Dynamics of transcendental Hénon maps

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.


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 C 2 . Recall that a polynomial Hénon map is a map of the form 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 C 2 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 C 2 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 C 2 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 C 2 , although there have been results showing holomorphic automorphisms of C 2 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 C 2 has a hyperbolic fixed point with a stable manifold which is dense in C 2 . Transcendental Hénon maps seems to be a natural class of holomorphic automorphisms of C 2 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 1 2 , 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 C 2 [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 C 2 , 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 C 2 , 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 C 2 of the form F k (z, w) := ( f k (z) + 1 2 w, 1 2 z), all having a hyperbolic fixed point at the origin. The family (F k ) is constructed inductively using Runge approximation in one variable to obtain an entire function f k+1 which is sufficiently close to f k on larger and larger disks, in such a way that the orbit of an open set U 0 ⊂ C 2 approaches the origin coming in along the stable manifold of F k and then goes outwards along the unstable manifold of F k , 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 C 2 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 : C 2 → C 2 there are two natural definitions of the Fatou set, which correspond to compactifying C 2 either with the onepoint compactification C 2 , or with P 2 . 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 ∞ = P 2 \ C 2 . For a general entire self-map of C 2 these two definitions can give two different Fatou sets (see Example 2.6). Notice that, if we compactify with C 2 , any open subset of C 2 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 C 2 with P 2 (which has the additional advantage of being a complex manifold). Section 2 is devoted to this argument.

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, C n ) to be normal. We denote by C n the one-point compactification of C n , and with the symbol ∞ we denote both the point at infinity and the constant map z → ∞.
∈ F there exists a subsequence ( f n k ) converging uniformly on compact subsets to f ∈ Hol(X, P n ). In other words, F is relatively compact in Hol(X, P n ).
A family F ⊂ Hol(X, C n ) is C n -normal if for every sequence ( f n ) ∈ F which is not divergent on compact subsets there exists a subsequence ( f n k ) converging uniformly on compact subsets to f ∈ Hol(X, C n ). This is equivalent to F being relatively compact in Hol(X, C n ) ∪ ∞ ⊂ C 0 (X, C n ).

Remark 2.2
When n = 1 the two definitions are equivalent.
A family F ⊂ Hol(X, C n ) is P n -normal if and only if it is equicontinuous with respect to the Fubini-Study distance on P n . This follows from the Ascoli-Arzelà theorem and from the fact that Hol(X, P n ) is closed in C 0 (X, P n ). One may think that, similarly, a family F ⊂ Hol(X, C n ) is C n -normal if and only if it is equicontinuous with respect to the spherical distance d C n on C n , but this is not the case, as the following example shows.

Example 2.3
For n ≥ 2, the family Hol(D, C n ) ∪ ∞ is not closed in C 0 (D, C n ). As a consequence, for a family F ⊂ Hol(D, C n ), being relatively compact in C 0 (D, C n ) is not equivalent to being C n -normal.
Proof Let n = 2. Let s n ≥ 0 be an increasing sequence of real numbers converging to 1 2 . Let α n ≤ π 2 be a decreasing sequence converging to 0. We define the Pac-man Let r n ≥ 0 be an increasing sequence converging to 1 2 . Let β n be a sequence decreasing to 0. We define the bait R n := D ∩ {z ∈ C : |Im z| ≤ β n , Re z ≥ r n }.
Clearly n∈N R n = z ∈ C : |Im z| = 0, which we call the slit S. Fig. 1 The mouth of Pac-man P n+1 is contained in bait R n We can choose the sequences (s n ), (α n ), (r n ), (β n ) in such a way that Notice that this implies that s n ≤ r n ≤ s n+1 for all n ∈ N. See Fig. 1 for an illustration of a single Pac-man P n+1 and two baits R n+1 and R n . Let b i > 0 be a sequence such that for all n ∈ N, x ∈ C 2 the following implication holds By using Runge approximation we can define a sequence of holomorphic functions f n : D(0, 1 + ) → C such that | f n | ≤ 1 2 n+1 on P 2n , and Re f n ≥ b n on R 2n , and a sequence of holomorphic functions g n : D(0, 1 + ) → C such that |g n | ≤ 1 2 n+1 on P 2n+1 , and Re g n ≥ b n on R 2n+1 . For all n ≥ 0, let H n : D(0, 1 + ) → C 2 be defined as H n (z) = n j=0 ( f n (z), g n (z)). Denote a n := max We claim that a n ≤ 1 2 n , and thus the sequence (H n ) converges uniformly on the disk D to a continuous mapping H : , then we have that Re n−1 i=0 g i (z) ≥ n−1 i=0 b i and |g n (z)| ≤ 1 2 n+1 , and thus by (1) both H n (z) and H n−1 (z) belong to the ball of radius 1 2 n+1 centered at ∞. If z ∈ R 2n , then we have that Re n−1 i=0 f i (z) ≥ n−1 i=0 b i and Re f n (z) ≥ b n , and thus by (1) both H n (z) and H n−1 (z) belong to the ball of radius 1 2 n+1 centered at ∞.

Lemma 2.4
If a family F ⊂ Hol(X, C n ) is P n -normal, then it is C n -normal.
Proof Let ( f n ) be a sequence in F. Since F is P n -normal there exists a subsequence ( f n k ) converging uniformly on compact subsets to a map f ∈ Hol(X, P n ). If there is is open, and this follows taking an affine chart around f (y) ∈ ∞ in such a way that ∞ = {z 1 = 0} and applying Hurwitz theorem to the sequence π 1 • f n . Thus, if the sequence ( f n ) is not diverging on compact subsets, the subsequence ( f n k ) converges uniformly on compact subsets to a map f ∈ Hol(X, C n ).
As a consequence of the previous discussion, for an entire map F : C n → C n we have two possible definitions of the Fatou set.
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 P n -Fatou set is always contained in the C n -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 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(C 2 ) by It follows immediately from the first coordinate that the forward orbit of any point (z 0 , w 0 ) converges to infinity, i.e. F n (z 0 , w 0 ) → ∞, hence the C 2 -Fatou set equals all of C 2 . Moreover, if |z 0 | < 1 then F n (z 0 , w 0 ) → [1 : 0 : 0], uniformly on compact subsets. Thus, the domain D × C is contained in a P 2 -Fatou component.
On the other hand, if the sequence N j increases sufficiently fast, then for 1 < |z 0 | ≤ 2 we have that F n (z 0 , w 0 ) → [0 : 1 : 0] ∈ ∞ , again uniformly on compact subsets. It follows that D × C is a P 2 -Fatou component. Therefore in this example the single C 2 -Fatou component contains infinitely many distinct P 2 -Fatou components.
In what follows, we will only consider P 2 -normality. We will call the P 2 -Fatou set simply the Fatou set. The Julia set is the complement of the Fatou set.

Periodic points
If g is a transcendental function or a polynomial Hénon map, then, for each k ≥ 1, the set Fix(g k ) is discrete. Clearly this statement is not satisfied for holomorphic automorphisms of C 2 . For example, one can consider any holomorphic conjugate of a rational rotation.
Consider a periodic orbit Since w j+1 = z j for each j, the first coordinate function of the Hénon map gives the following relations (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 When δ = −1 it is immediate that the set of solutions is discrete. When δ = −1 the solutions satisfy 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 1 2 . Then Fix(F k ) is discrete for all k ≥ 1. Proof Consider the entire function Since f is assumed to have order of growth strictly less than 1 2 , so does g, and Wiman's Theorem [32] implies that there exist radii r n → ∞ for which m(r n ) → ∞.
Suppose for the purpose of a contradiction that the solution set in C k 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, r n ) k . Then V also intersects the boundary ∂ D(0, r n ) k , say in a point (z 0 , . . . , z k−1 ). By the symmetry of the equations in (2) we may then well assume that |z 0 | = r n , and of course that |z j | ≤ r n for j = 1, . . . , k −1.
By Wiman's Theorem we may assume that |g(z 0 )| is arbitrarily large, and in particular that | f (z 0 )| > (1+|δ|)r n . 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. Recall that if f is an entire transcendental function, then Per 2 ( f ) has infinite cardinality by [6].
If δ = −1, let g(z) := f (z) 1+δ . By (4), Per 2 (F) coincides with which is non-empty and has infinite cardinality. If δ = −1, then 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 (z k ) in Per n (g) such that Corollary 3.7 (Non-empty Julia set) If δ = −1, then there exist infinitely many saddle points in Per 2 (F), and thus the Julia set of F is non-empty.
Proof We have seen that for all z 0 ∈ Per 2 (g), the point (z 0 , g(z 0 )) ∈ Per 2 (F). A computation using the explicit form for F gives , and hence f (g(z 0 )) · f (z 0 ), can be taken arbitrarily large by Theorem 3.6).

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
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 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.
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 : Note that q 0 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 |q n (z)| dominates the other terms in the form (2), so H (z, w) cannot vanish.
By Lemma 3.10 there exists d ∈ N such that |w j | < |z j | d for j sufficiently large. Hence for j sufficiently large It follows that where c > 0 is a constant. But since z j = w j it follows that for large enough j, all terms of the form

Classification of recurrent components
In this section we only assume that F is a holomorphic automorphism of C 2 with constant Jacobian δ.
Since the class of holomorphic automorphism of C 2 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 : → P 2 such that f n k → h uniformly on compact sets of for some subsequence n k → ∞.

Theorem 4.2 Let F be a holomorphic automorphism of C 2 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 C 2 . • 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 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 n k ) be a convergent subsequence of iterates on , with F n k (z) → w ∈ . We denote the limit of (F n k ) by g.

Lemma 4.3 The image g( ) is contained in C 2 .
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).

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 d w F has an eigenvalue of absolute value ≥ 1. Then the same is true for all iterates F n k . 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}.

Maximal rank 2
Theorem 4.6 Suppose that g has maximal rank 2. Then there exists a subsequence (m k ) so that F m k → 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.
Since F n k → g on U , we have that V ⊂ F n k (U ) for large k and the maps (F n k ) −1 converge to h uniformly on compact subsets of V . In particular V ⊂ . Replace n k by a subsequence so that n k+1 −n k ∞. We can then write F n k+1 −n k = F n k+1 •(F n k ) −1 on V . If we set m k := n k+1 − n k , then F m k → Id on V . Since we are in the Fatou set this implies that F m k → 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.
be a sequence of injective holomorphic mappings which are volume preserving. If G n converges to G uniformly on compact subsets, then G is holomorphic, injective and volume preserving.
Proof The map G is holomorphic and dG n n→∞ −→ dG, 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
Proof Assume that h( ) ⊂ C 2 . 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 ∈ K 1 , and let r > 0 be such that B( p, r ) ⊂ K 1 .
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 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 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 K 1 infinitely often. Thus, there exists a sequence We claim that the sets F k j (B( p, r )) must be pairwise disjoint. To see this, suppose that F k i (B( p, r )) ∩ F k j (B( p, r )) = ∅ for some i < j. Then clearly which contradicts (iii) and (iv). This proves the claim. Since F is volume preserving and the volume of K 4 \ K 3 is finite, we have a contradiction.

Corollary 4.10
The limit of any convergent subsequence (F n k ) 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 n k (K ). Hence F −n k ( p) ∈ \ K . Fix such k. By Theorem 4.6 there exists m > n k so that F m is close enough to the identity so that F m−n k ( 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.

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 n k ) is a convergent subsequence of iterates on such that F n k (z) → w ∈ . Replace n k by a subsequence so that n k+1 − n k ∞. Let (m k ) be a subsequence of (n k+1 − n k ) so that F m k converges, uniformly on compact subsets of . From now on we assume that g is the limit of the sequence (m k ).

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 n k ) converges to the map g : → C 2 of maximal rank 1, it follows that |δ| < 1.
We write := g( ). Notice that is a subset of ∩ C 2 , 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 C 2 , and let h : U → C 2 be a holomorphic map of maximal rank 1. Then for all w ∈ h(U ), the fiber h −1 (w) has no isolated points.
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 γ z : z → C 2 such that Moreover, if g has rank 1 at z, then γ z ( z ) is smooth and γ z = 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 → C 2 , 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)).
) 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 → C 2 , there exists a holomorphic injective map γ z : z → C 2 such that γ z ( z ) = g( z ).

Remark 4.16
For all z ∈ , there exists a unique surjective holomorphic map h z : U (z) → z such that g = γ z • h z on the neighborhood U (z). If g has rank 1 at z, then h z | 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 γ z (x) = γ w (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 π z : z → as π z (x) := [(x, z)].

Definition 4.17
We define a continuous map γ : → C 2 such that γ ( ) = in the following way: γ ([(x, z)]) = γ z (x). Notice that this is well defined. The map g : → C 2 can be lifted to a unique surjective continuous map g : → such that g = γ • g. Such map is defined on U (z) as g := π z • h z . Notice that if g has rank 1 at z then g| z = π z .

Lemma 4.18
The topological space is connected, second countable and Hausdorff.
Proof Since =ĝ( ), 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 γ z (x) = γ w (y), or γ z (x) = γ w (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 (π z ) z∈ gives the structure of a smooth Riemann surface. Let z, w ∈ such that π z ( z ) ∩ π w ( w ) = ∅. Then consider the map Let x ∈ z , y ∈ w such that π z (x) = π w (y). This means that γ z (x) = γ w (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 γ w • k = γ z . It is easy to see that k = π −1 w • π z on U .

Remark 4.19
With the complex structure just defined on , the maps γ and g are holomorphic.
Definition 4.20 Define R ⊂ as the set of points ζ ∈ such that there exists z ∈ with g(z) = ζ and rk z g = 1.

Lemma 4.21
The set \ R is discrete.
Proof Let w ∈ such that rk w g = 0. By the identity principle there exists a neighborhood V of w in w such that rk z (h w | w ) = 1 for all z ∈ V \ {w}. The result follows since g = π w • h w on U (w), and π w : w → π w ( w ) is a biholomorphism.

Lemma 4.22
There exists a unique holomorphic map F : → such that the following diagram commutes: Proof Let ζ ∈ R, and let z ∈ such that g(z) = ζ and rk z g = 1. Define on π z ( z ) the map F := g • F • π −1 z . 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.  ∈ g(N ), F m k (y) → y. Then, for all y ∈ g(N ), F m k ( g(y)) = g(F m k (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 : → C 2 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 m k ) 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 ⊂ × as the set of pairs (x, y) such that x = y and γ (x) = γ (y). Define the set C ⊂ as the set of x such that γ : → C 2 has rank 0 at x.
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 is a closed complex submanifold of . We will first consider the case where is biholomorphic to a circular domain and F is conjugate to an irrational rotation.

Lemma 4.26
If F is conjugate to an irrational rotation then there is at most one element ζ 0 ∈ 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 ζ 0 (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 ζ 0 ∈ and that ζ 0 ∈ C.
Then has a cusp at z := γ (ζ 0 ). 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 d z F with eigenvalue |λ 1 | = 1. Since F is strictly volume decreasing, the other eigenvalue λ 2 of d z F satisfies 0 < |λ 2 | < 1. Thus we obtain a forward invariant cone in T z (C 2 ), centered at the line T z ( ). 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 C 2 , which is an embedded complex submanifold of C 2 .
We claim that there exists a continuous function ϕ : V → (0, ∞), bounded from above and from below by compactness, such that Indeed, regardless of whether is a hyperbolic or Euclidean Riemann surface, there exists a conformal metric · on which is invariant under F. The function ϕ(z) := d z γ −1 satisfies (8).
Given > 0 and z ∈ V , we define the tangent cone C z ⊂ T z (C 2 ) by where v z is a unit tangent vector to z ∈ V and v ⊥ z is a unit vector orthogonal to v z .

Lemma 4.28
One can choose > 0 sufficiently small so that Proof This is a matter of linear algebra. Without loss of generality we may assume that v z = v F(z) = (1, 0). Thus, the cone field C z contains all vectors w = (w 1 , w 2 ) for which
Since |α(z)| is bounded on V , by choosing sufficiently small we can guarantee that for all z ∈ V we have from which it follows that Thus, d z F sends the cone C z strictly into C F(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 is a closed smooth one-dimensional embedded submanifold.
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 C 2
is the identity on a complex analytic curve γ with any kind of singularity at the origin, then d 0 G 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.
We immediately get stable manifolds transverse to , which imply as above that g : → 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 rk x 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

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 : → P 2 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 is recurrent then for every z ∈ the orbit is relatively compact in . 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) = 0 z , and h maps the fiber g −1 (z) ∩ U biholomorphically into a neighborhood of 0 z in the fiber L z [21, Proposition 3.3.2]. We can assume that d z h| L z = id L z . Notice that the map d z F j | L z : L z → L z acts as multiplication by δ j . The sequence (d F| 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 (z, w) → e iθ z, δ e iθ w .

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 1 2 . 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.

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 P 1 for which ∞ is a super-attracting fixed point. The first example of a Baker domain was given by Fatou [17], who considered the function (9) and showed that the right half-plane H is contained in an invariant Baker domain. For holomorphic automorphisms of C 2 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 C 2 . We call a point [ p : q : 0] ∈ ∞ an essential singularity at infinity if for all [s : t : 0] ∈ ∞ there exists a sequence (z n , w n ) of points in C 2 converging to [ p : q : 0] whose images F(z n , w n ) 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 Since f is transcendental, the function g must also be transcendental. Let (ζ n ) be a sequence of points in C converging to ∞ for which g(ζ n ) → s t . Then Here we consider the iteration of a map on C 2 analogous to (9), namely the transcendental Hénon map 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: 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 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 Notice that η α > 0, A α 1+A α = e −α , and that the domains R α are not nested.

Lemma 5.4 Each domain R α is forward invariant.
Proof Let (z 0 , w 0 ) ∈ R α . We claim that Re z 1 > Re w 1 + α + η α (Re w 1 ). Since Hence the claim follows if we show that This is the same as A α e −Re w 0 ≥ (1+ A α )e −Re z 0 , or equivalently e −α ≥ e −Re z 0 +Re w 0 . The latter is satisfied because Re z 0 − Re w 0 ≥ α (since η α is always positive).
We immediately obtain the following.

Corollary 5.5 The domain
It easily follows that then Re z n > Re z 0 + nα, and that Re w n > Re z 0 + (n − 1)α. Proof Let K be a compact subset of R α for some α > 0. Let M > 0. Then by Remark 5.6 we have that |z n | and |w n | converge to ∞ uniformly on K . Moreover, since for all n ≥ 0 we have that z n+1 − w n+1 = z n − w n + e −z n , it follows that

Lemma 5.7 All the orbits in R converge uniformly on compact subsets to the point
and thus z n w n 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.
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.
for all n > n and for all (z 0 , w 0 ) ∈ K . It follows that for every > 0 there exists C = C such that max{|z n |, |w n |} ≤ C · (1 + ) n (12) for every n ∈ N and for all (z 0 , w 0 ) ∈ K . Let β > 0. We claim that there exists an N > 0 so that u n < β on K for every n ≥ N . To prove this claim, let us suppose by contradiction that there exist a sequence (z k 0 , w k 0 ) k∈N in K and a strictly increasing sequence (n(k)) k∈N in N such that Then Re(z k n(k) ) ≤ −n(k) · β. It follows that By taking > 0 sufficiently small so that e β > (1 + ), we obtain, for k sufficiently large, which contradicts (12). The claim implies that there is a uniform bound from above for the functions u n on K , and that lim sup n→∞ u n ≤ 0, which completes the proof.

Lemma 5.10 Let H be a compact subset of A. Then there exists γ > 0 such that on H,
lim sup n→+∞ u n ≤ −γ.

Proposition 5.12
The set R is an absorbing domain.
Proof Let us assume, for the purpose of a contradiction, that U = A. Define 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) ∈ C 2 , L n (z, w) = ((n + 1)z − nw, nz − (n − 1)w).

Theorem 5.15
There exists a biholomorphism ψ : U → which conjugates F to the map L. In particular U is biholomorphic to H × C.
Proof We will construct the map ψ as the uniform limit on compact subsets of U of the maps Notice that for all n ≥ 1, the mapping ψ n is an injective volume-preserving holomorphic mapping. We have By Remark 5.6, on R α √ 2(n + 1)e −Re z n ≤ √ 2(n + 1)e −Re z 0 −nα , Hence for all (z 0 , w 0 ) ∈ R α , we have ∞ k=0 δ n (z 0 , w 0 ) < +∞, and the sequence (ψ n ) n≥0 converges uniformly on compact subsets of R to an injective volumepreserving (see Lemma 4.7) holomorphic mapping ψ : R → C 2 , satisfying 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 (z 0 , w 0 ) be in R α . Notice that the Euclidean distance d(R α , ) > 0. We claim that lim n→∞ ψ(z n , w n ) − (z n , w n ) = 0. Indeed, and the claim follows since Re z n+ j ≥ Re z 0 + (n + j)α. Hence, if n is large enough, ψ(z n , w n ) ∈ . 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 (x 0 , y 0 ) ∈ , and write (x n , y n ) := L n (x 0 , y 0 ). By definition of we have that β := Re(x 0 − y 0 ) > 0. Note that Re(x n − y n ) also equals β, and that Re(x n ), Re(y n ) → ∞. Let 0 < α < β. Recalling that η α (x) → 0 as x → +∞ it follows that (x n , y n ) ∈ R α for all n ∈ N sufficiently large. In fact, there exists an r > 0 and an N ∈ N such that R α contains the closed ball B((x n , y n ), r ) for all n ≥ N . We claim that Indeed, for all n ∈ N, Assume now that n ≥ N . Since B((x n , y n ), r ) ∈ R α we have that for all (x, y) ∈ B((x n , y n ), r ), where the last inequality follows from the fact that for all n ∈ N and (x, y) ∈ B((x n , y n ), r ) we have Re x ≥ Re x 0 + nα − r . This proves the claim. Let n ≥ N be such that ψ − Id ≤ r 2 on B((x n , y n ), r ). By Rouché's Theorem in several complex variables it follows that (x n , y n ) ∈ ψ(B ((x n , y n ), r )) ⊂ ψ(U ). Since is completely invariant under L and U is completely invariant under F, it follow that (x 0 , y 0 ) ∈ ψ(U ). 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 formf (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 f (z) = 1 + 2π cos(2π z).
Note that the set of critical points is invariant under translation by 1. By taking f acts as a translation by 1 on the sequence c n,+ .
For simplicity of notation we consider the map f obtained by conjugatingf with a real translation by α, so that the critical points c n,+ forf are mapped to critical points z n = 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 z n is a super-attracting fixed point. For all n ∈ Z denote by B n the immediate basin for g of the point z n . Clearly the basins B n are disjoint, B n+1 = B n + 1, and f (B n ) ⊂ B n+1 . If one shows that each B n is also a Fatou component for f , then clearly each B n is an escaping wandering domain for f .
There are two classical ways in one dimensional complex dynamics to show that each z n belongs to a different Fatou component for such a function f . One is by constructing curves in the Julia set that separate the points z n [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 : C 2 → C 2 which behave similarly to f, g, then use G to show that the norms of the differentials dG n 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 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 Hence the points P n = (n, n − 1) are all fixed points of G. Since the points z n are critical points of f , we can choose δ sufficiently small so that the fixed points P n = (n, n − 1) are attracting for G. Denote by A n the attracting basin of the point P n , which is biholomorphic to C 2 . It is easy to see that A n+1 = A n + (1, 1) and that F(A n ) = A n+1 . We claim that each A n is also a Fatou component for F, and thus each (A n ) is an escaping wandering domain. We first need a Lemma. Lemma 6.2 Let 0 < λ < 1, and let (ξ n : D → C) be a sequence of holomorphic functions satisfying |ξ n (z)| < λ n for all z ∈ D. Let 0 < β < 1. Then Proof It follows from the Cauchy estimates since the assumption on k is equivalent to 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 μ k for all n ∈ N. For each n we can choose an orthogonal projection π n onto a complex line through (0, 0) so that |π n (G n (P))| ≥ ∞ k=0 μ k .
Let ϕ : C → C 2 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 G n • ϕ grow exponentially fast for some point in the disk D(0, η), where η > 1 μ 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 Z n → 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 ψ n : 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 ψ n D(0,r ) < C · λ n .

By instead defining
for some large integer N ∈ N we may assume that ψ n D(0,r ) < λ n .
Writing ζ = ϕ −1 (P) ∈ D, we also have that for all n. Writing by (16) there is at least one k ∈ N for which |a k | ≥ μ k , and by (15), k ≥ n log λ (μr ) . Putting things together we get Let η > 1 μ . By Cauchy estimates this implies the existence of a z * ∈ D(0, η) for which log λ (μr ) > 1 and C > 0 is constant. Denote P * := ϕ(z * ). Since π n = 1 we get, up to changing C, It follows from F n = T n • G n that 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) We have that Since (z n , w n ) → [1 : 1 : 0], arguing as in (12) we obtain that for every > 0 there exists C > 0 such that and thus Hence if is small enough, this contradicts (18).

Oscillating wandering domain
We show in this section the existence of a transcendental Hénon map with an oscillating wandering domain biholomorphic to C 2 . 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 where a, b ∈ R, a < 1, are such that the origin is a saddle point. For example, since the eigenvalues of d 0 F are we can pick a = 1/2, and b = 1 to obtain λ s = 1− √ 2 2 , λ u = 1+ √ 2 2 . We will construct F as the uniform limit on compact subsets of a sequence of automorphisms F k (z, w) := ( f k (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 a sequence of points P n = (z n , w n ), where n = 0, 1, 2, . . . , sequences R k → ∞, 0 < k ≤ 1 2 k and β n → 0, a decreasing sequence θ k → 0, and strictly increasing sequences of integers {n k }, {n k } with n k < n k < n k+1 , such that the following five properties are satisfied: (iv) |z n | < R k − θ ρ for all 0 ≤ n < n k , and |z n k | > R k + 5θ k ; (v) P j ∈ B(0, 1 k ) for some j with n k−1 ≤ j ≤ n k , for all k ≥ 1. Here ρ := ρ(n) denotes the unique integer for which n ρ−1 ≤ n < n ρ , 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 F k converge uniformly on compact subsets to a holomorphic automorphism F(z, w) = ( f (z) + aw, az). By (ii), (iv) and (v) it follows that (P n ) 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(P j , β j ) have uniformly bounded Euclidean diameter (in fact, the diameter goes to zero), each ball B(P j , β j ) is contained in the Fatou set of F.

Lemma 7.2 If i = j, then P i and P j 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(P j , β j ). Since β n → 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 (P n ) enters and leaves a compact subset of N (0) infinitely often, there must be a subsequence (n j ) for which P n j converges to a point z ∈ N (0) \ {0}. But then P n j +k converges to F k (z) = z, which implies that P i and P j cannot be contained in the same Fatou component.
Let us now suppose that (i)-(v) hold for certain k, and let us proceed to define F k+1 and the points (P n ) n k <n≤n k+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 C 2 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 N 1 > 2N + 1, and a point x ∈ D( p) with G N 1 (x) ∈ D(q) such that G n (x) − G n ( p) < and G N 1 −n (x)−G −n (q) < for 0 ≤ n ≤ N , and G n (x) < when N < n < N 1 −N .
Using the Lambda Lemma we find a finite F k -orbit (Q j ) 0≤ j≤M , the new oscillation, which goes inward along the stable manifold of F k , reaching the ball B(0, 1 k+1 ), and then outwards along the unstable manifold of F k .
The second step of the proof relies upon another classical result: Using Runge approximation we find a map F k+1 connecting the previously constructed finite orbit (P n ) 0≤n≤n k with the new oscillation (Q j ) 0≤ j≤M via a finite orbit along which F k+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.

Finding the new oscillation
Lemma 7.5 There exist a finite F k -orbit (Q j ) := (z j , w j ) 0≤ j≤M intersecting the ball ) and a small enough θ k+1 > 0 such that the three disks are pairwise disjoint, and disjoint from the polynomially convex set Proof Let (ϕ s , ψ s ) : C → W s F k (0, 0) be linearizing coordinates for the stable manifold of the map F k . Let ζ ∈ C be a point with minimal absolute value such that It is easy to see that |ψ s (η)| < |ψ s (ζ )|, ∀η : |η| < |ζ |.
By taking an arbitrarily small perturbation of (s 0 , t 0 ), we can make sure that the discrete sequence (s i ) avoids the value v 0 a . Fig. 2 The first coordinates of the orbit (P n ) Consider arbitrarily small disks D 1 centered at (u 0 , v 0 ) and D 2 centered at (s 0 , t 0 ) transverse to the stable respectively the unstable manifold. It follows from the Lambda Lemma 7.3 that there exists a finite F k -orbit (Q j ) := (z j , w j ) 0≤ j≤M which intersects the ball B(0, 1 k+1 ) with Q 0 ∈ D 1 and Q M ∈ D 2 . If all the perturbations are chosen small enough, then the sequence (z j ) 0≤ j≤M avoids the disk D(z n k , θ k ) and the point w 0 a . Setting θ k+1 > 0 small enough completes the proof (Fig. 2).
To control the contraction we will need the following Lemma, a direct consequence of the formula of transcendental Hénon maps. Fix 0 < a < a < a < c < 1 and let N > 0 be such that c N · β n k <β 0 .
We now construct the contracting detour, starting at P n k and ending at Q 0 , obtaining a contraction by at least the factor c N . Choose a family of points {z j } 0≤ j≤N ⊂ C such that (1) z 0 = z n k , z N = w 0 a , (2) |z j | > |z n k | + 2 for all 1 ≤ j ≤ N − 1, (3) |z j − z i | > 2, for all 0 ≤ i = j ≤ N − 1.
By the choice of the points (z j ) and by Lemma 7.5 it follows that the disks D z j , θ k 0≤ j<N , D w 0 /a, θ k+1 , D z M , θ k+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: Define F k+1 (z, w) := ( f k+1 (z) + aw, az). It is easy to see that the sequences of points (P n ) 0≤n≤n k , (T j ) 1≤ j≤N , (Q j ) 0≤ j≤M form an F k+1 -orbit, that is, the contracting detour (T j ) 1≤ j≤N connects the old and new pieces of orbit. Set n k := n k + N + 1, n k+1 := n k + M + 1, so that P n k := Q 0 , and P n k+1 := F k+1 (Q M ). Define β n k + := β n k · c , ∀ 0 ≤ ≤ N , It is easy to see that, up to taking a smaller k+1 , the map F k+1 satisfies property (iii). Property (iv) follows from (21) and from A > R k+1 + 5θ k+1 . Since by construction the new piece of orbit (Q j ) intersects the ball B(0, 1 k+1 ), property (v) is satisfied. Thus Proposition 7.1 is proved, completing the proof of the existence of an oscillating wandering domain.

Complex structure
We will now prove that, by making the contracting detours sufficiently long, the oscillating wandering domains can be guaranteed to be biholomorphically equivalent to C 2 . We denote the Fatou component containing P 0 by .
Define the calibrated basin (P n ),(β n ) := n∈N F −n (B(P n , β n )), and notice that it is contained in .
Lemma 7. 7 We can guarantee that the calibrated basin (P n ),(β n ) is biholomorphic to C 2 .
Proof For all n ≥ 0, let H n ∈ Aut(C 2 ) be defined by H n (z) := P n + β n · z. For all m ≥ n ≥ 0, defineF m,n := H −1 m • F m−n • H n .
Then for all n ≥ 0, we have thatF n+1,n (B 2 ) ⊂ B 2 andF n+1,n (0) = 0. It is easy to see that the calibrated basin (P n ),(β n ) is biholomorphic to the set˜ := n∈NF −1 n,0 (B 2 ). If n ∈ N belongs to a contracting detour, then 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 := (d 0F0,n ) −1 •F 0,n converge to a biholomorphism from the calibrated basin to C 2 , see for example [33]. While we cannot guarantee that equation (22) with a 2 < a holds outside of the contracting detours, by making the contracting detours sufficiently long we can still ensure that n converges uniformly on compact subsets to a biholomorphism :˜ → C 2 , compare for example Theorem 1.3 of [29].
We will now show, again by using the plurisubharmonic method, that the wandering Fatou component can be forced to be equal to the calibrated basin. Consider the sequence of plurisubharmonic functions g n on C 2 given by g n (z, w) := log F n (z, w) − P n n .
Lemma 7. 8 We can guarantee that the functions g n converge to log a on (P n ),(β n ) \ {P 0 }.
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 = j k ∈ (n k−1 , n k ) with P j < 1 k . In particular the subsequence (P j k ) 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 j k (z, w) − P j k → 0 As a consequence, we have that (g j k ) is locally uniformly bounded from above and that for all (z, w) ∈ , lim sup k→∞ g j k (z, w) ≤ 0. Lemma 7. 9 We can guarantee that g j k → 0 on \ (P n ),(β n ) , uniformly on compact subsets.
Proof Recall that in the construction of the wandering Fatou component the radius θ j k is determined before the length of the contracting detour. By making the contracting detour sufficiently long, we can guarantee that j k is as large as we want. For points (z 0 , w 0 ) ∈ \ (P n ),(θ n ) we have that (z j k , w j k ) − P j k ≥ θ j k .
Thus, by making the contracting detours sufficiently long, we can guarantee that g j k > − k on \ (P n ),(θ n ) for any k 0. The conclusion follows.

Proposition 7.10
With the previous choices, the wandering Fatou component equals the calibrated basin (P n ),(θ n ) , and is thus biholomorphically equivalent to C 2 .
Proof Suppose by contradiction that = (P n ),(θ n ) . Let h be the upper-semicontinuous regularization of lim k→∞ g n k . Clearly h ≡ log a on (P n ),(θ n ) and h ≡ 0 on \ (P n ),(θ n ) . Then by [25,Prop 2.9.17] the function h is plurisubharmonic, and the submean value property at any ζ ∈ ∂ (P n ),(θ n ) gives a contradiction.
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