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 socalled 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 “NEWHOLITENew 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 nonzero 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 nontrivial 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 nontrivial 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
twodimensional 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 semiconjugacy 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 onedimensional 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 onedimensional 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 halfplane). 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 selfmap 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 onepoint 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 Pnnormal 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 Cnnormal 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 Pnnormal if and only if it is equicontinuous with
respect to the FubiniStudy 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 Cnnormal 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 Cnnormal.
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 Pacman
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 Pacman 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+1suacnhdthgant(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 Pnnormal, then it is Cnnormal.
Proof Let ( fn) be a sequence in F . Since F is Pnnormal 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 CnFatou set if the family of iterates
(F n) is Cnnormal near z. A point z ∈ Cn belongs to the PnFatou set if the family of
iterates (F n) is Pnnormal 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 PnFatou
set is always contained in the CnFatou 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 C2Fatou 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 P2Fatou 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 C2Fatou component contains infinitely many distinct P2Fatou components.
In what follows, we will only consider P2normality. We will call the P2Fatou 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 selfmap 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 nonempty 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 (Nonempty Julia set) If δ = −1, then there exist infinitely many saddle
points in Per2(F ), and thus the Julia set of F is nonempty.
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 pushforwards 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 < zd .
Proof If w > zd for arbitrarily large z and d, then wnqn(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 selfmap 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 welldefined 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
nondiscrete. 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)2w1.
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))2dz 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 onedimensional 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 selfintersections 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 1dimensional 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 hLz = 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 superattracting 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 halfplane 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 selfmap H of P2 \ {[0 : 1 : 0]} with a superattracting
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 semiconjugate 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
submean 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 volumepreserving
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 volumepreserving
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 biinfinite 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 superattracting 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 Gorbit 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(z2),
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(z2), 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
bthleatmaniψ 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) hD(z j ,θk ) is constantly equal to z j+1 − aw j for all 0 ≤ j < N ,
(3) hD(w0/a,θk+1) is constantly equal to z0 − awN ,
(4) hD(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.
Proof Suppose by contradiction that = (Pn ),(θn ). Let h be the uppersemicontinuous
regularization of limk→∞ gnk . Clearly h ≡ log a on (Pn ),(θn ) and h ≡ 0 on
\ (Pn ),(θn ). Then by [25, Prop 2.9.17] the function h is plurisubharmonic, and
the submean value property at any ζ ∈ ∂ (Pn ),(θn ) gives a contradiction.
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