On the dynamics of rational maps with two free critical points

Abstract

In this paper we discuss the dynamical structure of the rational family $(f_t)$ given by

$$\begin{aligned} f_t(z)=tz^{m}\left( \frac{1-z}{1+z}\right) ^{n}\quad (m\ge 2,~n\in \mathbb N,~t\in \mathbb C{\setminus }\{0\}). \end{aligned}$$

Each map $f_t$ has super-attracting immediate basins $\mathscr {A}_t$ and $\mathscr {B}_t$ about $z=0$ and $z=\infty$, respectively, and two free critical points. We prove that $\mathscr {A}_t$ (for $0<|t|\le 1$) and $\mathscr {B}_t$ (for $|t|\ge 1$) are completely invariant, and at least one of the free critical points is inactive. Based on this separation we draw a detailed picture of the structure of the dynamical and the parameter plane.

1 Introduction

Non-trivial rational families $(f_t)$ normally contain specific maps of different character with most interesting and unexpected Julia sets:

• totally disconnected Julia sets (Cantor sets) occur in any family $z\mapsto z^d+t$;

• Julia sets consisting of uncountably many (a Cantor set of) quasi-circles occur in the McMullen family $z\mapsto z^m+t/z^n$, which was introduced in [8]. The number of papers on various features of this family is legion; [3] marks the preliminary end of a long list of papers.

• Julia sets that are Sierpiński curves (Milnor and Tan Lei  [12] were the first to construct examples with this property) occur again in the McMullen family [16], the Morosawa-Pilgrim family $z\mapsto t\left( 1+\frac{(4/27)z^3}{1-z}\right)$ [4, 17], and the family $t\mapsto -\frac{t}{4}\frac{(z^2-2)^2}{z^2-1}$ [7].

• In any reasonable family, copies of the Mandelbrot sets of the families $z\mapsto z^d+t$ are dense in the bifurcation locus—the Mandelbrot set is universal [10].

Each of these families has just one free critical point (or several free critical points which have the same dynamical behaviour, this happens, for example, in the McMullen family; the quasi-conjugated family $F_t(z)=z^m(1+t/z)^d$ has just one free critical point). In contrast to this the rational maps

$$\begin{aligned} f_t(z)=tz^m\left( \frac{1-z}{1+z}\right) ^n\quad (m\ge 2,~n\in \mathbb N,~d=m+n,~t\ne 0) \end{aligned}$$

(1)

in the family under consideration have two free critical points. In this paper we will give a complete description of the parameter plane and the various dynamical planes. For basic notations and results the reader is referred to the texts [1, 2, 9, 11, 15].

2 Notation

The rational map (1) has

• two super-attracting fixed points 0 and $\infty$ with corresponding basins $\mathscr {A}_t$ and $\mathscr {B}_t$, respectively. Then $\mathscr {A}_t$, say, either is completely invariant or else has a single pre-image $\mathscr {A}_t^*$ that is mapped in a (n : 1)-manner onto $\mathscr {A}_t$, which will be written as

  $$\begin{aligned} \mathscr {A}_t^*\buildrel n:1 \over \longrightarrow \mathscr {A}_t; \end{aligned}$$

• two free critical points

  $$\begin{aligned} \qquad \alpha =-\frac{n}{m}+\sqrt{1+\left( \frac{n}{m}\right) ^2}\quad \mathrm{and}\quad \beta =-\frac{n}{m}-\sqrt{1+\left( \frac{n}{m}\right) ^2} \end{aligned}$$

  and critical values

  $$\begin{aligned} v^\alpha _t=f_t(\alpha )=tv^\alpha _1\quad \mathrm{and}\quad v^\beta _t=f_t(\beta )=t v^\beta _1; \end{aligned}$$

• two escape loci $\mathbf {\Omega }^\alpha$ and $\mathbf {\Omega }^\beta$, with $t\in \mathbf {\Omega }^\alpha$ and $t\in \mathbf {\Omega }^\beta$ if and only if $f^k_t(\alpha )\rightarrow 0$ and $f^k_t(\beta )\rightarrow \infty$, respectively, as $k\rightarrow \infty$;

• two residual sets $\mathbf {\Omega }_\mathrm{res}^\alpha$ and $\mathbf {\Omega }_\mathrm{res}^\beta$, with $t\in \mathbf {\Omega }_\mathrm{res}^\alpha$ and $t\in \mathbf {\Omega }_\mathrm{res}^\beta$ if and only if $v^\beta _t\in \mathscr {A}_t$ and $v^\alpha _t\in \mathscr {B}_t$, respectively.

The notation of the residual sets indicates that $\mathbf {\Omega }_\mathrm{res}^\alpha$ is related to $\mathbf {\Omega }^\alpha$ rather than $\mathbf {\Omega }^\beta$. The open sets $\mathbf {\Omega }^\alpha$ and $\mathbf {\Omega }^\beta$ are in a natural way sub-divided into

• $\mathbf {\Omega }_0^\alpha$ resp. $\mathbf {\Omega }_0^\beta$: $v^\alpha _t\in \mathscr {A}_t$ resp. $v^\beta _t\in \mathscr {B}_t$, and

• $\mathbf {\Omega }_k^\alpha$ resp. $\mathbf {\Omega }_k^\beta$: $f^k_t\left( v^\alpha _t\right) \in \mathscr {A}_t$, but $f^{k-1}_t\left( v^\alpha _t\right) \notin \mathscr {A}_t$ resp. $f^k_t\left( v^\beta _t\right) \in \mathscr {B}_t$, but $f^{k-1}_t\left( v^\beta _t\right) \notin \mathscr {B}_t$ ($k\ge 1)$.

Hitherto, $f_t$ is hyperbolic and the Fatou set of $f_t$ consists of the basins $\mathscr {A}_t$ and $\mathscr {B}_t$, and their pre-images, if any. However, there may and will be also other hyperbolic components. By $\mathbf {W}_k^\alpha$ and $\mathbf {W}^\beta _k$ we denote the open sets such that $\alpha$ and $\beta$ belongs to some (super-)attracting cycle of Fatou domains $U_1,\ldots ,U_k$, respectively, not containing 0 and $\infty$.

The bifurcation locus $\mathbf {B}$ of the family $(f_t)_{0<|t|<\infty }$ is the set of t such that the Julia set $\mathscr {J}_t$ does not move continuously over any neighbourhood of t, see McMullen [9]. In order that $t\in \mathbf {B}$ it is necessary and sufficient that at least one of the free critical points is active. Thus $\mathbf {B}=\mathbf {B}^\alpha \cup \mathbf {B}^\beta$, where $t\in \mathbf {B}^\alpha$ resp. $t\in \mathbf {B}^\beta$ means that $\alpha$ resp. $\beta$ is active. It is a priori not excluded that $\mathbf {B}^\alpha$ and $\mathbf {B}^\beta$ overlap. Although there is just one parameter plane, each point of this plane carries at least two pieces of information, so one could also speak of the $v^\alpha _t$- and $v^\beta _t$-plane.

We also set

$$\begin{aligned} Q_0(t)=v^\alpha _t=tv^\alpha _1\quad \mathrm{and}\quad Q_k(t)=f_t^k(v^\alpha _t)=f_t(Q_{k-1}(t))\quad (k\ge 1) \end{aligned}$$

and note that $Q_k$ is a rational function of degree $1+d+\cdots +d^k=\frac{d^{k+1}-1}{d-1}$ with a zero of order $\frac{m^{k+1}-1}{m-1}$ at the origin.

From

$$\begin{aligned} -1/f_t(-1/z)=f_{(-1)^{d+1}/t}(z)\quad (d=m+n) \end{aligned}$$

it follows that $f_t$ is conjugated to $f_{1/t}$ if d is odd, and to $f_{-1/t}$ if d is even, hence $t\in \mathbf {\Omega }^\alpha$ if and only if $(-1)^{d+1}/t\in \mathbf {\Omega }^\beta$, and this is also true for $\mathbf {\Omega }^\alpha _{k}$ and $\mathbf {\Omega }^\beta _{k}$, $\mathbf {\Omega }^\alpha _\mathrm{res}$ and $\mathbf {\Omega }^\beta _\mathrm{res}$, $\mathbf {W}_k^\alpha$ and $\mathbf {W}_k^\beta$, and $\mathbf {B}^\alpha$ and $\mathbf {B}^\beta$. This also indicates that the circle $|t|=1$ plays a distinguished role with strong impact on what follows.

Lemma 1

For every $m\ge 2$, $n\ge 1$ there exists some $r>0$, such that for $0<|t|\le 1$ the disc $\triangle _{r|t|}:|z|<r|t|$ contains $f_t(\overline{\triangle }_{r|t|}\cup [0,1])$, but does not contain $v^\beta _t$.

Proof

We will first consider $f_1$ and show that there exists some disc $\triangle _{r}:|z|<r$ such that $f_1(\overline{\triangle }_r\cup [0,1])\subset \triangle _r$ holds. This is easy to show if $n<m$ for $r=\frac{1}{3}$:

$$\begin{aligned} |f_1(z)|\le 3^{-m}2^n<\textstyle \frac{1}{3} \end{aligned}$$

holds if $|z|\le \frac{1}{3}$ and $m>n\ge 1$, and from

$$\begin{aligned} 0\le f_1(x)\le x^2\frac{1-x}{1+x}\le \textstyle \frac{1}{2}\left( 5\sqrt{5}-11\right) <\frac{1}{10}\quad (0\le x\le 1) \end{aligned}$$

the assertion follows.

We now consider the case $n\ge m$. Then $f_1(\overline{\triangle }_r)\subset \triangle _r$ holds as long as

$$\begin{aligned} g(r)=r^{m-1}\left( \frac{1+r}{1-r}\right) ^n<1, \end{aligned}$$

and $f_1$ maps [0, 1] into $\triangle _r$ provided

$$\begin{aligned} v^\alpha _1=\max _{0\le x\le 1}x^m\left( \frac{1-x}{1+x}\right) ^n<r. \end{aligned}$$

Since g is increasing this may be achieved if $g(v^\alpha _1)<1$ holds. To prove this we note that $\sqrt{1+\tau }-1=\frac{\tau }{2\sqrt{1+\theta \tau }}$ $(0<\theta <1,$ $\tau =\frac{m^2}{n^2}\le 1)$ implies $\frac{m}{2\sqrt{2}n}<\alpha <\frac{m}{2n}$, while from $\log \frac{1-x}{1+x}< -2x$ ($0<x<1$) it follows that

$$\begin{aligned} v^\alpha _1<\left( \frac{m}{2n}\right) ^me^{-2\frac{m}{2\sqrt{2}}}=\left( \frac{m}{2e^{\frac{1}{\sqrt{2}}} n}\right) ^m<\left( \frac{m}{4n}\right) ^m=\mu ^m. \end{aligned}$$

Moreover, from

$$\begin{aligned} \log \frac{1+x}{1-x}=2x\left( 1+\frac{1}{3}x^2+\frac{1}{5}x^4+\cdots \right) \le 2x\left( 1+\frac{x^2}{3}\frac{1}{1-x^2}\right) \le \textstyle 2x\left( 1+\frac{1}{45}\right) , \end{aligned}$$

which holds for $x=\left( \frac{m}{4n}\right) ^{m-1}\le \frac{1}{4}$, we obtain

$$\begin{aligned} \left( \frac{1+\mu ^m}{1-\mu ^m}\right) ^n=\left( \frac{1+\frac{m}{4}\frac{\mu ^{m-1}}{n}}{1-\frac{m}{4}\frac{\mu ^{m-1}}{n}}\right) ^n\le e^{\frac{23}{45}m\mu ^{m-1}}<\left( e^{(\frac{m}{4n})^{m-1}}\right) ^m. \end{aligned}$$

Thus $g(v^\alpha _1)<1$ follows from $\left( \frac{m}{4n}\right) ^{m-1}e^{(\frac{m}{4n})^{m-1}}\le \frac{1}{4}e^{\frac{1}{4}}<1$.

With this choice of $r\in (0,1)$ it is clear that $v^\beta _t$ belongs to $\triangle _r$ if |t| is small. For individual $0<|t|\le 1$, $f_t(z)=tf_1(z)$ maps $\overline{\triangle }_{r|t|}\cup [0,1]$ into $\triangle _{r|t|}$, while $v^\beta _t\notin \triangle _{r|t|}$ follows from $|v^\beta _t|=|t|/v^\alpha _1>|t|>r|t|$. $\Box$

3 The escape loci

The purpose of Lemma 1 is twofold. First of all it shows that the critical points $\alpha$ and $\beta$ cannot be simultaneously active, and the bifurcation sets $\mathbf {B}^\alpha$ and $\mathbf {B}^\beta$ are separated by the unit circle $|t|=1$. Secondly, the condition $v^\beta _t\notin \triangle _{r|t|}$ ($0<|t|\le 1)$ ensures that in an exhaustion $(D_\kappa )$ of $\mathscr {A}_t$ starting with $D_0=\triangle _{r|t|}$, $D_\kappa$ is simply connected as long as $\beta \notin D_\kappa$, and $f_t:D_{\kappa }\buildrel d:1 \over \longrightarrow D_{\kappa -1}$ has degree $d=m+n$. In particular, for $t\in \mathbf {\Omega }^\alpha _\mathrm{res}$ there exists some simply connected and forward invariant domain $D_\kappa \subset \mathscr {A}_t$ that contains $v^\beta _t$ (Figs. 1, 2).

We note some more simple consequences of Lemma 1; our focus is on the critical point $\alpha$ and the ${}^\alpha$-sets.

• $\{t:0<|t|\le 1\}\subset \mathbf {\Omega }^\alpha _0$;

• $\overline{\mathbf {\Omega }^\alpha _\mathrm{res}}\subset \mathbb D;$

• $\alpha$ is inactive on $0<|t|\le 1$;

• $\overline{\bigcup _{k\ge 1}(\mathbf {\Omega }^\alpha _k\cup \mathbf {W}^\alpha _k)}\subset \{t:1<|t|<T\}$ for some $T=T_{mn}>1$;

• $\mathbf {B}^\alpha \subset \{t:1<|t|<T\}$ for some $T=T_{mn}>1$.

The consequences for the dynamical planes are as follows.

Theorem 1

For $t\in \mathbf {\Omega }^\alpha _0$, the basin $\mathscr {A}_t$ is completely invariant, and any other Fatou component is simply connected. Moreover,

• for $t\in \mathbf {\Omega }^\alpha _0\cap \mathbf {\Omega }^\beta _0$ also $\mathscr {B}_t$ is completely invariant, the Julia set $\mathscr {J}_t=\partial \mathscr {A}_t=\partial \mathscr {B}_t$ is a quasi-circle, and $f_t$ is quasi-conformally conjugated to $z\mapsto z^d$;

• for $t\in \mathbf {\Omega }^\alpha _\mathrm{res}$, $\mathscr {A}_t$ is infinitely connected and the Fatou set consists of $\mathscr {A}_t$, $\mathscr {B}_t$, and the predecessors of $\mathscr {B}_t$ of any order.

Proof

To prove complete invariance of $\mathscr {A}_t$ we first assume $0<|t|\le 1$. Then $\mathscr {A}_t$ contains the interval [0, 1] by Lemma 1, hence is completely invariant. If, however, $|t|>1$, then $\mathscr {B}_t$ is completely invariant, and any other Fatou component is simply connected. Assuming $1\not \in \mathscr {A}_t$ ($t\in \mathbf {\Omega }^\alpha _0$, $|t|>1$) we obtain either $f_t:\mathscr {A}_t^*\buildrel n:1 \over \longrightarrow \mathscr {A}_t$ with $n=(n-1)+1$ critical points if $\alpha \in \mathscr {A}_t^*$ or else $f_t:\mathscr {A}_t\buildrel m:1 \over \longrightarrow \mathscr {A}_t$ with $m=(m-1)+1$ critical points if $\alpha \in \mathscr {A}_t$, this contradicting simple connectivity of both domains $\mathscr {A}_t$ and $\mathscr {A}_t^*$ by the Riemann–Hurwitz formula.

The first assertion is obvious since $\mathscr {B}_t$ shares the properties of $\mathscr {A}_t$ and $f_t$ is hyperbolic.

The second assertion follows from the Riemann-Hurwitz formula, since $f_t:\mathscr {A}_t\buildrel d:1 \over \longrightarrow \mathscr {A}_t$ has degree d and $r=(m-1)+(n-1)+1+1=d$ critical points 0, 1 (if $n>1$), $\alpha$, and $\beta$.$\Box$

Fig. 1

Left the $\alpha$-parameter plane for $\displaystyle f_t(z)=tz^2\frac{1-z}{1+z}$ displaying the unit circle, $\mathbf {\Omega }^\alpha$ (gray), $\mathbf {\Omega }^\alpha _\mathrm{res}$ and $\mathbf {\Omega }^\beta _\mathrm{res}$ (white, in and outside the unit circle), and $\mathbf {W}^\alpha$ (black). Right a neighbourhood of the origin displaying $\mathbf {\Omega }^\alpha _\mathrm{res}$ (gray) surrounded by points of $\mathbf {\Omega }^\alpha _0$ (white), $\mathbf {\Omega }^\beta _k$ ($k\ge 1$, white, small), and $\mathbf {W}^\beta$ (black)

Fig. 2

Left the parameter plane of $P_c(z)=cz^2(z+1)$. The escape region for $P_c$ (gray), the white region with slit, and the black regions correspond to $\mathbf {\Omega }^\alpha _\mathrm{res}$, $\mathbf {\Omega }^\beta \cap \mathbb D$, and $\mathbf {W}^\beta$, in case of $m=2$, $n=1$, respectively. The punctured disc $0<|t|<1$ corresponds to $\mathbb C{\setminus }[-2,0]$ in the c-plane. Right the parameter plane of $P_{-\frac{1}{2}(t+2+\frac{1}{t})}(z)$ in $-0.2<\mathrm{Re}\;t<0.25,$ $-0.25<\mathrm{Im}\;t<0.25$ (see also Fig. 1 right)

Theorem 2

$\mathbf {\Omega }^\alpha _0\cup \{0\}$, $\mathbf {\Omega }^\alpha _\mathrm{res}\cup \{0\}$, and the connected components of $\mathbf {\Omega }^\alpha _k$ $(k\ge 1)$ are simply connected domains. Riemann maps onto $\mathbb D$ are given by any branch of $\root m \of {E_0(t)}$, $\root m \of {E_\mathrm{res}(t)}$, and $\root n \of {E_{k}(t)}$, respectively.

For the proof we need two auxiliary results on the maps

$$\begin{aligned} \begin{array}{rcll} E_0(t)&{}=&{}t\left( \Phi _t(v^\alpha _t)\right) ^{m-1}&{}\left( t\in \mathbf {\Omega }^\alpha _0\right) ,\\ E_\mathrm{res}(t)&{}=&{}t\left( \Phi _t(v^\beta _t)\right) ^{m-1}&{}\left( t\in \mathbf {\Omega }^\alpha _\mathrm{res}\right) , \mathrm{~and}\\ E_k(t)&{}=&{}t^{\frac{1}{m-1}}\Phi _t\left( f^k(v^\alpha _t)\right) &{}\left( t\in \mathbf {\Omega }^\alpha _k,~k\ge 1\right) , \end{array}\end{aligned}$$

(2)

where $\Phi _t$ denotes the Böttcher function to the fixed point $z=0$. In the first step (Lemma 2) of the proof of Theorem 2 we will show that the functions (2) provide proper maps on $\mathbb D{\setminus }\{0\}$ and $\mathbb D$, respectively, which are only ramified over the origin. In the second step (Lemma 3) this will be used to show that the corresponding domains (with 0 included, if necessary) are simply connected.

The solution to Böttcher’s functional equation

$$\begin{aligned} \Phi _t(f_t(z))=t\Phi _t(z)^m\quad (\Phi _t(z)\sim z \mathrm{~as~}z\rightarrow 0) \end{aligned}$$

(3)

is locally given by

$$\begin{aligned} \Phi _t(z)=\lim _{k\rightarrow \infty }\root m^k \of {f_t^k(z)/t^{1+m+\cdots +m^{k-1}}}=t^{-\frac{1}{m-1}}\lim _{k\rightarrow \infty }\root m^k \of {f_t^k(z)}; \end{aligned}$$

it conjugates $f_t$ to $\zeta \mapsto \zeta ^m$. This conjugation holds throughout $\mathscr {A}_t$ in the third case, when $\Phi _t$ maps $\mathscr {A}_t$ conformally onto the disc $|z|<|t|^{-\frac{1}{m-1}}$; the maps $E_k$ are analytic and well-defined on the components of $\mathbf {\Omega }^\alpha _k$, $k\ge 1$.

In the first case the conjugation holds on some simply connected neighbourhood of $z=0$ that contains $z=0$ and $z=v^\alpha _t$, but does not contain $z=1$. The analytic continuation of $\Phi _t$ causes singularities at $z=1$ and its preimages under $f_t^k$, nevertheless $|\Phi _t(z)|$ is well-defined on $\mathscr {A}_t$ and $|\Phi _t(z)|\rightarrow |t|^{-\frac{1}{m-1}}$ as $z\rightarrow \partial \mathscr {A}_t$ holds anyway. Thus $E_0(t)=t\Phi _t(v^\alpha _t)^{m-1}$ is holomorphic on $\mathbf {\Omega }^\alpha _0$ and zero-free, with $E_0(t)\sim t(v^\alpha _t)^{m-1}=f_1(\alpha )^{m-1}t^m$ as $t\rightarrow 0$.

In the second case we construct an exhaustion $(D_\kappa )$ of $\mathscr {A}_t$ such that $f_t:D_\kappa \buildrel d:1 \over \longrightarrow D_{\kappa -1}$ has degree d and $D_\kappa$ is simply connected for $\kappa \le \kappa _0$ with $v^\beta _t\in D_{\kappa _0}$ and $\beta \in D_{\kappa _0+1}{\setminus } D_{\kappa _0}$. This is possible by Lemma 1, and the procedure applied to $t^{-\frac{1}{m-1}}\Phi _t(v^\alpha _t)$ on $\mathbf {\Omega }^\alpha _{0}$ also applies to $t^{-\frac{1}{m-1}}\Phi _t(v^\beta _t)$ on $\mathbf {\Omega }^\alpha _\mathrm{res}$.

Lemma 2

The functions in (2) are well-defined and provide proper maps from $\mathbf {\Omega }^\alpha _0\cup \{0\}$, $\mathbf {\Omega }^\alpha _\mathrm{res}\cup \{0\}$, and the connected components of $\mathbf {\Omega }^\alpha _k$ with $k\ge 1$, respectively, onto the unit disc $\mathbb D$.

Proof

To prove that $|E_0(t)|\rightarrow 1$ as $t\in \mathbf {\Omega }^\alpha _0$ tends to $\partial \mathbf {\Omega }^\alpha _0{\setminus }\{0\}$ we choose any disc $\triangle _r:|z|<r$ that is invariant under $f_t$ for every $t\in \mathbf {\Omega }^\alpha _0$. This is possible since $\mathbf {\Omega }^\alpha _0$ is contained in some disc $|t|<T$, hence we may choose $r<1$ such that $Tr^{m-1}\left( \frac{1+r}{1-r}\right) ^n=1$ holds. By $k=k(t)$ we denote the largest integer such that $f_t^k(v^\alpha _t)\not \in \triangle _r$. Then $k(t)\rightarrow \infty$ as $t\rightarrow \partial \mathbf {\Omega }^\alpha _0{\setminus }\{0\}$, and $|f_t^{k(t)}(v^\alpha _t)|\ge r$ implies

$$\begin{aligned} \liminf _{t\rightarrow \mathbf {\Omega }^\alpha _0{\setminus }\{0\}} |\Phi _t(v^\alpha _t)|\ge \lim _{t\rightarrow \mathbf {\Omega }^\alpha _0{\setminus }\{0\}} |t|^{-\frac{1}{m-1}}\root m^{k(t)} \of {r}=|t|^{-\frac{1}{m-1}}, \end{aligned}$$

while $|\Phi _t(z)|<|t|^{-\frac{1}{m-1}}$ is always true. Thus $E_0$ maps each connected component of $\mathbf {\Omega }^\alpha _0$ properly onto $\mathbb D{\setminus }\{0\}$. It follows that the origin is removable for (a zero of) $E_0$, and $\mathbf {\Omega }^\alpha _0\cup \{0\}$ is a domain which is mapped by $E_0$ properly with degree m onto the unit disc $\mathbb D$.

If $t\in \mathbf {\Omega }^\alpha _k$ for some $k\ge 1$, then again $|E_k(t)|$ tends to 1 as $t\rightarrow \partial \Omega$, where $\Omega$ is any component of $\mathbf {\Omega }^\alpha _k$. Thus $E_k$ is a proper map of $\Omega$ onto $\mathbb D$. We will prove that $E_k$ is ramified only over zero even for $k\ge 0$, that is $E_k'(t)=0$ implies $E_k(t)=0$. This is a well-known procedure, the idea of which is due to Roesch [13], and outlined in detail for the Morosawa-Pilgrim family $z\mapsto t\left( 1+\frac{(4/27)z^3}{1-z}\right)$ in [17, Lemma 2].

We take any $t_0\in \mathbf {\Omega }^\alpha _k$ and choose $\varepsilon >0$ such that for t sufficiently close to $t_0$, the closed disc $\triangle _{3\epsilon }:|w-v^\alpha _{t_0}|\le 3\varepsilon$ belongs to the Fatou component $D_{t_0}$ of $f_{t_0}$ containing $v^\alpha _{t_0}$ ($D_{t_0}$ is a predecessor of $\mathscr {A}_{t_0}$ of order $\ell \ge 0$). Furthermore let $\eta _t:\widehat{\mathbb C}\longrightarrow \widehat{\mathbb C}$ be any diffeomorphism such that $\eta _t(w)$ depends analytically on t, $\eta _t(w)=w$ holds on $|w-v^\alpha _{t_0}|\ge 3\varepsilon$ and $\eta _t(w)=w+(v^\alpha _t-v^\alpha _{t_0})$ on $|w-v^\alpha _{t_0}|<\varepsilon .$ Then $g_t=\eta _t\circ f_{t_0}:\widehat{\mathbb C}\longrightarrow \widehat{\mathbb C}$ is a quasi-regular map which equals $f_{t_0}$ on $\widehat{\mathbb C}{\setminus } f^{-1}_{t_0}(\triangle _{3\epsilon })$, and is analytic on $\widehat{\mathbb C}{\setminus } f_{t_0}^{-1}(A)$ with $A=\{w:\varepsilon \le |w-v^\alpha _{t_0}|\le 3\varepsilon \}$. To apply Shishikura’s qc-lemma [14] we need to know that $g_t$ is uniformly K-quasi-regular, that is, all iterates $g_t^p$ are K-quasi-regular with one and the same K. This is obviously true if the sets $f_{t_0}^{-p}(A)$ ($p=1,2,\ldots$) are visited at most once by any iterate of $g_t$. This is trivial if $k\ge 1$: the sets $f_{t_0}^{-p}(A)$ belong to different Fatou components, namely predecessors of $D_{t_0}$ of order p. If $k=0$ the argument is different. Let $\triangle _0:|z|<\delta$ be such that $f_{t_0}(\overline{\triangle }_0)\subset \triangle _0$ and set $\triangle _\nu =f_{t_0}^{-1}(\triangle _{\nu -1})$. Then choosing $\epsilon >0$ sufficiently small we have $A\subset \triangle _{\ell }{\setminus }\overline{\triangle _{\ell -1}}$ for some $\ell$ and $f_{t_0}^{-p}(A)\subset \triangle _{\ell +p}{\setminus }\overline{\triangle _{\ell +p-1}}$. By the above mentioned qc-lemma, $g_t$ is quasi-conformally conjugated to some rational function

$$\begin{aligned} R_t=h_t\circ g_t\circ h_t^{-1}. \end{aligned}$$

The quasi-conformal mapping $h_t:\widehat{\mathbb C}\longrightarrow \widehat{\mathbb C}$ is uniquely determined by the normalisation $h_t(z)=z$ for $z=0,\alpha ,1$, and depends analytically on the parameter t. Also $h_t$ is analytic on $\widehat{\mathbb C}{\setminus }\overline{\bigcup _{p\ge 0}f_{t_0}^{-p}(A)}$, which, in particular, contains the points 0, $v^\alpha _t$, and $v^\alpha _{t_0}$. We set $z_0=h_t(-1)$ to obtain $R_t(z)=a(t)z^m\left( \frac{1-z}{z-z_0}\right) ^n.$ Since $h_t(\alpha )=\alpha$, $R_t$ has a critical point at $z=\alpha$, and solving $R_t'(\alpha )=0$ for $z_0$ yields $z_0=-1$, thus

$$\begin{aligned} R_t(z)=a(t)z^m\left( \frac{1-z}{1+z}\right) ^n. \end{aligned}$$

From $R_t=h_t\circ \eta _t\circ f_{t_0}$ and $h_t(\alpha )=\alpha$, however, it follows that

$$\begin{aligned} a(t)v^\alpha _1=R_t(\alpha )=h_t\circ \eta _t\circ f_{t_0}(\alpha )=h_t\circ \eta _t\left( v^\alpha _{t_0}\right) =h_t\left( v^\alpha _t\right) , \end{aligned}$$

hence $R_t(z)=f_\tau (z)$ with $\tau =\tau (t)={h_t(v^\alpha _t)}/{v^\alpha _1}$ and $v^\alpha _\tau =h_t(v^\alpha _t);$ in particular, $\tau$ depends analytically on t. On some neighbourhood of $z=0$ we have

$$\begin{array}{rcl} \displaystyle (t_0/\tau )^{\frac{1}{m-1}}\Phi _{t_0}\circ h_t^{-1}\circ f_\tau &{}=&{}\displaystyle (t_0/\tau )^{\frac{1}{m-1}}\Phi _{t_0}\circ g_t\circ h_t^{-1}\\ &{}=&{}\displaystyle (t_0/\tau )^{\frac{1}{m-1}}\Phi _{t_0}\circ \eta _t\circ f_{t_0}\circ h_t^{-1}\\ &{}=&{}\displaystyle (t_0/\tau )^{\frac{1}{m-1}}\Phi _{t_0}\circ f_{t_0}\circ h_t^{-1}\\ &{}=&{}\displaystyle (t_0/\tau )^{\frac{1}{m-1}}t_0\left( \Phi _{t_0}\circ h_t^{-1}\right) ^{m}\\ &{}=&{}\displaystyle \tau \left( (t_0/\tau )^{\frac{1}{m-1}}\Phi _{t_0}\circ h_t^{-1}\right) ^{m}, \end{array}$$

hence $\phi _\tau =(t_0/\tau )^{\frac{1}{m-1}}\Phi _{t_0}\circ h_t^{-1}$ solves Böttcher’s functional equation

$$\begin{aligned} \phi _\tau \circ f_\tau (z)=\tau (\phi _\tau (z))^m. \end{aligned}$$

Since $\tau$ and $h_t$ depend analytically on t, this is also true for $h_t^{-1}$, which is not self-evident. Also from $h_t(g_t(z))=f_\tau (h_t(z))\sim \tau h_t(z)^m$ and $g_t(z)=f_{t_0}(z)\sim t_0z^m$ as $z\rightarrow 0$ it follows that $h_t(t_0z^m)\sim \tau h_t(z)^m$, hence $h_t(z)\sim \root m-1 \of {t_0/\tau }z$, $h^{-1}_t(z)\sim \root m-1 \of {\tau /t_0}z$ and $\phi _\tau (z)\sim \lambda z$ as $z\rightarrow 0$, with $\lambda ^{m-1}=1$. This implies $\phi _\tau =\lambda \Phi _\tau$ by uniqueness of the Böttcher coordinate, and from $\tau (t_0)=t_0$ and analytic dependence on t it follows that $\lambda =1$ and $\phi _\tau =\Phi _\tau$, hence

$$\begin{array}{rcl} E_k(\tau )&{}=&{}\tau ^{\frac{1}{m-1}}\Phi _\tau (Q_k(\tau ))=\tau ^{\frac{1}{m-1}}\Phi _\tau (f^k_\tau (v^\alpha _\tau ))\\ &{}=&{} t_0^{\frac{1}{m-1}}\Phi _{t_0}\circ h_t^{-1}(f^k_\tau (v^\alpha _\tau ))=t_0^{\frac{1}{m-1}}\Phi _{t_0}\circ f^k_{t_0}\circ h_t^{-1}(v^\alpha _\tau )\\ &{}=&{}t_0^{\frac{1}{m-1}}\Phi _{t_0}(f^k_{t_0}(v^\alpha _t))\quad \mathrm{if~} k\ge 1,~\quad \mathrm{and}\\ E_0(\tau )&{}=&{}t_0(\Phi _{t_0}(v^\alpha _t))^m. \end{array}$$

Since $t\mapsto \tau$ is locally univalent, $E_k$ is univalent at $t_0$ if and only if the map $t\mapsto t_0^{\frac{1}{m-1}}\Phi _{t_0}(f_{t_0}^k(v^\alpha _t))$ is univalent on some neighbourhood of $t_0$. If $k\ge 1$, $\Phi _{t_0}$ is univalent on $\mathscr {A}_{t_0}$, and $f_{t_0}^k$ is univalent on $|z-v^\alpha _{t_0}|<\delta$ provided $Q_k(t_0)=f^k_{t_0}(v^\alpha _{t_0})\ne 0$, while $f_{t_0}^k$ is n-valent at $v^\alpha _{t_0}$ if $Q_k(t_0)=0$. In case of $k=0$ we note that $\Phi _{t_0}$ is locally univalent on some forward invariant domain D that contains 0 and $v^\alpha _{t_0}$, and $v^\alpha _t=tv^\alpha _1\ne 0$ is trivially univalent.

$\Box$

The proof of Theorem 2 will be finished by

Lemma 3

Let h be a proper map of degree m of the domain D onto the unit disc $\mathbb D$, and assume that h is ramified exactly over zero, that is, $h'(z)=0$ implies $h(z)=0$. Then D is simply connected and h has a single zero on D.

Proof

Assume that h has zeros with multiplicities $m_\nu$ ($1\le \nu \le n)$. Then h has degree $d=m_1+\cdots +m_n$ and $r=d-n$ critical points. The Riemann-Hurwitz formula then yields $\# D-2=-d+r=-n$, hence $\#D=2-n$, which only is possible if $n=1$ and $\# D=1$.$\Box$

Remark

Each connected component of $\mathbf {\Omega }^\alpha _k$ contains a zero of $Q_k(t)=tf_1(Q_{k-1}(t))$ which is not a zero of $Q_{k-1}$, hence is a zero of $Q_{k-1}(t)-1$. Thus $\mathbf {\Omega }^\alpha _k$ consists of at most $\frac{d^k-1}{d-1}$ connected components.

4 The hyperbolic loci

The bifurcation locus $\mathbf {B}^\beta$ is contained in some annulus $\delta<|t|<1$, and this also holds for $\mathbf {W}^\beta$. Hence (super-)attracting cycles $U_1,\ldots ,U_k$ that contain the critical point $\beta$ may occur only if $\delta<|t|<1$.

Theorem 3

For $0<|t|<1$, $f_t$ is quasi-conformally conjugated to some polynomial

$$\begin{aligned} P_c(z)=cz^m(z+1)^n\quad (c=c_t\ne 0) \end{aligned}$$

with free critical point $-\frac{m}{m+n}$. The basin $\mathscr {A}_t$ is completely invariant, and simply connected if and only if $t\not \in \mathbf {\Omega }^\alpha _\mathrm{res}.$ For $t\not \in \mathbf {\Omega }^\beta _0$, the Fatou set consists of $\mathscr {A}_t$, the simply connected basin $\mathscr {B}_t$ and its pre-images and, additionally, of some (super-)attracting cycle of Fatou components and their pre-images if $t\in \mathbf {W}^\beta$; the cycle absorbs the critical point $\beta$.

Proof

To prove the second assertion we note that by Lemma 1 the pre-image D of the disc $\triangle =\triangle _{r|t|}$ is a simply connected Jordan domain that contains $\overline{\triangle }\cup [0,1]$, but does not contain $v^\beta _t$. Then $D_2=\widehat{\mathbb C}{\setminus }\overline{\triangle }$ is a backward invariant domain, and

$$\begin{aligned} f_t:D_1\buildrel d:1 \over \longrightarrow D_2\quad \left( D_1=f_t^{-1}(D_2)\right) \end{aligned}$$

is a polynomial-like mapping in the sense of [6], of degree $d=m+n$, hence is hybrid equivalent to some polynomial P of degree d. We may assume that the quasi-conformal conjugation $\psi _t$ with

$$\begin{aligned} \psi _t\circ f_t=P\circ \psi _t \end{aligned}$$

maps $\infty , 0,$ and $-1$ onto $0,\infty ,$ and $-1$, respectively. Thus P is given by $P(z)=P_c(z)=cz^m(z+1)^n$, and $\psi _t$, hence also $c=c_t$ depends analytically on t. $\Box$

Remark

We note that $D_2=D_2(|t|)=\{z:|z|>r|t|\}$ increases if |t| decreases, while $D_1=f_t^{-1}(\widehat{\mathbb C}{\setminus }\overline{\triangle }_{r|t|})=f_1^{-1}(\widehat{\mathbb C}{\setminus }\overline{\triangle }_r)$ is independent of t. Thus the conformal modulus $\mu (|t|)$ of $D_2(|t|){\setminus }\overline{D_1}$ satisfies $\mu (1)\le \mu (|t|)-\log \frac{1}{|t|}\le \log \frac{\inf _{z\in D_1}|z|}{r}$. The bifurcation locus of $P_c$ corresponds conformally to the bifurcation locus $\mathbf {B}^\beta$, and the hyperbolic components are just quasi-conformal images of the hyperbolic components of the quadratic family $z\mapsto z^2+\xi$.

For $t\in \mathbf {W}_k$, the multiplier map $t\mapsto \lambda _t$ is an algebraic function of t. This is easily seen by writing the equations $f^k_t(z)=z$ and $\lambda =(f^k_t)'(z)$ as polynomial equations $q_1(z,t)=0$ and $q_2(z,t,\lambda )=0$, and computing the resultant $R_f(t,\lambda )$ of $q_1$ and $q_2$ with respect to z. For example, in case of $k=1$, $m=2$, and $n=1$ we obtain

$$\begin{aligned} R_f(t,\lambda )=\left[ -2+14t-2t^2\right] +\left[ 1-10t+t^2\right] \lambda +2t\lambda ^2=0. \end{aligned}$$

For $P_c(z)=cz^2(z+1)$ we obtain in the same manner (multiplier $\mu$)

$$\begin{aligned} R_P(c,\mu )=9+2c-(c+6)\mu +\mu ^2=0. \end{aligned}$$

Since the quasi-conformal conjugation respects multipliers ($\lambda _t=\mu _{c_t})$, $c_t$ is an algebraic function of t by the identity theorem; in the present case we obtain $(1+2t+t^2+2tc)^2=0$ by computing the resultant of $R_f(t,\lambda )$ and $R_P(c,\lambda )$ with respect to $\lambda$, hence

$$\begin{aligned} t\mapsto c=c_t=-{\textstyle \frac{1}{2}}\left( t+2+\frac{1}{t}\right) \quad \left( \textstyle c=-\frac{9}{2}\leftrightarrow t=\frac{1}{2}\left( \sqrt{49}-\sqrt{45}\right) \right) \end{aligned}$$

maps $0<|t|<1$ conformally onto $\mathbb C{\setminus }[-2,0]$, see Fig. 2.

The following result was not explicitly stated but proved in [17]. The proof is an adaption of the procedure due to Douady [5], applied to the hyperbolic components of the quadratic family $R_t(z)=z^2+t$ with one free critical point. The occurrence of several critical points requirers a slightly more sophisticated argument. The present version applies to a wider class of functions like $R_t(z)=z^d+t$, $R_t(z)=z^m+t/z^n$, $R_t(z)=t\left( 1+\frac{((d-1)^{d-1}/d^d)z^d}{1-z}\right)$ ($d\ge 3$), $R_t(z)=-\frac{t}{4}\frac{(z^2-2)^2}{z^2-1}$, the present family, and many others, to show that the hyperbolic components are simply connected and are mapped properly onto the unit disc by the multiplier map $t\mapsto \lambda _t$.

Theorem 4

Let $(R_t)_{t\in T}$ be any family of rational maps that is analytically parametrised over some domain T. Suppose that each $R_t$ has a (super-)attracting cycle $U_0\buildrel m_1:1 \over \longrightarrow U_1\buildrel m_2:1 \over \longrightarrow \ldots \buildrel m_{n-1}:1 \over \longrightarrow U_{n-1}\buildrel m_{n}:1 \over \longrightarrow U_{n}=U_0,$ such that $R_t^n$ has a single critical point $c_t\in U_0$ of multiplicity $m-1$, where $m=m_1\cdots m_n$ is the degree of $R^n_t:U_0\buildrel m:1 \over \longrightarrow U_0$. Assume also that the multiplier $\lambda _t$ satisfies $|\lambda _t|\rightarrow 1$ as $t\rightarrow \partial T$. Then the multiplier map $t\mapsto \lambda _t$ provides a proper map $T\buildrel (m-1):1 \over \longrightarrow \mathbb D$ which is ramified just over $w=0$, and T is simply connected.