A transcendental Hénon map with an oscillating wandering Short ${\mathbb {C}}^2$

Abstract

Short ${\mathbb {C}}^2$’s were constructed in [5] as attracting basins of a sequence of holomorphic automorphisms whose rate of attraction increases superexponentially. The goal of this paper is to show that such domains also arise naturally as autonomous attracting basins: we construct a transcendental Hénon map with an oscillating wandering Fatou component that is a Short ${\mathbb {C}}^2$. The superexponential rate of attraction is not obtained at single iterations, but along consecutive oscillations.

Appendix: Calibrated basins and the plurisubharmonic method

As a further illustration of using the plurisubharmonic method to prove that an attracting basin equals the Fatou component containing the basin, we take a closer look at the calibrated basins constructed in [10]. Let $f_0, f_1,f_2 \ldots$ be a sequence of automorphisms of ${\mathbb {C}}^k$, all having an attracting fixed point at the origin. For all $j\ge 0$ there exists a radius $r_j>0$, a constant $0<\mu _j<1$ and a constant $C_j>0$ such that

$$\begin{aligned} \Vert f_j^n(z)\Vert \le C_j\mu _j^n\Vert z\Vert ,\quad \forall z\in B(0,r_j), n\ge 0. \end{aligned}$$

We can choose $n_j$ large enough to obtain

$$\begin{aligned} f_j^{n_j} (B(0, r_j)) \subset B(0,r_{j+1}), \end{aligned}$$

and $r_j\rightarrow 0$. We then define the calibrated basin $\Omega _{(r_j), (n_j)}$ by

$$\begin{aligned} \Omega _{(r_j),(n_j)} := \bigcup _{j \in {\mathbb {N}}} (f_{j-1}^{n_{j-1}}\circ \dots \circ f_0^{n_0})^{-1} (B(0, r_j)). \end{aligned}$$

It is easy to see that the calibrated basin may depend on the sequences $(r_j)$ and $(n_j)$ chosen.

Recall the following result from [10]:

Theorem 7

Fix the sequence $(r_j)$. For $n_0, n_1, \ldots$ sufficiently large, where each $n_j$ may depend on the choices of $n_0, \ldots , n_{j-1}$, the calibrated basin of the sequence $f_0^{n_0}, f_1^{n_1}, \ldots$ is biholomorphic to ${\mathbb {C}}^k$.

Recall that it may be necessary to replace the maps $f_j$ by large iterates: all assumptions (with $r_j = \frac{1}{2}$) are satisfied by the maps in [5], see Proposition 3, but in this case the calibrated basin is a Short ${\mathbb {C}}^k$, and hence not equivalent to ${\mathbb {C}}^k$.

One may wonder whether it is necessary to work with the calibrated basin instead of the basin that contains all points whose orbits converge to 0. It turns out that this may indeed be necessary: for suitable choices of the sequence $f_0, f_1, \ldots$ the full basin may not be open, even when replacing the maps with arbitrarily high iterates $f_0^{n_0}, f_1^{n_1}, \ldots$, see [10]. This raises another natural question: is the calibrated basin equal to the Fatou component ${\mathcal {F}}_0$ containing the origin, that is, the largest connected open set with locally uniform convergence to 0? Here we prove that this is indeed the case.

Theorem 8

Fix the sequence $(r_j)$. For $n_0, n_1, \ldots$ sufficiently large we have

$$\begin{aligned} {\mathcal {F}}_0 = \Omega _{(r_j),(n_j)}. \end{aligned}$$

Proof

Define

$$\begin{aligned} G_j(z) := \frac{\log \Vert f_j^{n_j} \circ \cdots \circ f_0^{n_0}(z)\Vert }{-n_j \log \mu _j}. \end{aligned}$$

Observe that, given a compact $K\subset {\mathcal {F}}_0$, for big j we have that $f_j^{n_j} \circ \cdots \circ f_0^{n_0}(K)\subset {\mathbb {B}}^k$, and thus that $G_j|_K\le 0$. Let

$$\begin{aligned} G = \limsup _{j\rightarrow \infty } G_j, \end{aligned}$$

and write $G^\star$ for the upper semi-continuous regularization of G. It follows that $G^\star$ is plurisubharmonic on ${\mathcal {F}}_0$.

If we choose $n_j$ sufficiently large we may assume that

$$\begin{aligned} \frac{\log r_{j+1}}{ n_j\log \mu _j}\longrightarrow 0, \end{aligned}$$

hence for any point $z \in {\mathcal {F}}_0{\setminus } \Omega _{(r_j),(n_j)}$ we have $G^\star (z) = G(z) = 0$.

On the other hand, let $z\in \Omega _{(r_j),(n_j)},$ and let $j\ge 0$ be large enough such that $z_j:=f_{j-1}^{n_{j-1}} \circ \cdots \circ f_0^{n_0}(z)\in B(0,r_j)$. Then

$$\begin{aligned} G_j(z)=\frac{\log \Vert f_j^{n_j}(z_j)\Vert }{-n_j\log \mu _j}\le \frac{\log C_j+\log \Vert z_j\Vert +n_j\log \mu _j}{-n_j\log \mu _j}. \end{aligned}$$

Choosing the $n_j$’s large enough we thus obtain $G(z)\le -1$ for all $z \in \Omega _{(r_j),(n_j)}$, which implies that $G^\star (z)\le -1$ for all $z \in \Omega _{(r_j),(n_j)}$. Since ${\mathcal {F}}_0$ is open and connected, it follows from the maximum principle that ${\mathcal {F}}_0{\setminus }\Omega _{(r_j),(n_j)}$ must be empty, which completes the proof. $\square$