Open Access

Adaptive trains for attracting sequences of holomorphic automorphisms

Complex Analysis and its Synergies20151:3

https://doi.org/10.1186/2197-120X-1-3

Received: 21 August 2014

Accepted: 7 December 2014

Published: 12 August 2015

Abstract

Consider a holomorphic automorphism acting hyperbolically on an invariant compact set. It has been conjectured that the arising stable manifolds are all biholomorphic to Euclidean space. Such a stable manifold is always equivalent to the basin of a uniformly attracting sequence of maps. The equivalence of such a basin to Euclidean space has been shown under various additional assumptions. Recently, Majer and Abbondandolo achieved new results by non-autonomously conjugating to normal forms on larger and larger time intervals. We show here that their results can be improved by adapting these time intervals to the sequence of maps. Under the additional assumption that all maps have linear diagonal part, the adaptation is quite natural and quickly leads to significant improvements. We show how this construction can be emulated in the non-diagonal setting.

1 Introduction

Let X be a complex manifold equipped with a Riemannian metric, and let f:XX be an automorphism that acts hyperbolically on some invariant compact subset KX. Let pK and write for the stable manifold of f through p. is a complex manifold, say of complex dimension m. In the special case where p is a fixed point, it is known that is biholomorphically equivalent to . It was conjectured by Bedford [5] that this equivalence holds for any pK.

Conjecture 1 (Bedford)

The stable manifold is always equivalent to .

The usual approach towards this problem is to translate it to a non-autonomous setting. Let f 0,f 1,… be a sequence of automorphisms of satisfying
(1)
for all and all z lying in the unit ball , where the constants 1>D>C>0 are independent of n. We will call a sequence (f n ) satisfying condition (1) uniformly attracting. We define the basin of attraction of the sequence (f n ) as
(2)

It was shown by Fornæss and Stensønes that a positive answer to the following conjecture implies a positive answer to Conjecture 1.

Conjecture 2.

The basin Ω is always biholomorphic to .

Here we present the following new results, both giving positive answers to Conjecture 2 under additional hypotheses.

Theorem 3.

Let (f n ) be a uniformly attracting sequence of automorphisms of , and suppose that the maps f n all have order of contact k. Assume further that the linear part of each map f n is diagonal and that

Then, the basin of attraction is biholomorphic to .

Theorem 4.

Let (f n ) be a uniformly attracting sequence of automorphisms of , and suppose that

Then, the basin of attraction is biholomorphic to .

In the next section, we will place these two results in a historical perspective. While Theorems 3 and 4 are significant improvements over previously known results, our main contribution lies in a new way of thinking about trains, as introduced by Abbondandolo and Majer in [2].

The authors would like to thank Alberto Abbondandolo, Leandro Arosio, John Erik Fornæss, Jasmin Raissy, Pietro Majer, and Liz Vivas for many stimulating discussions. The first author was supported by a SP3-People Marie Curie Actions grant in the project Complex Dynamics (FP7-PEOPLE-2009-RG, 248443).

2 A short history

Let f be an automorphism of a complex manifold X of dimension m, and let pX be an attracting fixed point. Define the basin of attraction by

The following result was proved independently by Sternberg [16] and Rosay-Rudin [13].

Theorem 5 (Sternberg, Rosay-Rudin).

The basin Ω is biholomorphic to .

If p is not attracting but hyperbolic, one considers the restriction of f to the stable manifold. This restriction is an automorphism of with an attracting fixed point at p. Moreover, its basin of attraction is equal to the entire stable manifold, which is therefore equivalent to . This naturally raised Conjecture 1. Equivalence to of generic stable manifolds was proved by Jonsson and Varolin in [10].

Theorem 6 (Jonsson-Varolin).

Let X, f, and K be as in Conjecture 1. For a generic point pK, the stable manifold through p is equivalent to .

Here, generic refers to a subset of K which has full measure with respect to any invariant probability measure on K. In fact, Jonsson and Varolin showed that Conjecture 1 holds for the so-called Oseledec points. The results of Jonsson and Varolin were extended by Berteloot et al. in [7]. More recently, the following was shown in [1]:

Theorem 7 (Abate-Abbondandolo-Majer).

The existence of Lyapunov exponents is enough to guarantee .

Shortly after the positive result of Jonsson and Varolin, Fornæss proved in [8] the following negative result, which led many people to believe that Conjectures 1 and 2 must be false. Consider a sequence of maps (f n ) n≥0 given by

where |a 0|<1 and |a n+1|≤|a n |2.

Theorem 8 (Fornæss).

The basin is not biholomorphic to . Indeed, there exists a bounded plurisubharmonic function on which is not constant.

We note that this result does not give a counterexample to Conjecture 2, as the sequence (f n ) violates the condition

on any uniform neighborhood of the origin. In Theorem 8, the rate of contraction is not uniformly bounded from below.

If the bounds C and D satisfy D 2<C, then it was shown by Wold in [17] that the basin of attraction is biholomorphic to . This result was generalized by Sabiini [14] to the following statement.

Theorem 9.

Let (f n ) be a sequence of automorphisms which satisfies the conditions in Conjecture 2, and suppose that D k <C for some . Suppose further that the maps f n all have order of contact k. Then, the basin of attraction is biholomorphic to .

Here, a map f has order of contact k if f(z)=l(z)+O(z k ), where l(z) is linear.

The following was recently shown in [2]:

Theorem 10 (Abbondandolo-Majer).

There exists a constant ε=ε(m)>0 such that the following holds: For any sequence (f n ) which satisfies the conditions in Conjecture 2 with

the basin is biholomorphic to .

It was noted in [2] that for maps with order of contact k, one should be able to obtain the same result if D k+ε <C, where ε=ε(m,k).

While this result only seems marginally stronger (indeed, ε-stronger), it is in fact a much deeper result. If D k <C one can ignore all but the linear terms, see Lemma 16 in the next section. But if D k C, one has to deal with the terms of order k, which is a major difficulty. In fact, looking at more classical results in local complex dynamics, the major difficulty in describing the behavior near a fixed point usually lies in controlling the lowest order terms which are not trivial. Theorems 10 and 3 may therefore be important steps towards a complete understanding of Conjecture 2.

Now that techniques have been found to deal with these terms of degree k, it is natural to ask whether the condition D k+ε <C can be pushed to the condition D k+1<C, since as long as D k+1<C is satisfied one can ignore terms of degree strictly greater than k. In Theorem 3, we show that we can indeed weaken the requirement to D k+1<C, at least in two complex dimensions and under the additional assumption that the linear parts of all the maps f n are diagonal.

Whether diagonality is a serious extra assumption or merely simplifies the computations remains to be seen, but we will see that some of the techniques we introduce to prove Theorem 3 can be applied without the diagonality assumption as well, leading to Theorem 4.

Since the computations in the general case are much more intensive than in the diagonal case, we have chosen not to prove Theorem 4 for maps with higher order of contact. We expect that the interested reader will be able to generalize Theorem 4 to maps with higher order of contact.

We conclude our historical overview with the following two results. The first was proved in [12].

Theorem 11 (Peters-Wold).

Let (f j ) be a sequence of automorphism of , each with an attracting fixed point at the origin. Then there exists a sequence of integers (n j ) such that the attracting basin of the sequence is equivalent to .

The next result, from [11], will be important to us not so much for the statement itself but for the ideas used in the proof. See also the more technical Lemma 17, which will be discussed in the next section.

Theorem 12 (Peters).

Let F be an automorphism of . Then there exists an ε>0 such that for any sequence (f n ) of automorphisms of which all fix the origin and satisfy for all n, one has that .

The combination of Theorems 11 and 12 provides a good idea for the techniques used to prove Theorems 3 and 4: If, after applying suitable changes of coordinates, we can find sufficiently long stretches in the sequence (f n ) that behave similarly, then we can show that the basin will be biholomorphic to . What is meant by behave similarly will be made more precise in later sections.

2.1 Applications

The theorems described above have been used to prove a number of results which are at first sight unrelated. The first example of such an application is of course the classical result of Fatou and Bieberbach which states that there exists a proper subdomain of which is biholomorphic to . Indeed, it is not hard to find an automorphism of with an attracting fixed point but whose basin of attraction is not equal to the entire . Proper subdomains of that are equivalent to are now called Fatou-Bieberbach domains.

It should be no surprise that for the construction of Fatou-Bieberbach domains with specific properties, it is useful to work with non-autonomous basins. Working with sequences of maps gives much more freedom than working with a single automorphism. Using Theorem 9, it is fairly easy to construct a sequence of automorphisms satisfying various global properties, while making sure that the attracting basin is equivalent to . We give two examples of results that have been obtained in this way.

Theorem 13 (Wold).

There exist countably many disjoint Fatou-Bieberbach domains in whose union is dense.

The question of whether such Fatou-Bieberbach domains exist was raised by Rosay and Rudin in [13].

Theorem 14 (Peters-Wold).

There exists a Fatou-Bieberbach domain Ω in whose boundary has Hausdorff dimension 4 and positive Lebesgue measure.

Here, we note that Fatou-Bieberbach domains with Hausdorff dimension equal to any h(3,4) were constructed by Wolf [19] using autonomous attracting basins. Hausdorff dimension 3 (and in fact C boundary) was obtained earlier by Stensønes in [15], who also used an iterative procedure involving a sequence of automorphisms of . See [18] for another application of non-autonomous basins to the theory of Fatou Bieberbach domains.

The last application we would like to mention is the Loewner partial differential equation. The link with non-autonomous attracting basins was made in [3], where Arosio used a construction due to Fornæss and Stensønes (see Theorem 20 below) to prove the existence of solutions to the Loewner PDE. See also [4] for the relationship between the Loewner PDE and non-autonomous attracting basins.

3 Techniques

3.1 The autonomous case

Essentially, the only available method for proving that a domain Ω is equivalent to is by explicitly constructing a biholomorphic map from Ω to . Let us first review how this is done for autonomous basins, before we look at how this proof can be adapted to the non-autonomous setting. We follow the proof in the appendix of [13], but see also the survey [6] written by Berteloot.

Given an automorphism F of with an attracting fixed point at the origin, one can find a lower triangular polynomial map G, and for any , a polynomial map X k of the form X k =Id+h.o.t., biholomorphic in a neighborhood of the origin, such that
(3)
Here, a polynomial map G=(G 1,…,G m ) is called lower triangular if
for i=1,…,m. Now, consider the sequence of holomorphic maps from Ω F to given by

Important here is that the lower triangular polynomial maps behave very similarly to linear maps. For example, it follows immediately by induction on m that the degrees of the iterates G n are uniformly bounded. One can also easily see that the basin of attraction of a lower triangular map with an attracting fixed point at the origin is always equal to the whole set .

Lemma 15.

Let G be a lower triangular. Then there exist a β>0 such that

for all .

Now let D>0 be such that F(z)<Dz for . Then there exists a such that D k+1·β<1. It follows from Equation 3 that, with this choice of k, the maps Φ n converge, uniformly on compact subsets of Ω F , to a biholomorphic map from Ω F to .

While the proof outlined above works elegantly, in the non-autonomous case it will often not be easy to control the maps X k for arbitrarily high k. However, it turns out that with a little more care it is sufficient to work with a much lower k that really only depends on the eigenvalues of D F(0). Let 0<C<D<1 be such that
and let k be such that D k+1<C. It turns out that it is sufficient to work with the maps X k . To see this, let C <C be such that D k+1<C . Then there exist an r>0 sufficiently small such that for one has
Let K be a relatively compact subset of the attracting basin. Then there exists an N such that . If r is chosen sufficiently small, then we easily see that for any mN and jm-N, we have that
It follows that

which is summable over n. Hence, the sequence (Φ n ) forms a Cauchy sequence and converges to a limit map Φ. Since D Φ n (0)=Id for all n, it follows that Φ is biholomorphic onto its image. To prove the surjectivity of Φ, one shows that for each there exists a compact KΩ F such that for all . The fact that follows since we are dealing with open maps.

3.2 Non-autonomous conjugation

In the non-autonomous setting, it is very rare that a single change of coordinates simplifies the sequence of maps. Instead, we use a sequence of coordinate changes.

Lemma 16.

Let (f n ) be a sequence of automorphisms that satisfies the hypotheses of Conjecture 2 (D k <C), and suppose that there exist uniformly bounded sequences (g n ) and (h n ), with h n =Id+h.o.t., such that the diagram

commutes as germs of order k. Then, .

If the maps (g n ) are all lower triangular polynomials then one still has that the basin of the sequence (g n ) is equal to . This simple fact was used in [11] to prove the following lemma, which for simplicity we state in the case m=2.

Lemma 17.

Let (f n ) be a sequence of automorphisms satisfying the conditions in Conjecture 2, and suppose that the linear part of each map f n is of the form

with |b n |2<ξ|a n | for some uniform constant ξ<1. Then, we can find bounded sequences (g n ) and (h n )as in Lemma 16. Moreover, the maps g n can be chosen to be lower triangular polynomials, and hence .

As was pointed out in [11], we can always find a non-autonomous change of coordinates such that the linear parts of the maps f n all become lower triangular. Let us explain the technique in the two-dimensional setting, where a matrix is lower diagonal if and only if [0,1] is an eigenvector. Using QR factorization, we can find, for any vector v 0, a sequence of unitary matrices (U n ) such that U 0 v 0=[0,1] and

with . Then by defining , we obtain a new sequence (g n ) whose basin is equivalent (by the biholomorphic map U 0) to the basin of the sequence (f n ). Notice that the linear parts of all the maps (g n ) are lower triangular. In this construction, we are free to choose the initial tangent vector v 0.

But while we may always assume that the linear parts are lower triangular, the condition that |b n |2<ξ|a n | for all in Lemma 17 is a strong assumption. In particular, there is no reason to think that one can change coordinates to obtain a sequence of lower triangular polynomial maps. Instead, we could aim for obtaining lower triangular polynomials on sufficiently large time intervals. Let us be more precise. Suppose that one can find an increasing sequence of integers p 1,p 2,… and change coordinates as in Lemma 16 to obtain that is equivalent to the basin of a sequence
(5)

where the U j are unitary matrices and the maps g n are all lower triangular. In the spirit of Theorem 11, one would expect that the basin of this new sequence is equal to as long as the sequence (p j ) is sparse enough. Indeed this is the case, as follows from the following lemma, proved by Abbondandolo and Majer in [2].

Lemma 18.

Suppose that the maps g n in the sequence given in Equation 5 are uniformly bounded lower triangular polynomials of degree k, and that
(6)

Then, the basin of the sequence in Equation (5) is equal to

The proof of Theorem 10 from [2] can now be sketched as follows. Define p j+1=k j +p j , and on each interval find a tangent vector v j which is contracted most rapidly by the maps . Next, find the non-autonomous change of coordinates by unitary matrices such that the maps g n as defined above all have lower triangular linear part. Then on each interval I j =[p j ,p j+1], the maps g n satisfy the conditions of Lemma 17 'on average.’ This is enough to find another non-autonomous change of coordinates after which the maps g n are lower triangular polynomial maps on each interval I j . Then it follows from Lemma 18 that the basin of the sequence (f n ) is equivalent to .

Now, we arrive at one of the main points presented in this article. In the argument of Abbondandolo and Majer, the intervals [p j ,p j+1] are chosen without taking the maps (f n ) into consideration; it is sufficient to make a simple choice such that Eq. 6 is satisfied. In the last section of this article, we will show that we can obtain stronger results if we instead let the intervals [p j ,p j+1] depend on the maps (f n ) or, to be more precise, on the linear parts of the maps (f n ).

3.3 Abstract basins

Let us discuss a construction due to Fornæss and Stensønes [9]. Let (f n ) now be a sequence of biholomorphic maps from the unit ball into , satisfying
for some uniform 1>D>C>0 as usual. We define the abstract basin of attraction of the sequence (f n ) as follows. Consider all sequences of the form
We say that
if there exists a j≥ max(k,l) such that x j =y j . This gives an equivalence relation , and we define

We refer to as the abstract basin of attraction, sometimes also called the tail space. We have now used the notation for both abstract and non-autonomous basins, but thanks to the following lemma this will not cause any problems.

Lemma 19.

Let (f n ) be a sequence of automorphisms of which satisfy the conditions in Conjecture 2. Then, the basin of attraction of the sequence (f n ) is equivalent to the abstract basin of the sequence .

Hence from now on, we allow ourselves to be careless and write for both kinds of attracting basins. Abstract basins were used by Fornæss and Stensønes to prove the following.

Theorem 20 (Fornæss-Stensønes).

Let f and p be as in Conjecture 1. Then, is equivalent to a domain in .

Remark 21.

Working with abstract basins can be very convenient. For example, Lemma 16 also holds for abstract basins which, in conjunction with Lemma 19, implies that in Diagram 4 we do not need to worry about whether the maps h n and g n are globally defined automorphisms. From the fact that the sequences (g n ) and (h n ) are uniformly bounded, it follows that their restrictions to some uniform neighborhood of the origin are biholomorphisms, which is all that is needed.

4 Definition of the trains

Let us go back to the proof of 10 by Abbondandolo and Majer. Instead of conjugating to lower triangular maps on the entire sequence (f n ), they introduced a partition of into intervals of rapidly increasing size and on each interval changed coordinates to either lower or upper triangular maps. These intervals were referred to as trains, and we will build on this terminology.

The main difference between the proof by Abbondandolo and Majer and the technique introduced in this paper is that we let the trains be determined by the sequence (f n ), or rather, by the linear parts of the maps. We will describe an algorithm to construct the partition

Each train [p j ,p j+1) will be headed by an interval [p j ,q j ) which we will call the engine. On the engine, we will have very good estimates of the linear maps. Our estimates are not nearly as strong on the interval [q j ,p j+1), but the good estimates on the engine will be used as a buffer to deal with estimates on the rest of the train. In fact, as soon as the buffer from the engine fails to be sufficient, we start a new train.

We proceed to define the trains explicitly, both in the diagonal and the general case.

4.1 The diagonal case

We assume here that the linear part of each map f n is of the form
We then define
(7)
We will define an increasing sequence p 0,q 0,p 1,… as follows. We set p 0=0 and q 0=2. We define the rest of the sequence recursively. Suppose that we have already defined p j and q j . We consider all pairs p j+1,q j+1 satisfying q j p j+1q j+1 and
(8)
Out of all such pairs, we choose q j+1 minimal, and then p j+1 such that

is maximal. We will write I j =[p j ,p j+1) and refer to this interval as the j th train. The interval [p j ,q j ) we will call the engine of the train I j .

In the diagonal case, we can immediately deduce several properties of the trains I j . First of all, from Equation 8, it is clear that

which will later allow us to apply Lemma 18.

Proposition 22.

On the engine of the train, we have
(9)
for q j np j , and moreover
(10)
Furthermore, for p j+1nmq j , we have
(11)

All three properties follow immediately from the definitions.

4.2 The general case

Our goal is now to emulate the above construction while dropping the assumption that the linear parts of the maps f n are diagonal. In the diagonal case, we eventually will obtain a sequence (g n ) where each map g n is either lower triangular or upper triangular, depending on whether n lies in an odd or even train. In the general case, this will be generalized as follows: for each train [p j ,p j+1), there will be a unit vector v j with respect to which the maps g n will be lower triangular. Let us be more precise. Suppose that the linear part of a map g has an eigenvector v. Let U be a 2×2 unitary matrix such that

We say that g is lower triangular with respect to v if U g U -1 is a lower triangular polynomial. Note that this definition is independent of the choice of U.

Looking back, we see that in the diagonal case, our maps g n are all lower triangular with respect to either [0,1] or [1,0]. In the general case, we allow the vector v to be any unit vector in , but v must remain constant on each train. This motivates the following definition.

Definition 23.

For k>x>1, recursively define an increasing sequence of integers p 1,q 1,p 2,q 2,… and a sequence of unit-vectors v 1,v 2,… as follows:

Set p 0=0, and v 0=(1,0). Define the rest of the sequences by induction:

Given p j , q j , and v j , consider all possible pairs of integers p j+1 and q j+1 such that q j+1>p j+1q j and for which
Out of all possible choices for p j+1 and q j+1, we choose q j+1 minimal, and then p j+1 such that

is minimal.

By composing with a unitary matrix U j+1, we may assume that is an eigenvector of . Let v j+1 be the other unit-length eigenvector of (unique up to multiplication by a unit-length scalar). This vector must exist, since the eigenvector has eigenvalue and we have

since . This implies that has two distinct eigenvalues and hence two distinct eigenvectors.

Our strategy to prove inequalities like those in Proposition 22 on an engine [p j ,q j ) is the following. We will work not only with the diagonal entries a n and b n in the current coordinates but first with the entries α n and β n , which are the diagonal entries in the coordinates that were used for the previous train. By the recursive definition of the train, we first obtain estimates for α n and β n , and then translate these to estimates on a n and b n in the new coordinates. In the translation from old to new coordinates, the off-diagonal entries, denoted by c n and γ n , will play an important role.

Given (p j ), (q j ), and (v j ) as defined above, we now apply a non-autonomous conjugation such that within each train [p j ,p j+1) all linear parts have [0,1] as an eigenvector. We know that v j is an eigenvector of . Pick a unitary matrix V j that transfers v j into [0,1]. Then, [0,1] will be eigenvector of . Next, for i=p j +1,…p j+1, pick a unitary matrix W i such that [0,1] is an eigenvector of . We can pick . Define
(12)
Now the composition can also be written as
(13)

We know that [0,1] is an eigenvector of each for p j mnp j+1. So these linear maps are lower triangular: write .

Then, we have
(14)
and
(15)
Write a n =a n+1,n , b n =b n+1,n and c n =c n+1,n . Note that and as in the diagonal case, and

Since all V j and W i are unitary matrices, we still have that for , and hence we know that Ca i D, Cb i D and c i D for all i.

Lemma 24.

In this new notation, our trains have the following properties:
  1. (i)
    On the engine of train j, we know that
     
  2. (ii)
    On the wagons of the train, we have:

    for q j mnp j+1.

     
  3. (iii)
    And for all j≥0, we have an estimate for the length of the j th engine:
     

Proof.

To prove the first statement, note that by (14) and (15) and the fact that U j is unitary, we have
Since v j and are distinct eigenvectors of , we obtain
And by definition of q j and p j , we know that

which completes (i).

For q j mnp j+1, we have:

by minimality of q j+1. The second half of statement (ii) follows from our choice of p j+1.

Finally, use part (i) to see that

which implies that . The case j=0 holds by definition.

Since is another eigenvector of , we now introduce notation that emphasizes the role of this vector. Pick a unitary matrix S j which maps to [0,1]. Then, pick unitary matrices T i for i=p j +1,…q j such that [0,1] is an eigenvector of . Note that we can take .

As before, define
(16)

Then, we have .

We know that [0,1] is an eigenvector of each for p j mnq j . Therefore, we can write .

Then, we have for p j mnq j :
and
Writing α n =α n+1,n , β n =β n+1,n and γ n =γ n+1,n as before, we again have
and

for all i.

Note that |a n ·b n |=| detD f n |=|α n ·β n | for all n. Since and are the eigenvalues of for two distinct eigenvectors, we must have , which implies that and .

This new notation allows us to derive additional inequalities for the engines of the trains:

Lemma 25.

For p j mnq j , we have:
  1. (i)

    ,

     
  2. (ii)

    , and

     
  3. (iii)

    .

     

Proof.

By the choice of p j and q j and the fact that
we obtain (i) and (ii):
and
Inequality (iii) is trivial if n=m. If m<n, we have:

since m<q j , which proves the right-hand side of inequality (iii).

By definition of q j , we have
since n-1<q j . Hence

which implies the left-hand side of inequality (iii).

We can now give an upper estimate for |γ n,m | on the engine as well:

Lemma 26.

For p j mnq j , we have

where is a constant depending only on C, D, and x.

Proof.

We already know that |γ i |<D for all i=p j ,…,q j -1, and we have:

where we used that |γ i |<D and |β i |>C.

By Lemma 25, we have , so and . Hence, we obtain

Using Lemmas 25 and 26, we can now find estimates for for p j nq j , which we can use to translate our information on α and β to information on a and b.

Lemma 27.

For p j nq j , we have
  1. (i)

    , and

     
  2. (ii)

    .

     

Proof.

We have
Using Lemma 26, we can now calculate that
Since (by Lemma 25), we have and therefore . Hence, we obtain
To prove inequality (ii), note that
By Lemma 25, we have that , which implies that we may remove from the maximum. Now, Lemma 26 implies that

We can now translate Lemma 25 to the following generalization of Proposition 22.

Theorem 28.

There exists a constant K 2=K 2(C,D,k,x) such that for each p j mnq j , we have
  1. (i)

    ,

     
  2. (ii)

    , and

     
  3. (iii)

    .

     

Proof.

For part (i), we can use Lemma 27 and Lemma 25 to see that
Part (ii) and (iii) follow from the same lemmas:

So we can set .

Remark 29.

When we continue the proof of the general case, we will work with the sequence the sequence instead of (f n ). The basins of these two sequences must be equivalent since at any time the composed functions differ only by multiplication with one last unitary matrix.

Hence, we will write f n rather than and . We will study the basin of the sequence where and all M j are unitary.

5 Completion of the proof in the diagonal case

Let us recall the statement of Theorem 3

Theorem 3.

Let (f n ) be a sequence of automorphisms of which satisfies the conditions in Conjecture 2, and suppose that the maps f n all have order of contact k. Assume further that the linear part of each map f n is diagonal. Then if D k+1<C, the basin of attraction is biholomorphic to .

The proof will be completed in two steps. In the first step, we will find a non-autonomous conjugation on each train by linear maps such that in each train the new maps all contract the same coordinate most rapidly. This means that we can apply Lemma 30 below on each train separately. In the second step, we worry about what happens at times p j where we switch from one train to the other. After these two steps, we construct the maps (g n ) and (h n ) on each of the trains using the following lemma, a special case of Lemma 17.

Lemma 30.

Suppose that each map of the sequence (f n ) has linear part of the form (z,w)(a n z,b n w), with |b n |≤|a n |. Then for any k≥2, we can find bounded sequences (g n ) and (h n ), with the maps g n lower triangular, such that diagram (4) commutes up to jets of order k.

Step 1: Directing the trains.

We assume that the automorphisms f 0,f 1,… satisfy the conditions of Theorem 3, and that we have constructed the increasing sequence p 0,q 0,p 1,… such that Proposition 22 holds. We will change coordinates with a sequence of linear maps l n of the form
(17)
We write for the new maps, and we immediately see that is of the form
(18)
where
(19)
Our goal is to define the maps (l n ) in such a way that on each interval I j =[p j ,p j+1) the maps satisfy the property (if j is odd), and (if j is even). In order for the higher order terms of the maps not to blow up, we also require that
(20)
In order to simplify the notation, we let j be odd; the results are analogous for j even. We assume that
and
and that the conditions in (20) are satisfied for n=p. Then for n+1I j , we recursively define
(21)
Similarly, we define
(22)
For convenience, we will often write

Lemma 31.

For pnq, we have
(23)

Proof.

Follows easily by induction on n.

Note that it follows immediately from the recursive definition of τ n+1 and the fact that θ n can never decrease, that for all n[p,r].

Lemma 32.

For pnr, we have that .

Proof.

For nq, this is guaranteed by the previous lemma. From q on it follows by induction on n that
(24)

for all qmn for which σ n,m ≤0. The statement in the lemma now follows from (11).

Finally, we need to check that θ r and τ r are large enough to satisfy the starting hypothesis for the next interval I j+1.

Lemma 33.

We have and .

Proof.

The estimate on θ r is immediate since θ n does not decrease with n. The estimate on τ p follows from
(25)

Our conclusion is the following:

Theorem 34.

Let be the sequence defined by
(26)
Then, is a bounded sequence of automorphisms, and . Moreover

for all nI j .

Proof.

The condition on the coefficients of the linear parts of the maps follows from the discussion earlier in this section. The fact that the sequence is bounded follows from the facts that the linear parts stay bounded, and that the higher order terms grow by at most a uniform constant.

To see that the two basins of attraction are biholomorphically equivalent (with biholomorphism l 0), note that . Since the entries of the diagonal linear maps l n+1 are always strictly greater than 1, it follows that the basin of the sequence is contained in the l 0 image of the basin of the sequence (f n ). The other direction follows from the fact that the coefficients of the maps l n grow at a strictly lower rate than the rate at which orbits are contracted to the origin by the sequence (f n ).

Step 2: Connecting the trains.

The following is a direct consequence of Lemma 17:

Theorem 35.

Let (f n ) be a bounded sequence of automorphisms of whose linear parts are of the form (z,w)(a n z,b n w) and whose order of contact is k. Suppose that |a n |≥|b n | for all n=0,1,…. Then there exists bounded sequences (h n ) and (g n ) such that g n h n =h n+1f n +O(k+1) for all n. Here, the maps g n can be chosen to be lower triangular maps, and the maps h n can be chosen to be of the form (z,w)Id+h.o.t..

In fact, even without the condition |a n |≥|b n |, we can always change coordinates such that the maps f n become of the form
(27)
Hence, we may assume that our maps are all of this form. We outline the proof of Theorem 35. Our goal is to find sequences (g n ) and (h n ) such that the following diagram commutes up to degree k.
Here, h n will be of the form
(29)
and g n will be of the form
(30)
We need that
(31)
It is clear that given the map h n+1 we can always choose g n such that h n is of the form (29), and that this map g n is unique. Hence, we can view α n as a function of α n+1. In fact, this function is affine and given by
(32)
or equivalently
(33)

Note that both coefficients of these affine maps are uniformly bounded from above in norm, and that . It follows that there exist a unique bounded orbit for this sequence of affine maps. In other words, we can choose a sequence (h n ) whose k-th degree terms are uniformly bounded. It follows directly that the maps g n are also uniformly bounded, which completes the proof.

Moreover, we note that we have much more flexibility if we have a sequence of intervals (I j ) (with j odd) and we want to find sequences (h n ) and (g n ) on each I j that are uniformly bounded over all j. In fact for each interval I j , we can start with any value for and inductively define the maps h n backwards. We merely need to choose uniformly bounded starting values .

The entire story works just the same for the intervals I j with j even. In this case, we have that |a n |≤|b n |, so we work with maps h n and g n of the form
(34)
and
(35)
The only matter to which we should pay attention is what happens when the different intervals are connected. Hence, let us look at the following part of the commutative diagram
We consider the case where j is odd and j+1 even; the other is similar. Imagine that we have constructed the maps h n and g n on the interval I j+1. Then, is of the form
(37)
As before, we can find such that the map given by is of the form
(38)

We note that while the map does depend on the coefficient , the coefficient (and equivalently also the map ) only depends on the coefficients of and not on those of . Hence we can find uniformly bounded maps h n on each of the intervals, and therefore also uniformly bounded maps g n , which are lower triangular for n in each interval [p j ,r j ) with j odd, and upper triangular when j is even. This completes the proof of Theorem 3.

6 Proof in the general case

In what follows, we drop our assumption of diagonality of the linear parts of the maps f n . Our approach will be different this time. We will not able to use linear maps l n as we did in the diagonal setting. The problem is that trying to keep bounded off-diagonal terms in the linear parts of the maps places strong restrictions on the maps l n , even if we use lower-triangular matrices instead of diagonal matrices.

Our solution will be to change coordinates to lower triangular maps g n immediately, but this comes at severe cost. The coefficients of the maps g n and h n will grow exponentially with n. What saves the day is that we can find sequences (g n ) and (h n ) for which the coefficients are bounded at the start and end of each train. This gives us estimates on the coefficients of all (g n ) and (h n ), which turn out to be sufficient under the additional assumption that

Our plan for proving Theorem 4 is the following: We use the trains defined in Definition 23, where we pick k such that and D k <C, and set . Then, we will analyze the conditions on the coefficients of the maps (g n ) and (h n ) that arise from the commutative diagram. We prove that it is possible to chose a specific sequence of maps for which the coefficients remain bounded at the start and end of each train. In the rest of the paper, we assume that we have made such a choice of maps.

In the second step, we find an estimate for the growth of the coefficients of (g n ) and (h n ) in the middle of trains and show that the basin of the sequence (g n ) is equal to . In the third step, we use the usual construction of the maps Φ n . Most work goes into proving that these maps converge on compact subsets to a holomorphic map Φ from the basin of the sequence (f n ) to the basin of the sequence . We end by showing that Φ is the required biholomorphism.

The choice of x we made seems rather random, but it turns out that the exact value of x does not influence our result. The bottleneck of our estimates on the quadratic coefficients of (g n ) and (h n ) depends on estimates on the wagons and the length of the trains. When x increases, the estimates on the wagons (Lemma 24) get weaker while the trains get longer. These effects cancel out.

Step 1: Exploring new coordinates

Instead of first conjugating with linear maps (l n ) as we did in the diagonal case, we will immediately conjugate (on each train j) with maps (for n[p j ,p j+1]) of the form
We will obtain a diagram of the form
which will commute up to degree two. We write
as usual. The maps g n will be lower triangular of the form

Fix j for now and drop the indices j in .

Note that . This gives us the following relations between the quadratic coefficients of h n and h n+1:

Here the r n ’s and s n ’s are linear functions with coefficients bounded by , and the constants x n and y n have the same bound. We will usually drop the variables in the functions r n and s n .

Given , we can now set
(39)

for p j n<p j+1. Then, will be zero for these values of n.

The values of up to can be studied using the relations above. We know that |a n | and |b n | are at most D<1. The value of is small on average, by the choice of our trains. Therefore, when we start with and work backwards, the quadratic constants up to should not get too big. This is made more precise in the lemma below.

For that lemma, pick , which is possible since we assumed that k<11/5.

Lemma 36.

For any δ>0, there exists a constant X=X(C,D,ε,δ)independent of j, such that if are all bounded by X, then for all n[p j ,p j+1], we have:

where Y=Y(C,D,ε)is a constant independent of both j and δ.

Proof.

We use backwards induction on n to prove the following slightly stronger statements one by one:

where we use the constants , and .

The case n=p j+1 is stated in the assumptions, so we can start our backward induction. Suppose that the first inequality holds for n+1. Then, we find that
Now use that and |b n | ε D ε/2·D ε/2 to obtain
for X as stated above. The other two cases are proved similarly and are left for the reader. For the upper bound on , , , and , we note that

We obtain the following similar estimate on the coefficients of g.

Corollary 37.

If the assumptions of Lemma 36 are satisfied and , then for all n[p j ,p j+1), we have that , and

Proof.

The first statement follows directly from the choice of d n in (39). For the second statement, we have
For n[p j ,p j+1-1), we similarly have
In the following lemma, we will give an upper bound for

which plays an important role in the above estimates. This allows us to estimate the quadratic terms at the beginning of train j.

Lemma 38.

We have

where K 3=K 3(C,D,k,x) is a constant independent of j.

Proof.

Suppose that p j stp j+1 and look at . We consider the cases tq j and tq j separately.

Case I: tq j .

We have
(40)
Now notice that
(41)
and since q j itp j+1, Lemma 24(ii) gives us also
(42)
Note that when t-i increases, the first estimate will increase while the second estimate will decrease. The estimates are equal when , which is when . Using estimate (41) for and the estimate (42) when , we get:
(43)
To estimate the second term in the product in (40), use Theorem 28 to see that , when p j iq j . And by Lemma 24, we know that . Combining these facts gives
Since , we obtain
(44)

Case II: tq j .

By Theorem 28, we have , and therefore

where we used that k<x 2 and . Setting now gives us the desired result.

Suppose we are now given and set . Recall that the M j are unitary matrices arising from the definition of the trains. A consequence of the fact that each map has at most six quadratic terms is the following.

Lemma 39.

If the coefficients of are bounded by R>0, then the coefficients of are bounded by 24R.

Proof.

We know that . Since M j is unitary, we obtain

The lemma follows.

The following is now an immediate consequence of Lemmas 36, 38, and 39.

Corollary 40.

If δ>0 is chosen sufficiently small and j 0 is sufficiently large, then for all jj 0, we have that if the coefficients of are bounded by X, then the coefficients of are bounded by X.

With this lemma, we now have all ingredients to actually define the sequence (h n ). Let be the compact set of all second-degree polynomials of the form h=I d+O(2) with second-degree coefficients bounded by X. By Lemmas 36 and 39 and Corollary 40, a choice of uniquely determines a map . For every i, we define
Each V i is a decreasing intersection of non-empty compact sets and hence non-empty and compact. By continuity of each ψ i , it follows that

Therefore, there exist an inverse orbit of the sequence ψ 0,ψ 1,…. That is, a sequence h 0,h 1,… in S with ψ i (h i+1)=h i . Thus, we have proved the following.

Theorem 41.

We can find sequences (h n ) and (g n ) whose coefficients are bounded by X at the start and end of every train j, where jj 0.

Step 2: Properties of these new coordinates.

What remains to show is that with the construction from the previous step gives a basin which is on the one hand equal to all of and on the other hand equivalent to the basin . In order to draw these conclusions, we will need finer estimates on the size of the coefficients of the maps h n and g n .

Lemma 42.

For n[p j ,p j+1), we have

where K 3=K 3(C,D,k,x) as in Lemma 38.

Proof.

First, note that
Now suppose p j nstp j+1. As in the proof of Lemma 38, we can distinguish several cases:
  1. (I)

    q j stp j+1

     
  2. (II)

    p j sq j tp j+1

     
  3. (III)

    p j stq j

     
In case (I), the proof of Lemma 38 gives us Equation (43), which implies that
(45)
For case (II), we can use Equation (43) again to see that
Next, Theorem 28 shows that
Combining these estimates gives
(46)
For the remaining case (III), we can once again apply Theorem 28:
(47)
On the other hand, we have the easy estimate
(48)
Note that as t-s increases, estimate (47) will decrease while estimate (48) will increase. When , we can therefore use estimate (48) to get
And when , estimate (47) shows that
since and . So we can conclude case (III) with
(49)
Finally, note that by Lemma 24. Combining this with Equations (45), (46), and (49) proves that

Lemmas 42 and 36 and Corollary 37 now give us an easy estimate on the coefficients of our maps and g n .

Corollary 43.

For n[p j ,p j+1] and jj 0, all quadratic coefficients of the maps g n and are bounded by
Since there are only finitely many functions for j<j 0, we can now pick a large constant Z such that the quadratic coefficients of all maps g n and are bounded by

Using these estimates, we can find :

Lemma 44.

The basin of the sequence as constructed is equal to .

Proof.

Let and write G j,i =G j-1G i as usual. For n[p j ,p j+1), we have , where L n is the linear part of f. Hence, we have
By Corollary 37, we know that
for i[p j ,p j+1). Therefore, we find
where the last step follows by Lemma 24(iii). Given we write R j =G j,0(z). Since M j is unitary, we obtain
Hence, we have the following two cases.
  1. (a)

    If : .

     
  2. (b)

    If : .

     
Let j 1 be such that . Then, we can derive from (a) that for jj 1 and , we have

Hence if for some jj 1, then the same holds for all larger j.

Suppose for the purpose of a contradiction that for each jj 1. Then for jj 1, we have:
Hence for all jj 1 sufficiently large, we have

This contradicts the assumption that for each jj 1. Hence, there exists j z j 1 such that for all jj z .

Now let n[p j ,p j+1) and jj z , and recall that
Using Lemma 42 and its proof, we find
and hence
For jj z and n[p j ,p j+1), we therefore find that

Since , we see that g n,1(z)→0 as n, which completes the proof.

Step 3: The biholomorphism.

What is left to show is that the basins of the sequences
and
are equivalent. We can simplify notation by composing the matrices M j with neighboring functions. Define:
(50)
(51)
(52)
Hence from now on, we need to study the basins of the sequences and . We note that the sequence still satisfies
for . The following diagram commutes up to degree two:
As usual, we define
It will also be convenient to write

Our goal is to show that the sequence (Φ n ) converges, uniformly on compact subsets of the basin , to an isomorphism from to . In order to accomplish this, we need a number of estimates.

Pick a constant λ<1 such that D k <λ C, and define the radius

All estimates for , and will be valid on .

Lemma 45.

For and , we have

Proof.

Writing , we obtain
And hence for two points , we get
Using our assumptions on λ<1, z and w, we have

Since the matrices M j are unitary, we immediate obtain the same estimates for the maps .

Lemma 46.

Let . If , we have

The proof follows immediately from our estimates on the coefficients of the maps h n .

Lemma 47.

There exists a constant M>0 such that for all and , we have

Proof.

Recall that the linear and quadratic parts of the map are exactly equal to . Therefore, we need to find estimates on the higher order terms of . Since the matrices M j are unitary, we can work with the f’s, g’s, and h’s instead.

Since the maps f n are uniformly attracting, it follows from the Cauchy estimates that the degree m terms of f n are bounded by .

A tedious but straightforward computation shows that the third-order part is bounded by
and the part of degree ≥4 is bounded by

Plugging in our bound on z, we obtain the required estimate.

Our next goal is to combine the estimates above to prove that φ n,m is close to on some neighborhood of the origin. To achieve this, we need to make sure that we can apply our estimates every step of the way. Therefore, we define a new constant

to give ourselves some wiggle room.

Since and , we have . Therefore increases exponentially and D s n <s n+1. So if , then . This is the point where we’ve really used that .

Write for . Next, let be minimal such that
(53)

For any mv(u) and zV u , we now have .

Lemma 48.

For all and , we have
  1. (i)

     
  2. (ii)

     

Proof.

We prove both statements simultaneously by induction on n. For n=0 the first statement is trivial, and the second statement follows immediately from Lemma 46.

Now assume that is such that both statements hold for all and . We fix and .

We have , so by our induction hypothesis, we know that the lemma also holds for n, m+1 and . Since , Lemma 46 shows that . This puts us into position to apply Lemma 45:
Since ws m r m , we can also apply Lemma 47 (and the triangle inequality) to obtain
Combining this with Lemma 46, we find that

Theorem 49.

The sequence (Φ n ) converges uniformly on each V u .

Proof.

First, fix η>0. Since the function is uniformly continuous on , we can find θ such that for all :
Now pick such that
Fix zV u and nN. For any mv(u), we have . In this setting, Lemma 48(i) shows that
(54)
And for m=v(u),v(u)+1,…,v(u)+N, Lemma 48(ii) gives us
and
If we apply up to in Equation (54) and use Lemma 45 repeatedly, we obtain:
which means that
Since both and lie in by Lemma 48(ii), the definition of θ gives:

For m,nN and zV u , we therefore have Φ n (z)-Φ m (z)<η, which proves the uniform convergence on V u .

Since , Theorem 49 shows that the maps Φ n converge uniformly on compact subsets to a map . We will now prove that this limit Φ maps biholomorphically onto . The maps Φ n are compositions of a number of global biholomorphisms, plus a holomorphic map which is injective on a small ball whose radius decreases with n. We first estimate the size of these radii

Lemma 50.

For and zr n , we have
  1. (i)

     
  2. (ii)

    is injective, and

     
  3. (iii)

    .

     

Proof.

All three statements follow from the estimates on the coefficients of h n found earlier and the fact that

Corollary 51.

The map Φ is a biholomorphism.

Proof.

Let and nv(u). Then for zV u , we have . By Lemma 50(ii) and the fact that all and are biholomorphisms, the map Φ n must be injective on V u . Therefore, is a uniform limit of biholomorphisms, and we know that D Φ n (0)=I for all n. By Hurwitz’ theorem, we can conclude that is also a biholomorphism. The statement follows since is the increasing union of the sets V u .

The final ingredient in the proof of Theorem 4 is to show that is surjective.

Lemma 52.

For , we have

Proof.

By Lemmas 50(iii) and 48(i), it follows that
for all . Now note that

By Lemma 50(i), and the fact that all and are biholomorphisms, the map φ m+n,m has a nonzero Jacobian on and therefore is an open mapping. As φ m+n,m (0)=0, the lemma follows.

Lemma 53.

The biholomorphism is surjective.

Proof.

Let . We will show that w lies in the image of Φ.

Since , we can find such that . Then, we have
By Lemma 52, we now find that
for any . This implies that
for any . By the compactness of and the uniform convergence of (Φ n ) on compact sets, we have that

With this lemma, we have proved Theorem 4.

Authors’ Affiliations

(1)
KdV Institute for Mathematics, University of Amsterdam

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© Peters and Smit; licensee Springer. 2015

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.