 0. We define the bait Rn := D ∩ {z ∈ C : |Im z| ≤ βn, Re z ≥ rn}. Clearly n∈N Rn = 1 z ∈ C : |Im z| = 0, 2 ≤ Re z ≤ 1 , which we call the slit S. We can choose the sequences (sn), (αn), (rn), (βn) in such a way that Pn ∩ Rn = ∅ D \ Pn+1 ⊂ Rn. n−1 i=0 Notice that this implies that sn ≤ rn ≤ sn+1 for all n ∈ N. See Fig. 1 for an illustration of a single Pac-man Pn+1 and two baits Rn+1 and Rn. Let bi > 0 be a sequence such that for all n ∈ N, x ∈ C2 the following implication holds By using Runge approximation we can define a sequence of holomorphic functions fn : D(0, 1 + ) → C such that | fn| ≤ 2n1+1 on P2n, and Re fn ≥ bn on R2n, and a sequence of holomorphic functions gn : D(0, 1 + ) → C such that |gn| ≤ 2n1+1 on P2n+1, and Re gn ≥n bn on R2n+1. For all n ≥ 0, let Hn : D(0, 1 + ) → C2 be defined as Hn(z) = j=0( fn(z), gn(z)). Denote an := max dC2 (Hn(z), Hn−1(z)). z∈D We claim that an ≤ 21n , and thus the sequence (Hn) converges uniformly on the disk Fix n ∈ N. If z ∈ P2n, then | fn(Dz)|→≤ C2n2+1suacnhdth|gant(Hz)(|D≤\ 2Sn1)+1⊂. TCh2uasnd( fHn((zS)), =gn({z∞))}. D to a continuous mapping H : 1 ≤ 21n , which implies that dC2 (Hn(z), Hn−1(z)) ≤ 2n . If z ∈ D \ (P2n ∪ R2n), then we 1 have that Re in=−01 gi (z) ≥ in=−01 bi and |gn(z)| ≤ 2n1+1 , and thus by (1) both Hn(z) and Hn−1(z) belong to the ball of radius 2n1+1 centered at ∞. If z ∈ R2n, then we have that Re in=−01 fi (z) ≥ in=−01 bi and Re fn(z) ≥ bn, and thus by (1) both Hn(z) and Hn−1(z) belong to the ball of radius 2n1+1 centered at ∞. Lemma 2.4 If a family F ⊂ Hol(X, Cn ) is Pn-normal, then it is Cn-normal. Proof Let ( fn) be a sequence in F . Since F is Pn-normal there exists a subsequence ( fnk ) converging uniformly on compact subsets to a map f ∈ Hol(X, Pn ). If there is a point x ∈ X such that f (x ) ∈ ∞, then f (X ) ⊂ ∞. Indeed, it suffices to show that f −1( ∞) is open, and this follows taking an affine chart around f (y) ∈ ∞ in such a way that ∞ = {z1 = 0} and applying Hurwitz theorem to the sequence π1 ◦ fn. Thus, if the sequence ( fn) is not diverging on compact subsets, the subsequence ( fnk ) converges uniformly on compact subsets to a map f ∈ Hol(X, Cn). As a consequence of the previous discussion, for an entire map F : Cn → Cn we have two possible definitions of the Fatou set. Definition 2.5 A point z ∈ Cn belongs to the Cn-Fatou set if the family of iterates (F n) is Cn-normal near z. A point z ∈ Cn belongs to the Pn-Fatou set if the family of iterates (F n) is Pn-normal near z. For a polynomial Hénon map, it follows from the existence of the invariant filtration that any forward orbit that converges to infinity must converge to the point [1 : 0 : 0] ∈ ∞. Thus, the two definitions of Fatou set coincide. By Lemma 2.4 the Pn-Fatou set is always contained in the Cn-Fatou set. If n > 1 the inclusion may be strict as the following example shows. Example 2.6 Given an increasing sequence N j ∈ N, consider the sequence of polynomials f j (z) = (z − 5( j − 1))N j , defined respectively on the disks D j = D(5( j −1), 2), where j ≥ 1. Given a sequence j 0, by Runge approximation (see e.g. Lemma 7.4) we can find an entire function f that is j -close to f j on D j for all j . Define the map F ∈ Aut(C2) by F (z, w) = (z + 5, w + f (z)). It follows immediately from the first coordinate that the forward orbit of any point (z0, w0) converges to infinity, i.e. F n(z0, w0) → ∞, hence the C2-Fatou set equals all of C2. Moreover, if |z0| < 1 then F n(z0, w0) → [1 : 0 : 0], uniformly on compact subsets. Thus, the domain D × C is contained in a P2-Fatou component. On the other hand, if the sequence N j increases sufficiently fast, then for 1 < |z0| ≤ 2 we have that F n(z0, w0) 0 : 1 : 0] ∈ ∞, again uniformly on compact subsets. It follows that D × C is →aP2[-Fatou component. Therefore in this example the single C2-Fatou component contains infinitely many distinct P2-Fatou components. In what follows, we will only consider P2-normality. We will call the P2-Fatou set simply the Fatou set. The Julia set is the complement of the Fatou set. 3 Invariant subsets 3.1 Periodic points If g is a transcendental function or a polynomial Hénon map, then, for each k ≥ 1, the set Fix(gk ) is discrete. Clearly this statement is not satisfied for holomorphic automorphisms of C2. For example, one can consider any holomorphic conjugate of a rational rotation. Consider a periodic orbit (z0, w0) → (z1, w1) → · · · → (zk , wk ) = (z0, w0) Since w j+1 = z j for each j , the first coordinate function of the Hénon map gives the following relations ⎧ ⎪⎪⎪⎪ f (z0) = z1 + δzk−1 ⎨ f (z1) = z2 + δz0 ⎪⎪⎪⎩⎪ ... (2) f (zk−1) = z0 + δzk−2. Lemma 3.1 If F is a transcendental Hénon map, then Fix(F ) and Fix(F 2) are discrete. Proof The fixed points (z, w) of F satisfy z = w and thus z = f (z) − δz. Since f is not linear the set of solutions is discrete. When k = 2 the system (2) gives f (z0) = (1 + δ)z1 f (z1) = (1 + δ)z0. ⎩ = z0 1+δ f (z0) . z1 = 1+δ When δ = −1 it is immediate that the set of solutions is discrete. When δ = −1 the solutions satisfy ⎧ ⎨ f f1(+z0δ) and again one observes a discrete set of solutions. Without making further assumptions it is not clear to the authors that Fix(F k ) is discrete when k ≥ 3. However, we can show discreteness when we assume that the function f has small order of growth. Proposition 3.2 Let F be a transcendental Hénon maps such that f has order of growth strictly less than 21 . Then Fix(F k ) is discrete for all k ≥ 1. Proof Consider the entire function g(z) := f (z) − f (0) z . Write m(r ) := |zin|=fr |g(z)|. Since f is assumed to have order of growth strictly less than 21 , so does g, and Wiman’s Theorem [32] implies that there exist radii rn → ∞ for which m(rn) → ∞. Suppose for the purpose of a contradiction that the solution set in Ck of the system (2) is not discrete. Then there exists an unbounded connected component V . Let n ∈ N be such that V intersects the polydisk D(0, rn)k . Then V also intersects the boundary ∂ D(0, rn)k , say in a point (z0, . . . , zk−1). By the symmetry of the equations in (2) we may then well assume that |z0| = rn, and of course that |z j | ≤ rn for j = 1, . . . , k − 1. By Wiman’s Theorem we may assume that |g(z0)| is arbitrarily large, and in particular that | f (z0)| > (1 + |δ|)rn. But this contradicts the first equation in (2), completing the proof. We now turn to the question whether transcendental Hénon maps always have periodic points. Definition 3.3 For a self-map h and for all n ≥ 1 we denote by Pern(h) the set of periodic points of h of minimal period n. Recall that if f is an entire transcendental function, then Per2( f ) has infinite cardinality by [6]. Proposition 3.4 If F (z, w) = ( f (z) − δw, z) is a transcendental Hénon map, then Fix(F ) = ∅ unless f (z) − z(δ + 1) = eh(z) for some holomorphic function h(z). The set Per2(F ) has infinite cardinality if δ = −1, and if δ = −1 the set Per2(F ) is not empty if and only if the set { f = 0} contains at least two points. Proof Let Z := { f − z(δ + 1) = 0} ⊂ C. Then Fix(F ) = {(z, z) ∈ C2 : z ∈ Z }. f (z) . By (4), Per2(F ) coincides with If δ = −1, let g(z) := 1+δ {(z0, g(z0)) ∈ C2 : z0 ∈ Per2(g)}, which is non-empty and has infinite cardinality. If δ = −1, then Per2(F ) = {(z, w) ∈ C2 : z, w ∈ Z , z = w}. Remark 3.5 Notice that the set Z has finite cardinality if and only if f (z)− z(δ +1) = p(z)eh(z), where p is a nonzero polynomial p and h is entire function. Thus in all other cases the sets Fix(F ) and Per2(F ) have infinite cardinality. If f : C → C is an entire transcendental function, we have additional information on the multiplier of repelling periodic points of period n ≥ 2. Indeed we have the following theorem [8, Theorem 1.2]: Theorem 3.6 Let g be a transcendental entire function and let n ∈ N, n ≥ 2. Then there is a sequence (zk ) in Pern(g) such that |(gn) (zk )| → ∞ as k → ∞. (5) Corollary 3.7 (Non-empty Julia set) If δ = −1, then there exist infinitely many saddle points in Per2(F ), and thus the Julia set of F is non-empty. Proof We have seen that for all z0 ∈ Per2(g), the point (z0, g(z0)) ∈ Per2(F ). A computation using the explicit form for F gives d(z0,g(z0)) F 2 = f (g(z0)) · f (z0) − δ f (z0) −δ f (g(z0)) −δ Since det d(z0,g(z0)) F 2 = δ2, tr d(z0,g(z0)) F 2 = f (g(z0)) · f (z0) − 2δ, the point (z0, g(z0)) is a periodic saddle point for | f (g(z0)) · f (z0)| sufficiently large (Observe that g (g(z0)) · g (z0), and hence f (g(z0)) · f (z0), can be taken arbitrarily large by Theorem 3.6). 3.2 Invariant algebraic curves It follows from a result of Bedford–Smillie [4] that a polynomial Hénon map does not have any invariant algebraic curve. Indeed, given any algebraic curve, the normalized currents of integration on the push-forwards of this curve converge to the (1, 1) current μ−, whose support does not lie on an algebraic curve. This type of argument is not available for transcendental dynamics. Here we present a different argument. Theorem 3.8 Let F be a holomorphic automorphism of the form F : (z, w) → ( f (z) − δw, z), where f is an entire function, and assume that F leaves invariant an irreducible algebraic curve {H (z, w) = 0}. Then f is affine. As we remarked earlier, the statement is known when f is a polynomial of degree at least 2, so we will assume that f is a transcendental entire function and obtain a contradiction. Let us first rule out the simple case where {H = 0} is given by a graph {z = g(w)} (the following argument actually works for an entire function g). In that case the invariance under F gives f (g(w)) − δw = g ◦ g(w). Writing f (z) = g(z) + h(z) gives h ◦ g(w) = δw, z = g( f (z) − δg(z)), | f (z j )| > |z j | j . which implies that g and h are invertible and thus affine. But then f is also affine and we are done. For a graph of the form {w = g(z)} we obtain the functional equation which again implies that the function g is affine, and then so is f . For the general case {H = 0}, where we may now assume that we are not dealing with a graph, we will use the following two elementary estimates. Lemma 3.9 There exist (z j , w j ) ∈ {H = 0}, with |z j | → ∞, for which Proof As we have already shown that {H = 0} is not a graph, it follows that {H = 0} intersects all but finitely many lines {z = c}. The result follows from the assumption that f is transcendental. We use two forms for the polynomial H : (1) H (z, w) = p(w)z N1 + (2) H (z, w) = q0(z) + N1−1 n=1 kq=0(z)w . N=20 αk, zk w . Note that q0 cannot vanish identically, because otherwise w is a factor of H and the zero set is not irreducible. Lemma 3.10 There exist d large enough so that if H (z, w) = 0 for |z| sufficiently large, then |w| < |z|d . Proof If |w| > |z|d for arbitrarily large |z| and d, then |w|n|qn(z)| dominates the other terms in the form (2), so H (z, w) cannot vanish. Proof of Theorem 3.8 By Lemma 3.9 there exist (z j , w j ) with z j → ∞, H (z j , w j ) = 0 and | f (z j )| > |z j | j . Let (z j , w j ) = F (z j , w j ) so that z j = f (z j ) − δw j and w j = z j . Since {H = 0} is invariant we have that H (z j , w j ) = 0. By Lemma 3.10 there exists d ∈ N such that |w j | < |z j |d for j sufficiently large. Hence for j sufficiently large |z j | = | f (z j ) − δw j | ≥ | f (z j )| − |δw j | ≥ | f (z j )|/2. (6) It follows that p w j z j N1 ≥ c z j N1 ≥ c z j N1−1 |z j | j , · 2 p(w j )(z j )N1 , which contradicts H (z j , w j ) = 0. where c > 0 is a constant. But since z j = w j it follows that for large enough j , all terms of the form αk, (z j )k (w j ) for k ≤ N1 − 1 will be negligible compared to 4 Classification of recurrent components In this section we only assume that F is a holomorphic automorphism of C2 with constant Jacobian δ. Definition 4.1 A point x ∈ C2 is recurrent if its orbit (F n(x )) accumulates at x itself. A periodic Fatou component is called recurrent if there exists a point z ∈ whose orbit (F n(z)) accumulates at a point w ∈ . Since the class of holomorphic automorphism of C2 with constant Jacobian is closed under composition, by replacing F with an iterate we can restrict to the case where is invariant. For an invariant Fatou component , a limit map h is a holomorphic function h : → P2 such that f nk → h uniformly on compact sets of for some subsequence nk → ∞. Theorem 4.2 Let F be a holomorphic automorphism of C2 with constant Jacobian δ and let be an invariant recurrent Fatou component for F . Then there exists a holomorphic retraction ρ from to a closed complex submanifold ⊂ , called the limit manifold, such that for all limit maps h there exists an automorphism η of such that h = η ◦ ρ. Every orbit converges to , and F | : → is an automorphism. Moreover, • If dim = 0, then is the basin of an attracting fixed point, and is biholomorphically equivalent to C2. • If dim = 1, either is biholomorphic to a circular domain A, and there exists a biholomorphism from to A × C which conjugates the map F to (z, w) → (z, δ j w). where θ is irrational, or there exists j ∈ N such that F j | = id , and there exists a biholomorphism from to × C which conjugates the map F j to • dim = 2 if and only if |δ| = 1. In this case there exists a sequence of iterates converging to the identity on . By a circular domain we mean either the disk, the punctured disk, an annulus, the complex plane or the punctured plane. For the polynomial Hénon maps case, see [5] and [19]. Let (F nk ) be a convergent subsequence of iterates on , with F nk (z) → w ∈ . We denote the limit of (F nk ) by g. Lemma 4.3 The image g( ) is contained in C2. Proof If there is a point x ∈ for which g(x ) belongs to the line at infinity ∞, then g( ) ⊂ ∞ (see e.g. the proof of Lemma 2.4), which gives a contradiction. Definition 4.4 We define the maximal rank of g as max p∈ rk(d p g). 4.1 Maximal rank 0 Lemma 4.5 Suppose that g has maximal rank 0. Then g( ) is the single point w, which is an attracting fixed point. Proof Since the maximal rank is 0, the map g is constant and must therefore equal w. Since F and g commute, the point w must be fixed. Suppose that the differential dw F has an eigenvalue of absolute value ≥ 1. Then the same is true for all iterates F nk . Hence they cannot converge to a constant map. So w must be an attracting fixed point. It follows that is the attracting basin of the point w, and the entire sequence F n converges to g. In this case the limit manifold is the point {w}. 4.2 Maximal rank 2 Theorem 4.6 Suppose that g has maximal rank 2. Then there exists a subsequence (mk ) so that F mk → Id on . Proof Let x be a point of maximal rank 2. There exist an open neighborhood U of x and an open neighborhood V of g(x ) such that g : U → V is a biholomorphism. Denote h := g−1 defined on V . Let V ⊂⊂ V be an open neighborhood of g(x ). Since F nk → g on U , we have that V ⊂ F nk (U ) for large k and the maps (F nk )−1 converge to h uniformly on compact subsets of V . In particular V ⊂ . Replace nk by a subsequence so that nk+1 −nk ∞. We can then write F nk+1−nk = F nk+1 ◦(F nk )−1 on V . If we set mk := nk+1 − nk , then F mk → Id on V . Since we are in the Fatou set this implies that F mk → Id on . It follows that every point p ∈ is recurrent and that F is volume preserving. The following fact is trivial but we recall it for convenience. Lemma 4.7 Let (Gn : ⊂ C2 → C2) be a sequence of injective holomorphic mappings which are volume preserving. If Gn converges to G uniformly on compact subsets, then G is holomorphic, injective and volume preserving. Proof The map G is holomorphic and d Gn n−→→∞ d G, and thus G is volume preserving. Thus by Hurwitz Theorem G is injective. Corollary 4.8 Suppose that g has maximal rank 2. Then if h is any limit map of F on , then either h is injective and h( ) ⊂ or h( ) ⊂ ∞. Proof Assume that h( ) ⊂ C2. Then by Lemma 4.7 the map h is holomorphic and injective. Arguing as in the proof of Theorem 4.6 we get that h( ) ⊂ . Proposition 4.9 If g has maximal rank 2 then each orbit (F n(z)) is contained in a compact subset of . Proof Let (K j ) be an exhaustion of by compact subsets such that K j ⊂K◦ j+1 for all j ∈ N. By passing to a subsequence of the exhaustion if needed, we may assume that F (K j ) ⊂⊂ K j+1. Let p ∈ . We can assume that p ∈ K1, and let r > 0 be such that B( p, r ) ⊂ K1. We may assume that if F n( p) ∈ K j then F n(B( p, r )) ⊂ K j+1. Indeed, suppose by contradiction that there exist j ∈ N and subsequence k such that F k ( p) ∈ K j for all k ∈ N, but F k (B( p, r )) ⊂ K j+k . (7) Then up to passing to a subsequence, the sequence F k converges uniformly on compact subsets of to a map h which satisfies h( ) ⊂ . Hence we have h(B( p, r )) ⊂⊂ , which contradicts (7). Similarly we may assume that if F n( p) ∈/ K j then F n(B( p, r )) ∩ K j−1 = ∅. Suppose by contradiction that the orbit of p is not contained in a compact subset of . Then the orbit of p is not contained in any K j . But since p is a recurrent point, the orbit of p must also return to K1 infinitely often. Thus, there exists a sequence k1 < l1 < m1 < k2 < l2 < m2 < · · · and a strictly increasing sequence (n j ), n j ≥ 3, such that (i) Each F k j ( p) lies in K4 \ K3 (ii) For k j < n < l j the points F n( p) lie outside of K3. (iii) Each Fl j ( p) lies outside Kn j , (iv) Each point F m j ( p) lies in K1, (v) For k j < n < m j the points F n( p) lie in Kn j+1−2. We claim that the sets F k j (B( p, r )) must be pairwise disjoint. To see this, suppose that F ki (B( p, r )) ∩ F k j (B( p, r )) = ∅ for some i < j . Then clearly F ki +n (B( p, r )) ∩ F k j +n(B( p, r )) = ∅, ∀n ∈ N. If l j − k j > mi − ki , then a contradiction is obtained since F k j +mi −ki (B( p, r )) ∩ K2 = ∅ due to (ii) while F mi (B( p, r )) ⊂ K2 due to (iv). If l j − k j < mi − ki , then a contradiction is obtained since Fl j (B( p, r )) ∩ Kn j −1 = ∅ due to (iii), while F ki +l j −k j (B( p, r )) ⊂ Kni+1−1 ⊂ Kn j −1 due to (v). Finally, if l j − k j = mi − ki , then F mi (B( p, r )) ∩ F j (B( p, r )) = ∅, which contradicts (iii) and (iv). This proves the claim. Since F is volume preserving and the volume of K4 \ K3 is finite, we have a contradiction. Corollary 4.10 The limit of any convergent subsequence (F nk ) is an automorphism of . Proof By Corollary 4.8 and Proposition 4.9 the limit h is a biholomorphic map h : → h( ) ⊂ . Suppose that p ∈ \ h( ). Let K be a compact subset of . Then for all large enough k, p ∈ \ F nk (K ). Hence F −nk ( p) ∈ \ K . Fix such k. By Theorem 4.6 there exists m > nk so that F m is close enough to the identity so that F m−nk ( p) ∈ \ K . This contradicts Proposition 4.9. In this case the limit manifold is the whole . Remark 4.11 It follows that the maximal rank of a limit map is independent of the chosen convergent subsequence. 4.3 Maximal rank 1 We now consider the case where the limit map g has maximal rank 1. By Remark 4.11 every other limit of a convergent subsequence on must also have maximal rank 1. Recall that (F nk ) is a convergent subsequence of iterates on such that F nk (z) → w ∈ . Replace nk by a subsequence so that nk+1 − nk ∞. Let (mk ) be a subsequence of (nk+1 − nk ) so that F mk converges, uniformly on compact subsets of . From now on we assume that g is the limit of the sequence (mk ). Remark 4.12 Notice that g(w) = w. Actually, if follows by the construction that there exists an open neighborhood N of z in such that g(N ) ⊂ and for all y ∈ g(N ), g(y) = y. Lemma 4.13 The map F is strictly volume decreasing. Proof By assumption the Jacobian determinant δ of F is constant. Since is recurrent we have |δ| ≤ 1. Since (F nk ) converges to the map g : → C2 of maximal rank 1, it follows that |δ| < 1. We write := g( ). Notice that is a subset of ∩ C2, and that, since F and g commute, the map F | : → is bijective. We need a Lemma. Lemma 4.14 Let U be an open set in C2, and let h : U → C2 be a holomorphic map of maximal rank 1. Then for all w ∈ h(U ), the fiber h−1(w) has no isolated points. Proof Assume by contradiction that q ∈ h−1(w) is isolated. If is small enough, then h(∂ B(q, )) is disjoint from w. Hence there exists a small ball B(w, δ) which is disjoint from h(∂ B(q, )), and if we restrict h to V := h−1(B(w, δ)) ∩ B(q, ), then h : V → B(w, δ) is a proper holomorphic map. Let ζ ∈ V be such that rkζ h = 1. Then the level set h−1(h(ζ )) contains a closed analytic curve in V . Such curve is not relatively compact in V , and this contradicts properness. It is not clear a priori that is a complex submanifold, but we will show, following a classical normalization procedure, that there exists a smooth Riemann surface such that the self-map F on can be lifted to a holomorphic automorphism F on . Note that such normalization procedure was used in a similar context in [31]. Lemma 4.15 For each point z ∈ there is an open connected neighborhood U (z) ⊂ , an affine disk z ⊂ through z and an injective holomorphic mapping γz : z → C2 such that (1) γz ( z ) is an irreducible local complex analytic curve which is smooth except possibly at γz (0) where it could have a cusp singularity, (2) γz ( z ) = g(U (z)). Moreover, if g has rank 1 at z, then γz ( z ) is smooth and γz = g| z . Proof If g has rank 1 at z, the result follows immediately from the constant rank Theorem. So suppose that g has rank 0 at z, and let z ⊂ be an affine disk through z on which g is not constant. By the Puiseux expansion of g : z → C2, it follows that, up to taking a smaller z , g( z) is an irreducible local complex analytic curve (with possibly an isolated cusp singularity at g(z)). Hence g( z ) is the zero set of a holomorphic function G defined in a open neighborhood V of g(z). Let U (z) be the connected component of g−1(V ) containing z. We claim that G ◦g vanishes identically on U (z), which implies that g(U (z)) = g( z ). If not, then (G ◦ g)−1(0) is a closed complex analytic curve in U (z) containing z . Pick a point q ∈ z where locally (G ◦ g)−1(0) = z . Then g−1(g(q)) is isolated at q since g is not constant on z , which gives a contradiction by Lemma 4.14. Finally, again by the Puiseux expansion of g : z → C2, there exists a holomorphic injective map γz : z → C2 such that γz ( z ) = g( z ). Remark 4.16 For all z ∈ , there exists a unique surjective holomorphic map hz : U (z) → z such that g = γz ◦ hz on the neighborhood U (z). If g has rank 1 at z, then hz | z = id. Consider the disjoint union z∈ z , and define an equivalence relation in the following way: (x , z) (y, w) if and only if γz (x ) = γw(y) and the images coincide locally near this point. Define as z∈ z / , endowed with the quotient topology, and denote π : z∈ z → the projection to the quotient. It is easy to see that the map π is open. For all z ∈ , define a homeomorphism πz : z → as πz (x ) := [(x , z)]. Definition 4.17 We define a continuous map γ : → C2 such that γ ( ) = in the following way: γ ([(x , z)]) = γz (x ). Notice that this is well defined. The map g : → C2 can be lifted to a unique surjective continuous map g : → such that g = γ ◦ g. Such map is defined on U (z) as g := πz ◦ hz . Notice that if g has rank 1 at z then g| z = πz . Lemma 4.18 The topological space is connected, second countable and Hausdorff. Proof Since = gˆ( ), and is connected, it follows that is connected. Since g is open, it follows also that is second countable. Let [(x , z)] = [(y, w)] ∈ . Then we have two cases. Either γz (x ) = γw(y), or γz (x ) = γw(y) but the images do not coincide locally near this point. In both cases there exist a neighborhood U ⊂ z of x and a neighborhood V ⊂ w of y such that π (U ) ∩ π (V ) = ∅. We claim that the collection of charts (πz )z∈ gives the structure of a smooth Riemann surface. Let z, w ∈ such that πz ( z ) ∩ πw( w) = ∅. Then consider the map πw−1 ◦ πz : πz−1(πz ( z ) ∩ πw( w)) → πw−1(πz ( z ) ∩ πw( w)). Let x ∈ z , y ∈ w such that πz (x ) = πw(y). This means that γz (x ) = γw(y) and the images coincide locally near this point. There exists an open neighborhood U ⊂ z of x , an open neighborhood V ⊂ w of y, and a unique biholomorphic function k : U → V such that γw ◦ k = γz . It is easy to see that k = πw−1 ◦ πz on U . Remark 4.19 With the complex structure just defined on holomorphic. , the maps γ and g are Definition 4.20 Define R ⊂ with g(z) = ζ and rkz g = 1. Lemma 4.21 The set \ R is discrete. as the set of points ζ ∈ such that there exists z ∈ Proof Let w ∈ such that rkw g = 0. By the identity principle there exists a neighborhood V of w in w such that rkz (hw| w ) = 1 for all z ∈ V \ {w}. The result follows since g = πw ◦ hw on U (w), and πw : w → πw( w) is a biholomorphism. Lemma 4.22 There exists a unique holomorphic map F : following diagram commutes: → such that the F F g g g g F . γ γ Proof Let ζ ∈ R, and let z ∈ such that g(z) = ζ and rkz g = 1. Define on πz ( z ) the map F := g ◦ F ◦ πz−1. This is well-defined and holomorphic away from the discrete closed set \ R, and can be extended holomorphically to the whole . The inverse of F is given by F −1, therefore F is an automorphism. Lemma 4.23 The Riemann surface (F mk ) converges to the identity. contains an open subset on which the sequence Proof By Remark 4.12 there exists an open neighborhood N of z such that g(N ) ⊂ and for all y ∈ g(N ), F mk (y) → y. Then, for all y ∈ g(N ), F mk (g(y)) = g(F mk (y)) → g(y). The set g(N ) is open. Lemma 4.24 Either there exists a j ∈ N for which F j = Id, or is biholomorphic to a circular domain, and the action of F is conjugate to an irrational rotation. Recall that by a circular domain we mean either the disk, the punctured disk, an annulus, the complex plane or the punctured plane. Proof Assume that F j = Id for all j ≤ 1. Since the holomorphic map γ : → C2 is nonconstant, it follows that the Riemann surface is not compact. Thus if is not a hyperbolic Riemann surface, then it has to be biholomorphic either to C or to C∗. In both cases, since F is an automorphism, it is easy to see that Lemma 4.23 implies that F is conjugate to an irrational rotation. If is hyperbolic, then the family (F n) is normal, and thus Lemma 4.23 implies that the sequence (F mk ) converges to the identity uniformly on compact subsets of . Thus the automorphism group of is non-discrete. Hence (see e.g. [16, p. 294]) the Riemann surface is biholomorphic to a circular domain and the action of F is conjugate to an irrational rotation. Definition 4.25 Define the set I ⊂ and γ (x ) = γ (y). Define the set C ⊂ rank 0 at x . × as the set of pairs (x , y) such that x = y as the set of x such that γ : → C2 has Since the map π is open, it follows immediately that the set I is discrete in and that the set C is discrete in . Our goal is to prove that the set will first consider the case where conjugate to an irrational rotation. is a closed complex submanifold of . We is biholomorphic to a circular domain and F is × Lemma 4.26 If F is conjugate to an irrational rotation then there is at most one element ζ0 ∈ C . The set I is empty, and thus the map γ : → is injective. Proof The set C is invariant by F . Since the action of F is conjugate to an irrational rotation, and since C is discrete it follows that C can only contain the center of rotation ζ0 (if there is one). Similarly, the set I is invariant by the map (x , y) → (F (x ), F (y)), but this contradicts the discreteness of I . Lemma 4.27 The set C ⊂ is empty. Proof If has no center of rotation, then there is nothing to prove. Suppose for the purpose of a contradiction that there is a center of rotation ζ0 ∈ and that ζ0 ∈ C . Then has a cusp at z := γ (ζ0). Notice that F (z) = z. Since F acts on as a rotation, it follows that the tangent direction of to z is an eigenvector of dz F with eigenvalue |λ1| = 1. Since F is strictly volume decreasing, the other eigenvalue λ2 of dz F satisfies 0 < |λ2| < 1. Thus we obtain a forward invariant cone in Tz (C2), centered at the line Tz ( ). Extending this cone to a constant cone field in a neighborhood of z, it follows that we obtain stable manifolds through z and all nearby points in , giving a continuous Riemann surface foliation near the point z, see the reference [23] for background on normal hyperbolicity. Since F acts on as a rotation, the stable manifolds through different points in must be distinct. However, this is not possible in a neighborhood of z. To see this, let h be a locally defined holomorphic function such that equals the zero set of h near z. We may assume that h vanishes to higher order only at z. Now consider the restriction of h to the stable manifold through z. Then h has a multiple zero at z, hence by Rouché’s Theorem, the number of zeroes for nearby stable manifolds is also greater than one. But since nearby stable manifolds are transverse to , and h does not vanish to higher order in nearby points, it follows that the restriction of h to nearby stable manifolds must have multiple single zeroes. Hence these nearby stable manifolds intersect in more than one point, giving a contradiction. Let V ⊂⊂ be open and invariant under the action of F , and let V be its image in C2, which is an embedded complex submanifold of C2. We claim that there exists a continuous function ϕ : V → (0, ∞), bounded from above and from below by compactness, such that ϕ(z) dz (F |V ) = ϕ(F (z)) . ϕ(z) := dz γ −1 Indeed, regardless of whether is a hyperbolic or Euclidean Riemann surface, there exists a conformal metric · on which is invariant under F . The function satisfies (8). Given > 0 and z ∈ V , we define the tangent cone Cz ⊂ Tz (C2) by Cz = w : | w, vz⊥ | ≤ ϕ(z)2| w, vz | , where vz is a unit tangent vector to z ∈ V and vz⊥ is a unit vector orthogonal to vz . Lemma 4.28 One can choose > 0 sufficiently small so that dz F (Cz ) ⊂⊂ CF(z). |w2| ≤ ϕ(z)2|w1|. Proof This is a matter of linear algebra. Without loss of generality we may assume that vz = vF(z) = (1, 0). Thus, the cone field Cz contains all vectors w = (w1, w2) for which (8) The vector dz F (w) is given by dz F (w) = (θ (z)w1 + α(z)w2, β(z)w2) , where |θ (z)| = ϕ(ϕF((zz))) . Since F has constant Jacobian determinant δ it follows that Since |α(z)| is bounded on V , by choosing sufficiently small we can guarantee that for all z ∈ V we have |θ (z)β(z)| = |δ|. from which it follows that ϕ(F (z))2|dz F (w)1| > |dz F (w)2|. Thus, dz F sends the cone Cz strictly into CF(z). Lemma 4.29 The image is contained in . Proof Since F is C 1, we can extend the invariant cone field to a neighborhood N (V ). Let z be a point whose forward orbit remains in N (V ), which holds in particular for all points in V . Then there exists a stable manifold W s (z), transverse to V , and these stable manifolds fill up a neighborhood of V . The forward iterates of F form a normal family on this neighborhood, which implies that V is contained in the Fatou set. Corollary 4.30 The map g : → is a holomorphic retraction. In particular a closed smooth one-dimensional embedded submanifold. is Proof On = g( ) ⊂ , g = kl→im∞ F mk = Id, which proves that g is a holomorphic retraction. In the case where F j equals the identity, it follows that F j equals the identity on . The next lemma, together with the fact that the Jacobian δ of F satisfies |δ| < 1, shows that there cannot be cusps and self-intersections in this case. Lemma 4.31 If a local biholomorphic map G of a neighborhood of the origin in C2 is the identity on a complex analytic curve γ with any kind of singularity at the origin, then d0G is the identity. Proof The curve γ will be the zero set of a holomorphic function φ which vanishes generically to first order on the curve, but vanishes to at least order 2 at the origin. Then we write G(z, w) = (z + A(z, w)φ, w + B(z, w)φ), which gives the result. g : → We immediately get stable manifolds transverse to , which imply as above that is a holomorphic retraction. Lemma 4.32 The retraction g has constant rank 1 on . Proof Since g is a retraction to the 1-dimensional manifold , there exists a neighborhood V of such that rkx g = 1 for all x ∈ V . Let x ∈ and N ≥ 0 be such that y = F N (x ) ∈ V . The result follows from the fact that F N has rank 2 and that g ◦ F N (x ) = F N ◦ g(x ). Remark 4.33 The stable manifolds of the points in fill up a neighborhood of . Since all orbits in the Fatou component get close to , this implies that the stable manifolds of the points in fill up the whole . Thus for any limit map h : → P2 we have that h( ) ⊂ . Moreover, by Lemma 4.24, the restriction h| is an automorphism of , and thus h( ) = . Notice that for all z ∈ the fiber g−1(z) coincides with the stable manifold W s (z). Hence h = h| ◦ g. Remark 4.34 In the view of the contents of this section if a Fatou component recurrent then for every z ∈ the orbit is relatively compact in . is Now we investigate the complex structure of . Notice that , being an open Riemann surface, is Stein. Proposition 4.35 If some iterate F j | is the identity, then there exists a biholomorphism : → × C which conjugates the map F j to (z, w) → (z, δ j w). Proof Let L be the holomorphic line bundle on given by Ker dg restricted to . Since is Stein, there exists a neighborhood U of and an injective holomorphic map h : U → L such that for all z ∈ we have that h(z) = 0z , and h maps the fiber g−1(z) ∩ U biholomorphically into a neighborhood of 0z in the fiber L z [21, Proposition 3.3.2]. We can assume that dz h|Lz = idLz . Notice that the map dz F j |Lz : L z → L z acts as multiplication by δ j . The sequence (d F |−Lnj ◦ h ◦ F nj ) is eventually defined on compact subsets of and converges uniformly on compact subsets to a biholomorphism : → L, conjugating F j to d F j |L : L → L. The line bundle L is holomorphically trivial since is a Stein Riemann surface. If no iterate of F is the identity on , then is a biholomorphic to a circular domain A, and by the same proof as in [5, Proposition 6] there exists a biholomorphism from to A × C which conjugates the map F to 4.4 Transcendental Hénon maps So far in this section we have not used that the maps we study here are of the form F (z, w) = ( f (z) − δw, z). Absent of any further assumptions we do not know how to use this special form to obtain a more precise description of recurrent Fatou components. We do however have the following consequence of Proposition 3.2 and Theorem 4.2. Corollary 4.36 Let F be a transcendental Hénon map, and assume that f has order of growth strictly smaller than 21 . Then any recurrent periodic Fatou component of rank 1 must be the attracting basin of a Riemann surface ⊂ which is biholomorphic to a circular domain, and f acts on as an irrational rotation. 5 Baker domain A Baker domain for a transcendental function f in C is a periodic Fatou component on which the iterates converge to the point ∞, an essential singularity for f . Notice that, in contrast, every polynomial p in C can be extended to a rational function of P1 for which ∞ is a super-attracting fixed point. The first example of a Baker domain was given by Fatou [17], who considered the function f (z) = z + 1 + e−z (9) and showed that the right half-plane H is contained in an invariant Baker domain. For holomorphic automorphisms of C2 the definition of essential singularity at infinity has to be adapted in order to be meaningful. Definition 5.1 Let F be a holomorphic automorphism of C2. We call a point [ p : q : 0] ∈ ∞ an essential singularity at infinity if for all [s : t : 0] ∈ ∞ there exists a sequence (zn, wn) of points in C2 converging to [ p : q : 0] whose images F (zn, wn) converge to [s : t : 0]. Remark 5.2 If F is a transcendental Hénon map, then any point [ p : q : 0] in ∞ is an essential singularity at infinity. To see this, it is in fact sufficient to consider only points on the line {( pζ, qζ )} when p = 0. Define the entire function g(ζ ) = f ( pζ ) − f (0) − δqζ pζ . Since f is transcendental, the function g must also be transcendental. Let (ζn) be a sequence of points in C converging to ∞ for which g(ζn) → st . Then F ( pζn, qζn ) = ( pζn · g(ζn) + f (0), pζn) → [s : t : 0]. Here we consider the iteration of a map on C2 analogous to (9), namely the transcendental Hénon map F (z, w) := (e−z + 2z − w, z), and we show that it admits an invariant Fatou component U on which the iterates tend to the point [1 : 1 : 0] on the line at infinity. Remark 5.3 Notice that by Remark 5.2 the point [1 : 1 : 0] is an essential singularity at infinity for the map F , and this implies that F cannot be extended, even continuously, to the point [1 : 1 : 0]. The situation is radically different for a polynomial Hénon map H , for which the escaping set I∞ := {(z, w) ∈ C2 : H n(z, w) → ∞} is a Fatou component on which all orbits converge to the point [1 : 0 : 0], and the map H extends to a holomorphic self-map H of P2 \ {[0 : 1 : 0]} with a super-attracting fixed point at [1 : 0 : 0]. For a transcendental function f in C it is known that Baker domains are simply connected (proper) domains of C and by results of Cowen [11] (see also [9, Lemma 2.1]) the function f is semi-conjugate to one of the following automorphisms: (1) z → λz ∈ Aut(H), where λ > 1, (2) z → z ± i ∈ Aut(H), (3) z → z + 1 ∈ Aut(C). In our case we show that on the Fatou component U the map F is conjugate to the linear map L ∈ Aut({Re(z − w) > 0}) given by L(z, w) := (2z − w, z). We show that the conjugacy maps U onto the domain {Re(z − w) > 0}, proving that U is biholomorphic to H × C. We begin by constructing an appropriate forward invariant domain R. For each α > 0 we define the domain where Rα := {(z, w) : Re z > Re w + α + ηα(Re w)}, e−α ηα(x ) := 1 − e−α · e−x := Aα · e−x . Notice that ηα > 0, 1+AαAα = e−α, and that the domains Rα are not nested. Lemma 5.4 Each domain Rα is forward invariant. Proof Let (z0, w0) ∈ Rα. We claim that Re z1 > Re w1 + α + ηα(Re w1). Since w1 = z0, Re z1 − Re w1 = Re e−z0 + Re z0 − Re w0 > α + ηα(Re w0) + Re (e−z0 ) ≥ α + ηα(Re w0) − e−Re z0 . Hence the claim follows if we show that ηα(Re w0) ≥ ηα(Re z0) + e−Re z0 . This is the same as Aαe−Re w0 ≥ (1+ Aα)e−Re z0 , or equivalently e−α ≥ e−Re z0+Re w0 . The latter is satisfied because Re z0 − Re w0 ≥ α (since ηα is always positive). We immediately obtain the following. Corollary 5.5 The domain R := Rα α>0 is forward invariant. Remark 5.6 Let (z0, w0) ∈ Rα. Since w1 = z0 we have that Re z1 − Re z0 > α + ηα(Re z0). (10) It easily follows that then Re zn > Re z0 + nα, and that Re wn > Re z0 + (n − 1)α. Lemma 5.7 All the orbits in R converge uniformly on compact subsets to the point [1 : 1 : 0] on the line at infinity. Proof Let K be a compact subset of Rα for some α > 0. Let M > 0. Then by Remark 5.6 we have that |zn| and |wn| converge to ∞ uniformly on K . Moreover, since for all n ≥ 0 we have that zn+1 − wn+1 = zn − wn + e−zn , it follows that zn − wn = z0 − w0 + n−1 j=0 e−z j , and thus wznn converges to 1 uniformly on K . The domain R is therefore contained in an invariant Fatou component U . Our next goal is to show that R is an absorbing domain for U , i.e. U = A := F −n(R). n∈N un(z0, w0) := −Re (zn) . n It is immediate that A is contained in U . In order to show that U is not larger than A, we will for the first time use the plurisubharmonic method referred to in the introduction. Definition 5.8 Define the sequence of pluriharmonic functions (un : U → R)n≥1 by Lemma 5.9 The functions un are uniformly bounded from above on compact subsets of U , and Proof Let K be a compact subset of U . Since U is a Fatou component, on K the orbits converge uniformly to [1 : 1 : 0]. So for every > 0 there exists n ≥ 0 such that lim sup un ≤ 0. n→+∞ zn zn−1 zn wn < 1 + , By taking large, > 0 sufficiently small so that eβ > (1 + ), we obtain, for k sufficiently which contradicts (12). The claim implies that there is a uniform bound from above for the functions un on K , and that |zn(k)+1| > C (1 + )n(k)+1, k lim sup un ≤ 0, n→∞ for all n > n and for all (z0, w0) ∈ K . It follows that for every C = C such that max{|zn|, |wn|} ≤ C · (1 + )n for every n ∈ N and for all (z0, w0) ∈ K . Let β > 0. We claim that there exists an N > 0 so that un < β on K for every n ≥ N . To prove this claim, let us suppose by contradiction that there exist a sequence (z0k, w0)k∈N in K and a strictly increasing k sequence (n(k))k∈N in N such that > 0 there exists for all k ∈ N. Then Re(znk(k)) ≤ −n(k) · β. It follows that un(k) z0k, w0k ≥ β k k k zn(k)+1 = e−zn(k) + 2znk(k) − wn(k) k k ≥ e−zn(k) − 2znk(k) − wn(k) ≥ en(k)·β − 3C (1 + )n(k). which completes the proof. (11) (12) Lemma 5.10 Let H be a compact subset of A. Then there exists γ > 0 such that on H , Proof Let (z0, w0) ∈ Rα for some α > 0. Then by Remark 5.6 we have, lim sup un ≤ −γ . n→+∞ Re z0 n un(z0, w0) < − − α, ∀ n ≥ 1, H ⊂ F −n1 (Rα1 ) ∪ · · · ∪ F −nk (Rαk ). lim sup un ≤ max{−α1, . . . , −αk }. n→+∞ which implies that lim supn→+∞ un(z0, w0) ≤ −α. Let now H be a compact subset in A. Then there exist α1, . . . , αk > 0 and n1, . . . , nk ∈ N such that Lemma 5.11 On U \ A, lim supn→+∞ un = 0. Proof Let (z0, w0) ∈ U \ A. Suppose by contradiction that there exists α > 0 and N ∈ N such that un(z0, w0) ≤ −α for all n ≥ N , that is, Re zn n ≥ α, ∀ n ≥ N . Let n0 ≥ 0 be such that ηα/3(Re zn) < α/3 for all n > n0. For all 0 < β < α there are arbitrarily large n such that Re zn+1 − Re zn ≥ β. Setting β := 2α/3, there exists n ≥ n0 such that Re zn+1 − Re zn > 2α/3 > α/3 + ηα/3(Re zn), which implies that (zn+1, wn+1) ∈ Rα/3. Proposition 5.12 The set R is an absorbing domain. Proof Let us assume, for the purpose of a contradiction, that U = A. Define u(z) := lim sup un(z) n→+∞ and let u be its upper semicontinuous regularization. Then by [25, Prop 2.9.17] the function u is plurisubharmonic. By Lemmas 5.10 and 5.11 the function u is strictly negative on A, and constantly equal to zero on U \ A. But then u contradicts the sub-mean value property at boundary points ζ ∈ ∂ A. Remark 5.13 It is easy to see that for all n ∈ Z, (z, w) ∈ C2, Ln(z, w) = ((n + 1)z − nw, nz − (n − 1)w). Definition 5.14 We denote by is biholomorphic to H × C. the domain {(z, w) ∈ C2 : Re (z − w) > 0}, which Theorem 5.15 There exists a biholomorphism ψ : U → the map L. In particular U is biholomorphic to H × C. which conjugates F to Proof We will construct the map ψ as the uniform limit on compact subsets of U of the maps ψn := L−n ◦ F n : U → C2. Notice that for all n ≥ 1, the mapping ψn is an injective volume-preserving holomorphic mapping. We have δn(z0, w0) := L−n−1(F n+1(z0, w0)) − L−n(F n(z0, w0)) = = L−n L−1 F (zn, wn) − (zn, wn) L−n(0, −e−zn ) = (−ne−zn , (−n − 1)e−zn ) ≤ √2(n + 1)e−Re zn . By Remark 5.6, on Rα √2(n + 1)e−Re zn ≤ √2(n + 1)e−Re z0−nα, Hence for all (z0, w0) ∈ Rα, we have k∞=0 δn(z0, w0) < +∞, and the sequence (ψn)n≥0 converges uniformly on compact subsets of R to an injective volumepreserving (see Lemma 4.7) holomorphic mapping ψ : R → C2, satisfying L ◦ ψ = ψ ◦ F. (13) Since R is an absorbing domain, ψ extends to an injective volume-preserving holomorphic mapping (still denoted by ψ ) defined on the whole Fatou component U and still satisfying (13). We claim that ψ (U ) = . We first show that ψ (U ) ⊂ . Let (z0, w0) be in Rα. Notice that the Euclidean distance d(Rα, ) > 0. We claim that lim n→∞ ψ (zn, wn) − (zn, wn) = 0. j=0 ∞ Indeed, ψ (zn, wn) − (zn, wn) ≤ ψ j+1(zn, wn ) − ψ j (zn, wn ) = δ j (zn, wn) ≤ ( j + 1)e−Re (zn+ j ), and the claim follows since Re zn+ j ≥ Re z0 + (n + j )α. Hence, if n is large enough, ψ (zn, wn ) ∈ . Since is completely invariant under L and U is completely invariant under F , it follows that ψ (U ) ⊂ . What is left is to show that ⊂ ψ (U ). Let (x0, y0) ∈ , and write (xn, yn ) := Ln(x0, y0). By definition of we have that β := Re(x0 − y0) > 0. Note that Re(xn − yn) also equals β, and that Re(xn), Re(yn) → ∞. Let 0 < α < β. Recalling that ηα(x ) → 0 as x → +∞ it follows that (xn, yn ) ∈ Rα for all n ∈ N sufficiently large. In fact, there exists an r > 0 and an N ∈ N such that Rα contains the closed ball B((xn, yn ), r ) for all n ≥ N . We claim that lim n→∞ ψ − Id B((xn,yn),r) = 0. Indeed, for all n ∈ N, ψ − Id B((xn,yn),r) ≤ ψ j+1 − ψ j B((xn,yn),r) δ j B((xn,yn),r) ( j + 1) e−Re (π1◦F j ) B((xn,yn),r). Assume now that n ≥ N . Since B((xn, yn ), r ) ∈ Rα we have that for all (x , y) ∈ B((xn, yn ), r ), e−Re π1(F j (x,y)) ≤ e−Re x− jα ≤ e−Re x0−(n+ j)α+r , where the last inequality follows from the fact that for all n ∈ N and (x , y) ∈ B((xn, yn ), r ) we have Re x ≥ Re x0 + nα − r . This proves the claim. Let n ≥ N be such that ψ − Id ≤ r2 on B((xn, yn ), r ). By Rouché’s Theorem in several complex variables it follows that (xn, yn ) ∈ ψ (B((xn, yn ), r )) ⊂ ψ (U ). Since is completely invariant under L and U is completely invariant under F , it follow that (x0, y0) ∈ ψ (U ). 6 Escaping wandering domain Definition 6.1 Let F be a transcendental Hénon map. A Fatou component wandering domain if it is not preperiodic. A wandering domain is a (1) is escaping if all orbits converge to the line at infinity, (2) is oscillating if there exists an unbounded orbit and an orbit with a bounded subsequence, (3) is orbitally bounded if every orbit is bounded. For polynomials in C it is known that wandering domains cannot exist [30]. For transcendental functions there are examples of escaping wandering domains. [7] uses for example the function f (z) = z + λ sin(2π z) + 1 for suitable λ. There are also examples of oscillating wandering domains [10,15], and it is an open question whether orbitally bounded wandering domains can exist. It follows from the existence of the filtration that a polynomial Hénon map does not admit any escaping or oscillating wandering domain. In the remainder of the paper we will give examples of both escaping and oscillating wandering domains for transcendental Hénon maps. The existence of orbitally bounded wandering domains is an open question for both polynomial and transcendental Hénon maps. We start with the escaping case, and we will be inspired by the construction of escaping wandering domains for transcendental functions. Similar to [7] we will use functions of the form f˜(z) = z + sin(2π z) + λ with appropriate values of λ. It will be convenient for us to take the constant λ such that we obtain an escaping orbit, all consisting of critical points. Note that The critical points of f˜ are therefore the points that satisfy f˜ (z) = 1 + 2π cos(2π z). cos(2π z) = −2π1 . A computation gives that f˜ has two distinct bi-infinite sequences of critical points cn,+ := α + n cn,− := −α + n with sin(2π cn,+) = + with sin(2π cn,−) = − 1 1 − 4π 2 1 1 − 4π 2 for the appropriate value of 41 < α < 21 given by solving (14). (14) Note that the set of critical points is invariant under translation by 1. By taking λ = 1 − f˜ acts as a translation by 1 on the sequence cn,+. For simplicity of notation we consider the map f obtained by conjugating f˜ with a real translation by α, so that the critical points cn,+ for f˜ are mapped to critical points zn = n for f . Thus, f will act on Z as a translation by 1, and f commutes with translation by 1. Consider the function g(z) := f (z) − 1, which commutes with f and with the translation by 1. For the function g each point zn is a super-attracting fixed point. For all n ∈ Z denote by Bn the immediate basin for g of the point zn. Clearly the basins Bn are disjoint, Bn+1 = Bn + 1, and f (Bn) ⊂ Bn+1. If one shows that each Bn is also a Fatou component for f , then clearly each Bn is an escaping wandering domain for f . There are two classical ways in one dimensional complex dynamics to show that each zn belongs to a different Fatou component for such a function f . One is by constructing curves in the Julia set that separate the points zn [12, p. 183], and the other one is by showing that any two commuting maps which differ by a translation have the same Julia set [2, Lemma 4.5]. The first kind of argument is typically unavailable in higher dimensions. The proof we present is an ad hoc version of the second argument. We will define transcendental Hénon maps F, G : C2 → C2 which behave similarly to f, g, then use G to show that the norms of the differentials d Gn and d F n at points on the boundaries of the attracting basins for G grow exponentially in n, and finally use the latter information to imply that those boundaries must be contained in the Julia set for F . Since the attracting basins are disjoint for G, the corresponding sets are disjoint for F , giving a sequence of wandering domains. Let us define the Hénon map F (z, w) = ( f (z) + δ(z − 1) − δw, z), where f is as constructed before and δ > 0 is some constant to be suitably chosen later. Since f commutes with translation by 1, the map F commutes with translation by (1, 1). Moreover, on the sequence of points (n, n −1) the map F acts as a translation by (1, 1). We also define the map G(z, w) := F (z, w) − (1, 1). Hence the points Pn = (n, n − 1) are all fixed points of G. Since the points zn are critical points of f , we can choose δ sufficiently small so that the fixed points Pn = (n, n − 1) are attracting for G. Denote by An the attracting basin of the point Pn, which is biholomorphic to C2. It is easy to see that An+1 = An + (1, 1) and that F ( An) = An+1. We claim that each An is also a Fatou component for F , and thus each ( An) is an escaping wandering domain. We first need a Lemma. ξn(k)(0) < k! · βk k < logλ(β) . λn < βk . Lemma 6.2 Let 0 < λ < 1, and let (ξn : D → C) be a sequence of holomorphic functions satisfying for all z ∈ D. Let 0 < β < 1. Then for Proof It follows from the Cauchy estimates since the assumption on k is equivalent to Theorem 6.3 Let F be a transcendental Hénon map that commutes with a translation T = T(1,1). Write G := T −1 ◦ F, assume that G has an attracting fixed point Q, and denote its basin of attraction by A. Then A is also a Fatou component of F . Proof Notice that (T k (Q)) is an F -orbit converging to the point [1 : 1 : 0] on the line at infinity. Since F n = T n ◦ Gn for all n ∈ N, it follows that the F -orbits of all points in A converge to [1 : 1 : 0]. Hence A is contained in a Fatou component of F . We show that equals A by proving that all boundary points of A are contained in the Julia set of F . Without loss of generality we may assume that the point Q is the origin. Let P ∈ ∂ A. Since P ∈/ A, its G-orbit avoids some definite ball centered at (0, 0), hence there exists 0 < μ < 1 for which for all n ∈ N. For each n we can choose an orthogonal projection πn onto a complex line through (0, 0) so that Gn( P) ≥ μk |πn(Gn( P))| ≥ k=0 μk . Let ϕ : C → C2 be an affine embedding for which P ∈ ϕ(D) and ϕ(0) ∈ A. We will show that for any choice of ϕ the derivatives of Gn ◦ ϕ grow exponentially fast for some point in the disk D(0, η), where η > μ1 is chosen independently of ϕ. This will imply that some point in ϕ(D(0, η)) is contained in the Julia set of F . Since ϕ is arbitrary, one can choose ϕ(D(0, η)) to be contained in arbitrarily small neighborhoods of P, giving a sequence of points Zn → P belonging to the Julia set of F . Since the latter is closed, the statement of the theorem follows. We consider the sequence of maps ψn : C → C defined by Let r > 0 be such that ϕ(D(0, r )) ⊂⊂ A. Then there exist C > 0 and λ < 1 so that ψn := πn ◦ Gn ◦ ϕ. ψn D(0,r) < C · λn. ψn = πn ◦ Gn+N ◦ ϕ. ψn D(0,r) < λn. |ψn(ζ )| ≥ ψn(z) = ∞ k=0 μk ak zk , ∞ k=0 |ak | ≥ μk , (15) (16) for some large integer N ∈ N we may assume that By defining the maps ξn(z) := ψn(r · z) we obtain a sequence of maps (ξn : D → C) that satisfy the conditions of Lemma 6.2, which we apply with β := μ · r . Hence which implies that ξn(k)(0) < k!μ r k k for k < ψn(k)(0) < k!μk for k < logλ(μr ) n logλ(μr ) , . Writing ζ = ϕ−1( P) ∈ D, we also have that for all n. Writing n and by (15), k ≥ logλ(μr) . Putting things together we get ψn(k)(0) ≥ k!μk n for some k ≥ logλ(μr ) . Let η > μ1 . By Cauchy estimates this implies the existence of a z∗ ∈ D(0, η) for which ψn(z∗) > ηk−1μk k! ≥ C (k − 1)! 1 where = (ημ) logλ(μr) > 1 and C > 0 is constant. Denote P∗ := ϕ(z∗). Since πn = 1 we get, up to changing C , It follows from F n = T n ◦ Gn that dP∗ (Gn) ≥ C dP∗ (F n) ≥ C n. n. We now show that a point in ϕ(D(0, η)) is contained in the Julia set of F . Assume for the purpose of a contradiction that ϕ(D(0, η)) is contained in the Fatou component . It follows that F n → [1 : 1 : 0] uniformly on a neighborhood U of P∗. Consider the affine chart of P2 defined as [z : w : t ] → (w/z, t /z) defined on {z = 0}. On {z = 0} ∩ {t = 0} such chart has the expression H (z, w) = (w/z, 1/z). The sequence (H ◦ F n : U → C2) is defined for n large enough and converges uniformly to the point (1, 0) ∈ C2. Hence, denoting P∗ := (z0, w0), d(zn,wn) H ◦ d(z0,w0)(F n) = d(z0,w0)(H ◦ F n) → 0. We have that Since (zn, wn) → [1 : 1 : 0], arguing as in (12) we obtain that for every exists C > 0 such that > 0 there (d(zn,wn) H )−1 = 0 zn 2 −zn −wn zn . 1 (d(zn,wn) H )−1 1 1 ≥ C (1 + )n , and thus d(zn,wn) H ◦ d(z0,w0)(F n) ≥ d(z0,w0)(F n) (d(zn,wn) H )−1 C n ≥ C (1 + )n . Hence if is small enough, this contradicts (18). (17) (18) 7 Oscillating wandering domain We show in this section the existence of a transcendental Hénon map with an oscillating wandering domain biholomorphic to C2. Notice that, up to a linear change of variables, any transcendental Hénon map can be written in the alternative form F (z, w) = ( f (z)+aw, az), where f is a transcendental function and a = 0. We will consider maps of the form F (z, w) := ( f (z) + aw, az), f (z) := bz + O(|z|2), where a, b ∈ R, a < 1, are such that the origin is a saddle point. For example, since the eigenvalues of d0 F are λ = b ± √b2 + 4a2 2 , we can pick a = 1/2, and b = 1 to obtain λs = 1−2√2 , λu = 1+2√2 . We will construct F as the uniform limit on compact subsets of a sequence of automorphisms Fk (z, w) := ( fk (z) + aw, az) with the same value for a and b. We note that for transcendental Hénon maps F (z, w) = ( f (z) + aw, az), G(z, w) = (g(z) + aw, az) and any subset A ⊂ C we have f − g A = F − G A×C. Proposition 7.1 There exists a sequence of entire maps Fk (z, w) = ( fk (z) + aw, az), fk (z) = bz + O(|z|2), k = 0, 1, 2, . . . a sequence of points Pn = (zn, wn), where n = 0, 1, 2, . . . , sequences Rk → ∞, 0 < k ≤ 21k and βn → 0, a decreasing sequence θk → 0, and strictly increasing sequences of integers {nk }, {nk } with nk < nk < nk+1, such that the following five properties are satisfied: (i) Fk − Fk−1 D(0,Rk−1)×C ≤ k for all k ≥ 1; (ii) Fk ( Pn) = Pn+1 for all 0 ≤ n < nk ; (iii) Fk (B( Pn, βn)) ⊂⊂ B( Pn+1, βn+1) for all 0 ≤ n < nk , where βn ≤ θ2ρ for all 0 ≤ n ≤ nk ; (iv) |zn| < Rk − θρ for all 0 ≤ n < nk , and |znk | > Rk + 5θk ; (v) Pj ∈ B(0, k1 ) for some j with nk−1 ≤ j ≤ nk , for all k ≥ 1. Here ρ := ρ(n) denotes the unique integer for which nρ−1 ≤ n < nρ , using n−1 = 0. Before proving this proposition, let us show that it implies the existence of a wandering Fatou component. By (i), the maps Fk converge uniformly on compact subsets to a holomorphic automorphism F (z, w) = ( f (z) + aw, az). By (ii), (iv) and (v) it follows that ( Pn) is an oscillating orbit for F , that is, it is unbounded and it admits a subsequence converging to the origin. Since by (iii) for all j ∈ N the iterate images of B( Pj , β j ) have uniformly bounded Euclidean diameter (in fact, the diameter goes to zero), each ball B( Pj , β j ) is contained in the Fatou set of F . Lemma 7.2 If i = j , then Pi and Pj are in different Fatou components of F , and hence F has an oscillating wandering domain. Proof For all j ∈ N, denote j the Fatou component containing B( Pj , β j ). Since βn → 0 as n → ∞, by (iii) and by identity principle it follows that all limit functions on each j are constants. Assume j > i and let k = j − i . Let N (0) be a neighborhood of the origin that contains no periodic points of order less than or equal to k. Since the orbit ( Pn) enters and leaves a compact subset of N (0) infinitely often, there must be a subsequence (n j ) for which Pn j converges to a point z ∈ N (0) \ {0}. But then Pn j +k converges to F k (z) = z, which implies that Pi and Pj cannot be contained in the same Fatou component. We will prove Proposition 7.1 by induction on k. We start the induction by letting R0 := 1, θ0 := 1, β0 := 21 and P0 := (z0, w0) with |z0| > 6. We set F0(z, w) := (bz + aw, az) and n0 := 0. Let us now suppose that (i)-(v) hold for certain k, and let us proceed to define Fk+1 and the points ( Pn)nk<n≤nk+1 . This is done in two steps. The first step relies upon the classical Lambda Lemma 7.3: Lemma 7.3 (Lambda Lemma, see e.g. [27]) Let G be a holomorphic automorphism of C2 with a saddle fixed point at the origin. Denote by W s (0) and W u (0) the stable and unstable manifolds respectively. Let p ∈ W s (0) \ {0} and q ∈ W u (0) \ {0}, and let D( p) and D(q) be holomorphic disks through p and q, respectively transverse to W s (0) and W u (0). Let > 0. Then there exists N ∈ N and N1 > 2N + 1, and a point x ∈ D( p) with G N1 (x ) ∈ D(q) such that Gn(x ) − Gn( p) < and G N1−n(x )−G−n(q) < for 0 ≤ n ≤ N , and Gn (x ) < when N < n < N1−N . Using the Lambda Lemma we find a finite Fk -orbit (Q j )0≤ j≤M , the new oscillation, which goes inward along the stable manifold of Fk , reaching the ball B(0, k +11 ), and then outwards along the unstable manifold of Fk . The second step of the proof relies upon another classical result: Lemma 7.4 (Runge approximation) Let K ⊂ C be a polynomially convex compact subset (recall that K is polynomially convex if and only if C \ K is connected). Let h ∈ O(K ), and let { pi }0≤i≤q be a set of points in K . Then for all > 0 there exists an entire holomorphic function f ∈ O(C) such that f − h K ≤ and such that f ( pi ) = h( pi ), f ( pi ) = h ( pi ) ∀ 0 ≤ i ≤ q. Using Runge approximation we find a map Fk+1 connecting the previously constructed finite orbit ( Pn)0≤n≤nk with the new oscillation (Q j )0≤ j≤M via a finite orbit along which Fk+1 is sufficiently contracting. The contraction neutralizes possible expansion along the new oscillation (Q j )0≤ j≤M , and we refer to this connecting orbit as the contracting detour. 7.1 Finding the new oscillation B(0, k +11 ) and a small enough θk+1 > 0 such that the three disks Lemma 7.5 There exist a finite Fk -orbit (Q j ) := (z j , w j )0≤ j≤M intersecting the ball D(znk , θk ), D w0/a, θk+1 , D z M , θk+1 are pairwise disjoint, and disjoint from the polynomially convex set K := D(0, Rk ) ∪ D zi , θk+1 . 0≤i<M fPorlodooffLtheet (mϕas p,ψFsk ). :LCet ζ→∈WCFsbk(e0a, 0p)oibnet wliniteharmiziinnigmcaoloabrdsionlautteesvfaolruethseusctha bthleatmani|ψ s (ζ )| = a |znk | − 4θk . Setting (u0, v0) := (ϕs (ζ ), ψ s (ζ )), by definition we have |ψ s (η)| < |ψ s (ζ )|, ∀η : |η| < |ζ |. Moreover, the forward orbit (ui , vi ) satisfies and since |ui | = |vi+1| it follows that a |v0| a = |znk | − 4θk . |vi | = |ψ s (λis · ζ )| < |v0|, ∀i ≥ 1, |ui | < |znk | − 4θk , ∀i ≥ 0. Similarly, we find a point (s0, t0) in the unstable manifold WFuk (0, 0) whose backwards orbit (s−i , t−i ) satisfies |s−i | < |s0| and for which |s0| = |znk | − 2θk . By taking an arbitrarily small perturbation of (s0, t0), we can make sure that the discrete sequence (si ) avoids the value va0 . Fig. 2 The first coordinates of the orbit (Pn) Consider arbitrarily small disks D1 centered at (u0, v0) and D2 centered at (s0, t0) transverse to the stable respectively the unstable manifold. It follows from the Lambda Lemma 7.3 that there exists a finite Fk -orbit (Q j ) := (z j , w j )0≤ j≤M which intersects the ball B(0, k +11 ) with Q0 ∈ D1 and Q M ∈ D2. If all the perturbations are chosen small enough, then the sequence (z j )0≤ j≤M avoids the disk D(znk , θk ) and the point w0 . Setting θk+1 > 0 small enough completes the proof (Fig. 2). a 7.2 Connecting the orbits via the contracting detour By continuity of Fk there exist constants 0 < β˜ j ≤ θk2+1 for all 0 ≤ j ≤ M such that Fk (B(Q j , β˜ j )) ⊂⊂ Fk (B(Q j+1, β˜ j+1)), 0 ≤ j < M. To control the contraction we will need the following Lemma, a direct consequence of the formula of transcendental Hénon maps. Lemma 7.6 Let f ∈ O(C) and let F (z, w) := ( f (z) + aw, az). Choose a , a such that 0 < a < a < a < 1, and let D1 ⊂⊂ D2 ⊂ C two open disks. Then there exists α > 0 such that if we have f − s D2 ≤ α, for some constant s ∈ C, then a (z, w) − (z , w ) ≤ F (z, w) − F (z , w ) ≤ a (z, w) − (z , w ) , ∀z, z ∈ D1. Fix 0 < a < a < a < c < 1 and let N > 0 be such that cN · βnk < β˜0. (20) We now construct the contracting detour, starting at Pnk and ending at Q0, obtaining a contraction by at least the factor cN . Choose a family of points {z j }0≤ j≤N ⊂ C such that (1) z0 = znk , z N = wa0 , (2) |z j | > |znk | + 2 for all 1 ≤ j ≤ N − 1, (3) |z j − zi | > 2, for all 0 ≤ i = j ≤ N − 1. Let Rk+1 > 0 be such that {T j }1≤ j≤N ⊂ C2 defined as Rk+1 > |z j | + θk , ∀0 ≤ j ≤ N , and Rk+1 > |z j | + θk+1, ∀0 ≤ j ≤ M. (21) Define w0 := wnk and w j := az j−1 for all 1 ≤ j ≤ N , and consider the points T j := z j , w j . By the choice of the points (z j ) and by Lemma 7.5 it follows that the disks D z j , θk 0≤ j<N , D w0/a, θk+1 , D z M , θk+1 are pairwise disjoint, and disjoint from the polynomially convex set K . Let H denote the union of such disks. We define a holomorphic function on the polynomially convex set K ∪ H in the following way: (1) h coincides with fk on K , (2) h|D(z j ,θk ) is constantly equal to z j+1 − aw j for all 0 ≤ j < N , (3) h|D(w0/a,θk+1) is constantly equal to z0 − awN , (4) h|D(zM ,θk+1) is constantly equal to some value A > Rk+1 + 5θk+1. By the Runge approximation Lemma 7.4 there exists a function fk+1 ∈ O(C) such that (1) fk+1(0) = h(0) = 0, fk+1(0) = h (0) = b, (2) fk+1(z j ) = h(z j ) for all 0 ≤ j < nk , (3) fk+1(z j ) = h(z j ) for all 0 ≤ j ≤ M , (4) fk+1(z j ) = h(z j ) for all 0 ≤ j ≤ N , (5) fk+1 − h K ∪H < k+1. (1) D(z j , θ2k ) ⊂⊂ D(z j , θk ) for all 0 ≤ j < N , (2) D(w0/a, θk2+1 ) ⊂⊂ D(w0/a, θk+1), (3) D(z M , θk2+1 ) ⊂⊂ D(z M , θk+1). where k+1 ≤ 2k1+1 , and k+1 is also smaller than the minimum of the constants α given by Lemma 7.6 for the following pairs of disks: Define Fk+1(z, w) := ( fk+1(z) + aw, az). It is easy to see that the sequences of points ( Pn)0≤n≤nk , (T j )1≤ j≤N , (Q j )0≤ j≤M nk := nk + N + 1, nk+1 := nk + M + 1, so that Pnk := Q0, and Pnk+1 := Fk+1(Q M ). Define βnk + := βnk · c , ∀ 0 ≤ ≤ N , βnk + := β˜ , ∀ 0 ≤ ≤ M. It is easy to see that, up to taking a smaller k+1, the map Fk+1 satisfies property (iii). Property (iv) follows from (21) and from A > Rk+1 + 5θk+1. Since by construction the new piece of orbit (Q j ) intersects the ball B(0, k +11 ), property (v) is satisfied. Thus Proposition 7.1 is proved, completing the proof of the existence of an oscillating wandering domain. We will now prove that, by making the contracting detours sufficiently long, the oscillating wandering domains can be guaranteed to be biholomorphically equivalent to C2. We denote the Fatou component containing P0 by . We have constructed an orbit ( Pn) and a sequence of radii (βn) such that Define the calibrated basin F (B( Pn, βn)) ⊂⊂ B( Pn+1, βn+1). (Pn),(βn) := F −n(B( Pn, βn)), Lemma 7.7 We can guarantee that the calibrated basin to C2. Proof For all n ≥ 0, let Hn ∈ Aut(C2) be defined by Hn(z) := Pn + βn · z. For all m ≥ n ≥ 0, define F˜m,n := Hm−1 ◦ F m−n ◦ Hn. (Pn),(βn) is biholomorphic Then for all n ≥ 0, we have that F˜n+1,n(B2) ⊂ B2 and F˜n+1,n(0) = 0. It is easy to see that the calibrated basin (Pn),(βn) is biholomorphic to the set ˜ := n∈N F˜n−,01(B2). If n ∈ N belongs to a contracting detour, then a c x ≤ a F˜n+1,n(x ) ≤ c x , ∀x ∈ B2, where we can assume that a 2 < a . If this was the case for every n then it would follow immediately that the maps n := (d0 F˜0,n)−1 ◦ F˜0,n gn(z, w) := log F n(z, w) − Pn . n Lemma 7.8 We can guarantee that the functions gn converge to log a on (Pn),(βn) \ { P0}. Proof Take sufficiently many contractions that are sufficiently close to multiplication by a. Recall from induction hypothesis (v) in Proposition 7.1 that for every k there exists an integer j = jk ∈ (nk−1, nk ) with Pj < k1 . In particular the subsequence ( Pjk ) is bounded. Since every limit function of (F n) on the Fatou component is constant, it follows that for all (z, w) ∈ we have F jk (z, w) − Pjk → 0 lim sup g jk (z, w) ≤ 0. k→∞ As a consequence, we have that (g jk ) is locally uniformly bounded from above and that for all (z, w) ∈ , Lemma 7.9 We can guarantee that g jk → 0 on subsets. \ (Pn),(βn), uniformly on compact Proof Recall that in the construction of the wandering Fatou component the radius θ jk is determined before the length of the contracting detour. By making the contracting detour sufficiently long, we can guarantee that jk is as large as we want. For points (Pn ),(θn ) we have that (z jk , w jk ) − Pjk ≥ θ jk . (Pn ),(θn ) for any k Thus, by making the contracting detours sufficiently long, we can guarantee that 0. The conclusion follows. Proposition 7.10 With the previous choices, the wandering Fatou component equals the calibrated basin (Pn ),(θn ), and is thus biholomorphically equivalent to C2. 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Leandro Arosio, Anna Miriam Benini, John Erik Fornæss, Han Peters. Dynamics of transcendental Hénon maps, Mathematische Annalen, 2018, 1-42, DOI: 10.1007/s00208-018-1643-6