Relation between Hénon maps with biholomorphic escaping sets

Abstract

Let H and F be two Hénon maps with biholomorphically equivalent escaping sets, then there exist affine automorphisms $A_1$ and $A_2$ in ${\mathbb {C}}^2$ such that

$$\begin{aligned} F=A_1\circ H \circ A_2 \end{aligned}$$

in ${\mathbb {C}}^2$.

1 Introduction

In the complex plane the simplest examples of holomorphic dynamical systems with non-trivial dynamical behaviour are the polynomial maps of degree greater than or equal to 2. The linear polynomials, in other words, the automorphisms of the complex plane, generate trivial dynamics. In contrast, the class of polynomial automorphisms of ${\mathbb {C}}^2$ is large and possesses rich dynamical features. A dynamical classification of these maps was given by Friedland–Milnor [7]. They showed that any polynomial automorphism of ${\mathbb {C}}^2$ is conjugate to one of the following maps:

$\bullet$ an affine map;

$\bullet$ an elementary map, i.e., the maps of the form $(x,y)\mapsto (ax+b, sy+p(x))$ with $as\ne 0$, where p is a polynomial in single variable of degree strictly greater than one;

$\bullet$ a finite composition of Hénon maps, where Hénon maps are the maps of the form

$$\begin{aligned} (x,y)\mapsto (y, p(y)-\delta x) \end{aligned}$$

(1.1)

with $\delta \ne 0$ and p a polynomial in single variable of degree $d\ge 2$.

The degree of a single Hénon map H of the form (1.1) is defined to be the degree of the polynomial p. The degree of composition of Hénon maps $H_n\circ \cdots \circ H_1$ is defined to be $d_n\cdots d_1$ where $d_i=\deg (H_i)$, for $1\le i \le n$.

Hénon maps are generalization of classical real quadratic Hénon maps introduced by astronomer Michel Hénon. The relevance of these maps in complex dynamics became apparent in the above-mentioned classification theorem of Friedland–Milnor. Moreover it turned out that the composition of Hénon maps are the only dynamically non-trivial polynomial automorphisms in ${\mathbb {C}}^2$, which naturally drew attention of many foremost researchers towards these maps. The pioneering work on Hénon maps was done by Bedford–Smillie [1,2,3], Fornæss–Sibony [8] and Hubbard–Oberste–Vorth [9, 10].

As in the case of polynomials in the complex plane, the orbit of any point in ${\mathbb {C}}^2$ under the iterations of a Hénon map H (or more generally, finite composition of Hénon maps) either diverges to infinity or always remains bounded. The collection of points $I_H^+ \subseteq {\mathbb {C}}^2$ which escape to infinity are called the escaping set of H and the collection of points $K_H^+\subseteq {\mathbb {C}}^2$ whose orbits remain bounded are called the non-escaping set of H. The union of escaping set $I_H^+$ and the interior of non-escaping set $K_H^+$ is the largest set of normality of the sequence of maps $\{H^n\}_{n\ge 1}$, that is, $I_H^+ \cup {\textrm{int}}K_H^+$ is the Fatou set of H. The escaping sets and the interiors of non-escaping sets of Hénon maps can be thought of as analogues of the unbounded components and the bounded components of Fatou sets of polynomials in the complex plane. However, the non-escaping sets of Hénon maps are not bounded in ${\mathbb {C}}^2$. The common boundary set $J_H^+$ of the escaping set $I_H^+$ and the non-escaping set $K_H^+$ is the Julia set of H.

The present article addresses a rigidity property of Hénon maps of the form (1.1). To what extent do the escaping sets of Hénon maps determine the Hénon maps? In other words, if the escaping sets of a pair of Hénon maps H and F of degree d are biholomorphically equivalent, then are these two Hénon maps closely related?

This question is first studied in a recent work of Bonnot–Radu–Tanase [4], where they prove that H and F coincide, for $d=2$. Further, they produce examples to show that for $d\ge 3$, H and F might not be even conjugate to each other. In this article we establish a precise relation between H and F of any degree $d\ge 2$ with biholomorphic escaping sets (Theorem 1.2).

The rigidity question raised here is conceived based on an explicit description of analytic structure of the escaping sets given by Hubbard–Oberste–Vorth in [9]. A convenient description of the escaping set $I_H^+$ of a given Hénon map H is given in terms of logarithmic rate of escape function, the so-called Green’s function $G_H^+$ of the Hénon map H. One can show that $I_H^+$ is precisely where the Green’s function is strictly positive. Further, $G_H^+: I_H^+ \rightarrow {\mathbb {R}}_+$ is a pluri-harmonic submersion and the level sets of $G_H^+$ are three dimensional manifolds, which are naturally foliated by copies of ${\mathbb {C}}$. The Green’s function $G_H^+$ is inextricably related to the Böttcher function $\phi _H^+$ of H, which is one of the key ingredients in describing the analytic structure of $I_H^+$. The Böttcher functions of Hénon maps can be considered as analogues of Böttcher functions of polynomials in ${\mathbb {C}}$ near infinity and they are defined in appropriate neighbourhoods of a point at infinity [0 : 1 : 0] in ${\mathbb {P}}^2$. To be a bit more precise, for $R>0$ sufficiently large, $\phi _H^+$ is defined on the open set $V_R^+=\left\{ (x,y)\in {\mathbb {C}}^2: |y|> \max \left\{ |x|, R\right\} \right\} \subseteq I_H^+$ by means of approaching the point [0 : 1 : 0] by the n-fold iteration $H^n$, then returning back by appropriate $d^n$-th root of the mapping $y\mapsto y^{d^n}$ and finally taking the limit as $n\rightarrow \infty$. Consequently, the range of $\phi _H^+$ lies inside ${\mathbb {C}}{\setminus } \bar{{\mathbb {D}}}$, where ${\mathbb {D}}$ is the unit disc in ${\mathbb {C}}$ and $G_H^+\equiv \log \left|\phi _H^+\right|$ in $V_R^+$. Although the Böttcher function $\phi _H^+$ does not extend analytically to $I_H^+$, it extends along curves in $I_H^+$ starting in $V_R^+$ and defines a multi-valued analytic map in $I_H^+$. Let ${\tilde{I}}_H^+$ be the covering of $I_H^+$ obtained as the Riemann domain of $\phi _H^+$ and let $\phi _H^+$ lifts as a single-valued holomorphic function ${\tilde{\phi }}_H^+: {\tilde{I}}_H^+ \rightarrow {\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}$. Further, ${\tilde{I}}_H^+$ is biholomorphically equivalent to the domain $\{(z,\zeta ): z\in {\mathbb {C}}, |\zeta |>1\}$. By construction $\zeta ={\tilde{\phi }}_H^+$ and thus in this new coordinate the level sets of ${\tilde{\phi }}_H^+$ simply straightens out. The Hénon map H lifts as a map ${\tilde{H}}: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})\rightarrow {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ and one can write down ${\tilde{H}}$ explicitly (see 2.6).

The following result relies on two main ingredients, also used by Bonnot–Radu–Tanase in [4]: The above-mentioned explicit description of the covering ${\tilde{I}}_H^+$ of $I_H^+$ and a method given by Bousch in [5].

Theorem 1.1

Let $H(x,y)=(y,p_H(y)-\delta _H x)$ and $F(x,y)=(y,p_F(y)-\delta _F x)$ be a pair of Hénon maps, where $p_H(y)=y^d+\sum _{i=0}^{d-2} a_{i}^H y^i$ and $p_F(y)=y^d+\sum _{i=0}^{d-2} a_{i}^F y^i$. Let $I_H^+$ and $I_F^+$ be escaping sets of H and F, respectively and let $I_H^+$ and $I_F^+$ be biholomorphically equivalent. Then $\beta p_H(y)=\alpha p_F(\alpha y)$, for some $\alpha , \beta \in {\mathbb {C}}$ with $\alpha ^{d+1}=\beta$ and $\beta ^{d-1}=1$. Further, we have $\delta _H=\gamma \delta _F$, with $\gamma ^{d-1}=1$. Therefore,

$$\begin{aligned} F\equiv L \circ B \circ H \circ B, \end{aligned}$$

(1.2)

where $B(x,y)=(\gamma \alpha \beta ^{-1}x, \alpha ^{-1}y)$ and $L(x,y)=(\gamma ^{-1}\beta x,\beta y )$, for all $(x,y)\in {\mathbb {C}}^2$.

Now note that $p_H$ and $p_F$ in Theorem 1.1 are monic and centered (next to highest coefficients vanish), whereas for an arbitrary Hénon map the associated polynomial in one variable is not necessarily monic and centered. However, it is not hard to see that up to conjugation any arbitrary Hénon map is of the form as in Theorem 1.1. Here goes a brief justification. Up to conjugation by an affine automorphism of ${\mathbb {C}}$, any polynomial in one variable is a monic and centered polynomial of the same degree. In particular, there exists an affine automorphism $\sigma _H$ of ${\mathbb {C}}$ such that $\sigma _H^{-1}\circ {p}_H\circ \sigma _H={\hat{p}}_H$ in ${\mathbb {C}}$, where ${\hat{p}}_H$ is monic and centered. Thus if we consider the affine automorphism $A_H(x,y)=(\sigma _H(x),\sigma _H(y))$, for $(x,y)\in {\mathbb {C}}^2$, then $A_H^{-1}\circ {H} \circ A_H={\hat{H}}$, where ${\hat{H}}(x,y)=(y,{\hat{p}}_H(y)-{\delta }_H x)$, for all $(x,y)\in {\mathbb {C}}^2$ with $\deg ({{\hat{p}}_H})=\deg (p_H)$. Clearly, $A_H(I_H^+)=I_{{\hat{H}}}^+$. Similarly, there exist an affine map $A_F$ and a Hénon map ${\hat{F}}$ such that $A_F^{-1}\circ F \circ A_F={\hat{F}}$ and $A_F(I_F^+)=I_{{\hat{F}}}^+$, where ${\hat{F}}(x,y)=(y,{\hat{p}}_F(y)-{\delta }_F x)$, for all $(x,y)\in {\mathbb {C}}^2$ with ${\hat{p}}_F$ monic and centered. Therefore, once we prove Theorem 1.1, the following result is obtained immediately.

Theorem 1.2

Let H and F be two Hénon maps with biholomorphically equivalent escaping sets, then there exist affine automorphisms $A_1$ and $A_2$ in ${\mathbb {C}}^2$ such that

$$\begin{aligned} F=A_1\circ H \circ A_2 \end{aligned}$$

in ${\mathbb {C}}^2$.

2 Preliminaries

Let

$$\begin{aligned} H(x,y)=(y,p(y)-\delta x) \end{aligned}$$

(2.1)

be a Hénon map, where p is a monic and centered polynomial in single variable of degree $d\ge 2$ and $\delta \ne 0$. In this section, we see a few fundamental definitions and a couple of known results on Hénon maps pertaining to the theme of the present article.

Filtration: For $R>0$, let

$$\begin{aligned} V^+_R&= \{ (x,y) \in {\mathbb {C}}^2: \vert x \vert< \vert y \vert , \vert y \vert> R \},\\ V^-_R&= \{ (x,y) \in {\mathbb {C}}^2: \vert y \vert < \vert x \vert , \vert x \vert > R \},\\ V_R&= \{ (x, y) \in {\mathbb {C}}^2: \vert x \vert , \vert y \vert \le R \}. \end{aligned}$$

This is called a filtration. For a given Hénon map H, there exists $R > 0$ sufficiently large such that

$$\begin{aligned} H(V^+_R) \subset V^+_R, \; H(V^+_R \cup V_R) \subset V^+_R \cup V_R \end{aligned}$$

and

$$\begin{aligned} H^{-1}(V^-_R) \subset V^-_R, \; H^{-1}(V^-_R \cup V_R) \subset V^-_R \cup V_R. \end{aligned}$$

Escaping and non-escaping sets: The set

$$\begin{aligned} I^{+}_H =\left\{ (x, y) \in {\mathbb {C}}^2: \Vert H^n(x,y)\Vert \rightarrow \infty \text { as } n\rightarrow \infty \right\} \end{aligned}$$

is called the escaping set of H and the set

$$\begin{aligned} K_H^+ = \left\{ (x, y) \in {\mathbb {C}}^2:\;\text {the sequence}\; \left\{ H^{n}(x, y) \right\} _{n\ge 1} \; \text {is bounded} \right\} \end{aligned}$$

is called the non-escaping set of H. One can prove that $K_H^+ \subset V_R \cup V^-_R$ and

$$\begin{aligned} I_H^+={\mathbb {C}}^2 {\setminus }K^{+}_H = \bigcup _{n=0}^{\infty } H^{- n}(V^{+}_R). \end{aligned}$$

(2.2)

Any Hénon map H can be extended meromorphically to ${\mathbb {P}}^2$ with an isolated indeterminacy point [0 : 1 : 0]. In fact, one can prove that the points in $I_H^+$ under iteration of H converges to the point [0 : 1 : 0] uniformly (on compacts).

Green’s function: The Green’s function of H is defined to be

$$\begin{aligned} G^{+}_H(x, y):= \lim _{n \rightarrow \infty } \frac{1}{d^n} \log ^+ \left\| H^{ n}(x, y)\right\| , \end{aligned}$$

for all $(x,y)\in {\mathbb {C}}^2$, where $\log ^+(t)=\max \{\log t,0\}$. It turns out that $G_H^+$ is non-negative everywhere in ${\mathbb {C}}^2$, plurisubharmonic in ${\mathbb {C}}^2$, pluriharmonic on ${\mathbb {C}}^2 {\setminus }K^{+}_H$ and vanishes precisely on $K^{+}_H$. Further,

$$\begin{aligned} G^{+}_H \circ H = d G^{+}_H \end{aligned}$$

in ${\mathbb {C}}^2$. The Green’s function $G^{+}_H$ has logarithmic growth near infinity, i.e., there exist $R>0$ and $L>0$ such that

$$\begin{aligned} \log ^+ |y |-L\le G_H^+ (x,y) \le \log ^+ |y |+L, \end{aligned}$$

for $(x,y)\in \overline{V_R^+ \cup V_R}$.

Böttcher function: Let $H^n(x,y)=\left( {(H^{n})}_1(x,y),{(H^{n})}_2(x,y)\right)$, for $(x,y)\in {\mathbb {C}}^2$. Note that $y_n={(H^{n})}_2(x,y)$ is a polynomial in x and y of degree $d^{n}$. The function

$$\begin{aligned} \phi _H^+(x,y):=\lim _{n\rightarrow \infty }y_{n}^{\frac{1}{d^{n}}}=y.\frac{{y_1}^{\frac{1}{d}}}{y}.\cdots .\frac{y_{n+1}^{\frac{1}{d^{n+1}}}}{y_{n}^{\frac{1}{d^{n}}}}\cdots , \end{aligned}$$

for $(x,y)\in V_R^+$, defines an analytic function from $V_R^+$ to ${\mathbb {C}}{\setminus } \bar{{\mathbb {D}}}$ and is called the Böttcher function of the Hénon map H. Further,

$$\begin{aligned} \phi _H^+\circ H(x,y)={\left( \phi _H^+(x,y)\right) }^d, \end{aligned}$$

for all $(x,y)\in V_R^+$ and

$$\begin{aligned} \phi _H^+(x,y)\sim y \text { as } \Vert (x,y)\Vert \rightarrow \infty \end{aligned}$$

in $V_R^+$. Comparing the definitions of Green’s function and Böttcher functions it follows instantly that

$$\begin{aligned} G_H^+=\log |\phi _H^+|\end{aligned}$$

in $V_R^+$.

For a detailed treatment of the above discussion, the inquisitive readers can see [1,2,3, 9, 11].

Now we present one of the technical ingredients required for the proof of Theorem 1.1. A series of change of coordinates we see here is a part of the standard theory of Hénon maps. However, most of the results presented here are paraphrasing of a couple of lemmas appearing in the beginning of the appendix of [4]. However the genesis of these results goes back to [6, 9].

For $M>0$ sufficiently large, let

$$\begin{aligned} U_R^+=\left\{ (x,y)\in V_R^+: |\phi _H^+(x,y)|> M \max \{R, |x |\}\right\} . \end{aligned}$$

One can easily check that $U_R^+ \subseteq V_R^+$ and $H(U_R^+)\subseteq U_R^+$.

Lemma 2.1

[11, Lemma 7.3.7],[6, Prop. 2.2] There exists a holomorphic function $\psi _H$ on $U_R^+$ such that

• $\psi _H \circ H (x,y)=( {\delta }/{d}) \psi _H(x,y) + Q \left( \phi _H^+(x,y)\right)$, for all $(x,y)\in U_R^+$, where Q is a monic polynomial of degree $d+1$;

• the map $\Phi _H=(\psi _H, \phi _H^+):U_R^+ \rightarrow {\mathbb {C}}^2$ is an injective holomorphic map.

The map $(x,y) \mapsto (x,\zeta )=(x,\phi _H^+(x,y))$ maps $U_R^+$ biholomorphically to $\left\{ (x,y)\in {\mathbb {C}}^2: |y|> M \max \{R, |x |\}\right\} .$ Let $(x,\zeta ) \mapsto (x, y(x,\zeta ))$ be the inverse map. It turns out that

$$\begin{aligned} \psi _H(x,y)=\zeta \int _0^{x} \frac{\partial y}{\partial \zeta }(u,\zeta )du, \end{aligned}$$

for $(x,y)\in U_R^+$ and $Q(\zeta )$ is the polynomial part of the power series expansion of

$$\begin{aligned} \zeta ^d \int _0^{y(0,\zeta )} \frac{\partial y}{\partial \zeta }(u,\zeta ^d)du \end{aligned}$$

(2.3)

(see proof of [11, Lemma 7.3.7] for details).

Next we consider the following change of coordinate near $p=[0:1:0]$:

$$\begin{aligned} T:\left( x,y(x,\zeta )\right) =(x,y) \mapsto \left( \frac{x}{y}, \frac{1}{y}\right) =(t,w). \end{aligned}$$

Note that $T^2={\textrm{Id}}$ and H takes the following form in (t, w)-coordinate:

$$\begin{aligned} (t,w) \mapsto (x,y) \mapsto (y, p(y)-\delta x) \mapsto \left( \frac{y}{p(y)-\delta x}, \frac{1}{p(y)-\delta x}\right) \\ =\left( \frac{w^{d-1}}{w^d p(1/w)-\delta t w^{d-1}},\frac{w^{d}}{w^d p(1/w)-\delta t w^{d-1}}\right) . \end{aligned}$$

Lemma 2.2

With the above notations, near $p=[0:1:0]$ we have

$$\begin{aligned} {1}/{\zeta }= w(1+ w \alpha (t,w)), \end{aligned}$$

(2.4)

where $\alpha (t,w)$ is a power series in t, w.

Proof

Note that

$$\begin{aligned} \frac{1}{\zeta }=\frac{1}{\phi _H^+(x,y)}=\lim _{n\rightarrow \infty }{\left[ \frac{1}{{(H^n(x,y))}_2}\right] }^{\frac{1}{d^n}}=\lim _{n\rightarrow \infty }{\left[ {(TH^n T^{-1}(t,w))}_2\right] }^{\frac{1}{d^n}}. \end{aligned}$$

Equivalently,

$$\begin{aligned} \frac{1}{\zeta }= & {} w. {\left[ \left( \frac{w^d}{w^d p(\frac{1}{w})-\delta t w^{d-1}}\right) /{w^d}\right] }^{\frac{1}{d}}. {\left[ \left( \frac{w_1^d}{w_1^d p(\frac{1}{w_1})-\delta t_1 w_1^{d-1}}\right) /{w_1^d}\right] }^{\frac{1}{d^2}}\cdots \\= & {} w. {\left[ \frac{1}{w^d p(\frac{1}{w})-\delta t w^{d-1}}\right] }^{\frac{1}{d}}.{\left[ \frac{1}{w_1^d p(\frac{1}{w_1})-\delta t_1 w_1^{d-1}}\right] }^{\frac{1}{d^2}} \cdots =w X(t,w) \end{aligned}$$

(here the $d^n$-th roots, for $n\ge 1$, are taken to be the principal branches of roots). One can check that the above series converges. Further, since $\left|{\phi _H^+(x,y)}/{y}\right|$ is bounded in $V_R^+$, it follows that X(t, w) is also bounded in its domain of definition, which is a subset of $\{|t|<1\}\times \left\{ 0< |w|<{1}/{R}\right\}$. Thus X(t, w) has a power series expansion. Also note that $X(t,0)=1$, for all t. So $X(t,w)=1+ w \alpha (t,w)$, where $\alpha (t,w)$ is a power series in t and w. Therefore, finally we obtain (2.4). $\square$

Next we consider the following change of variables: $(t,w) \mapsto (r,s)= ({x}/{\zeta }, {1}/{\zeta })$, which is a biholomorphism and is tangent to the identity (see [4, Lemma 4.4]). Therefore,

$$\begin{aligned} t=r+ T_2(r,s) \text { and } w=s+ S_2(r,s), \end{aligned}$$

where $T_2$ and $S_2$ both are power series with monomials of degree at least 2.

Lemma 2.3

With the above notations, we have

$$\begin{aligned} y= & {} \zeta \left( 1+{C}/{\zeta }+ U\left( {x}/{\zeta }, {1}/{\zeta }\right) \right) , \end{aligned}$$

where $C\in {\mathbb {C}}$ and U is a power series in two variables with monomials of degree at least 2.

Proof

By Lemma 2.2, we have ${1}/{\zeta }= w(1+w \alpha (t,w))$. Thus ${y}/{\zeta }= (1+w \alpha (t,w))$. Therefore, by replacing $t=r+ T_2(r,s) \text { and } w=s+ S_2(r,s)$, we get

$$\begin{aligned} y= & {} \zeta \left( 1+(s+ S_2(r,s))(\alpha (r+ T_2(r,s), s+ S_2(r,s) ))\right) \\= & {} \zeta \left( 1+s \beta (r,s)+ S_2(r,s) \beta (r,s) \right) , \end{aligned}$$

where $\beta (r,s)=\alpha (r+ T_2(r,s), s+ S_2(r,s))$. Now since $(r,s)= ({x}/{\zeta }, {1}/{\zeta })$, we have

$$\begin{aligned} y= \zeta \left( 1+{C}/{\zeta }+ U\left( {x}/{\zeta }, {1}/{\zeta }\right) \right) , \end{aligned}$$

where $C\in {\mathbb {C}}$ is the constant term of the power series expansion of $\beta$ and U is a power series in two variables without constant term. $\square$

Therefore,

$$\begin{aligned} y(x,\zeta )= & {} \left( \zeta +C+\frac{A^0_{-1}}{\zeta }+\cdots \right) +x\left( \frac{A_{-1}^1}{\zeta ^{m_1}}+\cdots \right) \\{} & {} \quad + x^2\left( \frac{A_{-1}^2}{\zeta ^{m_2}}+\cdots \right) + \cdots + x^n\left( \frac{A_{-1}^n}{\zeta ^{m_n}}+\cdots \right) +\cdots , \end{aligned}$$

where $m_1=1$, $m_n=n-1$, for $n\ge 2$ and $A_{-1}^0, A_{-1}^1,A_{-1}^2$ and so on are in ${\mathbb {C}}$.

The following lemma is one of the main takeaways from this section.

Lemma 2.4

With the above notations, we have

$$\begin{aligned} y(0,\zeta )=\zeta + \frac{D_1}{\zeta }+ \frac{D_2}{\zeta ^2}\cdots , \end{aligned}$$

and

$$\begin{aligned} \zeta (0,y)=y+ \frac{L_1}{y}+ \frac{L_2}{y^2}\cdots , \end{aligned}$$

where $D_i$’s and $L_i$’s are constants. Further, $Q(\zeta )$ is given by the polynomial part of $\zeta ^d y(0,\zeta )$ and therefore, Q is centered and monic.

Proof

Since p is monic and centered in (2.1), it follows from [9, Prop. 5.2] that

$$\begin{aligned} \zeta (0,y)=y+ \frac{L_1}{y}+ \frac{L_2}{y^2} \cdots . \end{aligned}$$

Now since $y=y(0,\zeta (0,y))$,

$$\begin{aligned} y=y(0,\zeta (0,y))= & {} \zeta (0,y)+D_0+ \frac{D_1}{\zeta (0,y)}+\cdots = \left( y+\frac{L_1}{y}+\frac{L_2}{y^2}+\cdots \right) +D_0\\{} & {} \quad + D_1{\left( y+\frac{L_1}{y}+\frac{L_2}{y^2}+\cdots \right) }^{-1}+\cdots . \end{aligned}$$

Thus $D_0=0$ and consequently, we have

$$\begin{aligned} y(0,\zeta )=\zeta +\frac{D_1}{\zeta }+ \frac{D_2}{\zeta ^2}+\cdots . \end{aligned}$$

The last assertion follows immediately from (2.3). $\square$

As we shall see in Section 3 and in Section 4, the particular forms of the power series representations of $y(0,\zeta )$ and $\zeta (0,y)$ as obtained in Lemma 2.4 and the knowledge of analytic structure of the escaping sets of Hénon maps are the key players in unravelling the relation between any pair of Hénon maps with biholomorphic escaping sets.

Here we give a brief description of analytic structure of Hénon maps. For a detailed account of analytic structure of escaping sets of Hénon maps the inquisitive readers can look at [6, 9, 11]. For any Hénon map H, it turns out that the Riemann surface of the Böttcher function $\phi _H^+$ is a covering space of the escaping set $I_H^+$ and it is isomorphic to ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$. For a Hénon map H of degree d, the fundamental group of $I_H^+$ is

$$\begin{aligned} {\mathbb {Z}}\left[ {1}/{d}\right] =\left\{ {k}/{d^n}: k,n \in {\mathbb {Z}}\right\} \end{aligned}$$

and the covering ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ of $I_H^+$ arises corresponding to the subgroup ${\mathbb {Z}} \subseteq {\mathbb {Z}}\left[ {1}/{d}\right]$. Therefore, $I_H^+$ is a quotient of ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ by some discrete subgroup of the automorphisms of ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$ isomorphic to ${{\mathbb {Z}}\left[ {1}/{d}\right] }/{{\mathbb {Z}}}$. For each element $\left[ {k}/{d^n}\right] \in {\mathbb {Z}}[{1}/{d}]/{\mathbb {Z}}$, there exists a unique deck transformation $\gamma _{{k}/{d^n}}$ from ${\mathbb {C}}\times ({\mathbb {C}}{\setminus }\bar{{\mathbb {D}}})$ to ${\mathbb {C}}\times ({\mathbb {C}}{\setminus }\bar{{\mathbb {D}}})$ such that

$$\begin{aligned} \gamma _{{k}/{d^n}} \begin{bmatrix} z\\ \zeta \end{bmatrix} =\begin{bmatrix} z+ \frac{d}{\delta } \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q(\zeta ^{d^l})-Q\left( {\left( e^{\frac{2 k\pi i}{d^n}}\zeta )\right) }^{d^l} \right) \right) \\ e^{\frac{2 k\pi i}{d^n}}\zeta \end{bmatrix}, \end{aligned}$$

(2.5)

for $n\ge 0$ and $k\ge 1$. Further, H lifts as a holomorphic map

$$\begin{aligned} {\tilde{H}}(z,\zeta )=\left( ({\delta }/{d})z+Q(\zeta ), {\zeta }^d \right) \end{aligned}$$

(2.6)

from ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$ to ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$.

3 A brief idea of the proof of the main theorem

In this section we sketch the main idea of the proof of Theorem 1.1.

Let $H(x,y)=(y,p_H(y)-\delta _H x)$ and $F(x,y)=(y,p_F(y)-\delta _F x)$ be a pair of Hénon maps of degree d. Thus the fundamental groups of both $I_H^+$ and $I_F^+$ are ${\mathbb {Z}}[{1}/{d}]$. Let a be a biholomorphism between $I_H^+$ and $I_F^+$. Any biholomorphism between $I_H^+$ and $I_F^+$, which induces identity as an isomorphism between the fundamental groups of $I_H^+$ and $I_F^+$, can be lifted as a biholomorphism from ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ to ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$. Now since any group isomorphism of ${\mathbb {Z}}[{1}/{d}]$ is of the form $x\mapsto \pm d^s x$, for some $s\in {\mathbb {Z}}$, the map a up to pre-composition with some n-fold iterates of F (or $F^{-1}$), i.e., the map $F^{\pm n}\circ a$, for some $n\in {\mathbb {N}}$, induces identity map between the fundamental groups of the escaping sets $I_H^+$ and $I_F^+$. Thus, it is harmless to assume that a lifts as a biholomorphism A from ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$ to ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$. Therefore, $\pi \circ A=a\circ \pi$ and consequently, the fiber of any point $p\in I_H^+$ of the natural projection map $\pi _H: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}) \rightarrow I_H^+$ maps into the fiber of the point $a(p)\in I_F^+$ of the projection map $\pi _F: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}) \rightarrow I_F^+$ by the biholomorphism A. Fiber of any point of the projection maps $\pi _H$ and $\pi _F$ are captured by the group of deck transformation of ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$, which is isomorphic to ${\mathbb {Z}}[{1}/{d}]/{\mathbb {Z}}$. Thus, it turns out that if $(z,\zeta )\in {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ is in the fiber of any point $p\in I_H^+$ (or $I_F^+$), then any other point in the same fiber will be of the form $\gamma _{{k}/{d^n}}(z,\zeta )$, for $k\ge 1$ and $n\ge 0$, where $\gamma _{{k}/{d^n}}$ is the deck transformation corresponding to $[{k}/{d^n}]\in {\mathbb {Z}}[{1}/{d}]/{\mathbb {Z}}$. An explicit description of this fibers can be written down (see (2.5)). Now adapting an idea of Bousch ( [5]), one can narrow down the possible forms of A by comparing the fibers of p and a(p). In fact, in our case, A has a very simple form $(u,v)\mapsto (u+\gamma , \alpha v)$, with $|\alpha |=1$ and $\gamma \in {\mathbb {C}}$. Thus we obtain an explicit expression for the image of the foliation ${\tilde{\phi }}_H^+=\zeta _0$ under the map A. These foliations come down to the corresponding escaping sets and induce rigidity on escaping sets. As a consequence, one expects a close relation between H and F, which is validated in our main theorem.

The above-mentioned idea was employed by Bonnot–Radu–Tanase in [4] in establishing the relation between H and F with biholomorphic escaping sets when $\deg (H)=\deg (F)=2$. They attempts to extract the relation between the coefficients of $p_H$ and $p_F$ by directly comparing the fibers of p and a(p). Although their approach works for lower degrees (for $d=2,3$), since precise computations can be carried out in these cases but as degree increases, to extract the relation between the coefficients of $p_H$ and $p_F$ seems very difficult by performing such direct calculations.

We take a different approach towards this problem. For two Hénon maps H and F with biholomorphic escaping sets, we first establish the relation between coefficients of $Q_H$ and $Q_F$ (see Section 3). The explicit relation between $Q_H$ and $Q_F$ is used to obtain a neat relation between the polynomials $p_H$ and $p_F$, namely, $\beta p_H(y)=\alpha p_F(\alpha y)$, for all $y\in {\mathbb {C}}$ with $\alpha ^{d+1}=\beta$ and $\beta ^{d-1}=1$ (see Section 4). To establish the relation between H and F, we are yet to investigate the relation between the Jacobians $\delta _H$ and $\delta _F$. It turns out that $\delta _H=\gamma \delta _F$, with $\gamma ^{d-1}=1$, thanks to [4]. It is noteworthy that all our calculations in Section 3 and in Section 4 are performed under the assumption that $\delta _H=\delta _F$. In Section 5, we outline how to handle the case when $\delta _H$ and $\delta _F$ are different.

4 Relation between $Q_H$ and $Q_F$

Let H and F be two Hénon maps as in Theorem 1.1 with bihholomorphic escaping sets $I_H^+$ and $I_F^+$, respectively. As indicated in the Introduction, for now we assume $\delta _H=\delta _F=\delta$. Later (in Section 6), we handle the case when $\delta _H\ne \delta _F$.

Lifting biholomorphisms between escaping sets: Recall from Section 2 that the fundamental group of escaping set of any Hénon map of degree d is ${\mathbb {Z}}[{1}/{d}]$ and ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$ is the covering of the escaping set corresponding to the subgroup ${\mathbb {Z}}\subseteq {\mathbb {Z}}[{1}/{d}]$. Let $\pi _H: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}) \rightarrow I_H^+$ and $\pi _F: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}) \rightarrow I_F^+$ be the covering maps. Now note that a biholomorphism a from $I_H^+$ to $I_F^+$ can be lifted as an automorphism A of ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ if and only if the induced group isomorphism

$$\begin{aligned} \pi _1(a): {\mathbb {Z}}\left[ {1}/{d}\right] \rightarrow {\mathbb {Z}}\left[ {1}/{d}\right] \end{aligned}$$

is $\pm {\textrm{Id}}$ (identity maps). It is easy to see that any group isomorphism of ${\mathbb {Z}}[{1}/{d}]$ is of the form $x\mapsto \pm d^{s} x$, for some $s\in {\mathbb {Z}}$. Thus, there exists $n\in {\mathbb {N}}$ such that $F^{\pm n}\circ a$ induces identity on ${\mathbb {Z}}[{1}/{d}]$ and thus lifts as an automorphism of ${\mathbb {C}} \times \left( {\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}\right)$. Therefore, without loss of generality, we assume that a lifts as an automorphism A of ${\mathbb {C}} \times \left( {\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}\right)$.

It follows from (2.5) that for H there exists a polynomial $Q_H$ of degree $d+1$ such that for each element $\left[ {k}/{d^n}\right] \in {\mathbb {Z}}[{1}/{d}]/{\mathbb {Z}}$, with $n\ge 0$ and $k\ge 1$, there exists a unique deck transformation $\gamma _{{k}/{d^n}}$ from ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }}\bar{{\mathbb {D}}})$ to ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }}\bar{{\mathbb {D}}})$ of the form

$$\begin{aligned} \gamma _{{k}/{d^n}} \begin{bmatrix} z\\ \zeta \end{bmatrix} =\begin{bmatrix} z+ \frac{d}{\delta } \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_H(\zeta ^{d^l})-Q_H\left( {\left( e^{\frac{2 k\pi i}{d^n}}\zeta )\right) }^{d^l} \right) \right) \\ e^{\frac{2 k\pi i}{d^n}}\zeta \end{bmatrix}. \end{aligned}$$

(4.1)

The same holds for the Hénon map F. Thus if ${\tilde{p}}=(z,\zeta )\in {\mathbb {C}}\times \left( {\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}\right)$ lies in the $\pi _H^{-1}(p)$ for some $p\in I_H^+$, then

$$\begin{aligned} \pi _H^{-1}(p)=\{\gamma _{{k}/{d^n}}(z,\zeta ): n\ge 0, k\ge 1\}. \end{aligned}$$

Form of lifts: First note that any automorphism A of ${\mathbb {C}}\times ({\mathbb {C}}{\setminus } \bar{{\mathbb {D}}})$ is of the form:

$$\begin{aligned} A(z,\zeta ) = (A_1(z,\zeta ), A_2(z,\zeta ))= (\beta (\zeta ) z+ \gamma (\zeta ), \alpha \zeta ), \end{aligned}$$

(4.2)

where $|\alpha |=1$ and $\beta , \gamma$ are holomorphic maps from ${\mathbb {C}}{\setminus } \bar{{\mathbb {D}}}$ to ${\mathbb {C}}^*$ and ${\mathbb {C}}$, respectively (see [5, Section 3]).

Note that since $a\circ \pi _H= \pi _F \circ A$, if $(z,\zeta ), (z',\zeta ') \in \pi _H^{-1}(p)$, for some $p\in I_H^+$, then $A(z,\zeta ), A(z',\zeta ') \in \pi _F^{-1}(a(p))$. Now let $(z,\zeta )$ and $(z',\zeta ')$ be in the same fiber of $\pi _H$, then using (4.1), we have

$$\begin{aligned} {(\zeta '/\zeta )}^{d^n}=1 \end{aligned}$$

and

$$\begin{aligned} z'=z+\frac{d}{\delta } \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_H(\zeta ^{d^l})-Q_H\left( {\left( e^{\frac{2 k\pi i}{d^n}}\zeta )\right) }^{d^l} \right) \right) , \end{aligned}$$

for some $n\in {\mathbb {N}}$. Therefore the difference between the first coordinates of A, i.e.,

$$\begin{aligned} A_1(z',\zeta ')-A_1(z,\zeta )&=\left( \beta (\zeta ')-\beta (\zeta )\right) u\nonumber \\&\quad +\beta (\zeta ')\frac{d}{\delta } \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_H(\zeta ^{d^l})-Q_H\left( {\left( e^{\frac{2 k\pi i}{d^n}}\zeta )\right) }^{d^l} \right) \right) \nonumber \\&\quad +\left( \gamma (\zeta ')-\gamma (\zeta )\right) . \end{aligned}$$

(4.3)

Now since $A(z,\zeta )$ and $A(z',\zeta ')$ are in the same fiber of $\pi _F$, the difference $A_1(z,\zeta )-A_1(z',\zeta ')$ is a function of $\alpha \zeta$ and $\alpha \zeta '$. Thus it follows from (4.3) that $\beta (\zeta )=\beta (\zeta ')$, i.e.,

$$\begin{aligned} \beta (\zeta )=\beta \left( \zeta . e^\frac{2\pi i k}{d^n}\right) , \end{aligned}$$

for all $k\ge 1$ and for all $n\ge 0$. Therefore, $\beta (\zeta )\equiv \beta$ in ${\mathbb {C}}$. Thus it follows form (4.3) that

$$\begin{aligned} A_1(z',\zeta ')-A_1(z,\zeta )=\Delta _H(\zeta ,\zeta ')+\gamma (\zeta ')-\gamma (\zeta ), \end{aligned}$$

(4.4)

where

$$\begin{aligned} \Delta _H(\zeta ,\zeta ')=\beta \frac{d}{\delta } \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_H(\zeta ^{d^l})-Q_H\left( {\left( e^{{2 k\pi i}/{d^n}}\zeta )\right) }^{d^l} \right) \right) . \end{aligned}$$

On the other hand since $A(z,\zeta )$ and $A(z',\zeta ')$ are in the same fiber of $\pi _F$,

$$\begin{aligned} A_1(z',\zeta ')-A_1(z,\zeta )= & {} \Delta _F(\alpha \zeta ,\alpha \zeta ') \nonumber \\= & {} \frac{d}{\delta } \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_F(\alpha ^{d^l} \zeta ^{d^l})-Q_F\left( {\left( e^{\frac{2 k\pi i}{d^n}}\alpha \zeta )\right) }^{d^l} \right) \right) . \nonumber \\ \end{aligned}$$

(4.5)

Since $\gamma$ is a holomorphic function on ${\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}$, comparing (4.4) and (4.5), it follows that for a fixed $\zeta \in {\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}$, the modulus of the difference between $\Delta _H(\zeta ,\zeta ')$ and $\Delta _F(\alpha \zeta ,\alpha \zeta ')$ is uniformly bounded for all $v'=e^{{2 k\pi i}/{d^n}}\zeta$ with $n\ge 0$ and $k\ge 1$. Note that for $\zeta '=e^{{2 k\pi i}/{d^n}}\zeta$, the difference $\Delta _H(\zeta ,\zeta ')-\Delta _F(\alpha \zeta ,\alpha \zeta ')$ is a polynomial of degree $(d+1)d^{n-1}$ and it can be written as

$$\begin{aligned}&\beta \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_H\left( \zeta ^{d^l}\right) -Q_H\left( {\left( e^{\frac{2 k\pi i}{d^n}}\zeta \right) }^{d^l} \right) \right) \\&\quad - \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left( Q_F\left( \alpha ^{d^l} \zeta ^{d^l}\right) -Q_F\left( {\left( e^{\frac{2 k\pi i}{d^n}}\alpha \zeta \right) }^{d^l} \right) \right) \\&= \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left[ \beta Q_H\left( \zeta ^{d^l}\right) -Q_F\left( \alpha ^{d^l} \zeta ^{d^l}\right) \right] \\&\quad - \sum _{l=0}^{n-1} {\left( \frac{d}{\delta }\right) }^l\left[ \beta Q_H\left( {\left( e^{\frac{2 k\pi i}{d^n}}\zeta \right) }^{d^l} \right) - Q_F\left( {\left( e^{\frac{2 k\pi i}{d^n}}\alpha \zeta \right) }^{d^l} \right) \right] \\&=\frac{d^{n-1}}{\delta ^{n-1}} \left[ \beta {\zeta }^{d^{n-1}(d+1)}-{\left( \alpha \zeta \right) }^{d^{n-1} (d+1)} -\beta {\left( e^{\frac{2 \pi i}{d^n}}\zeta \right) }^{d^{n-1}(d+1)}+{\left( \alpha \zeta e^{\frac{2 \pi i}{d^n}}\right) }^{d^{n-1}(d+1)}\right] \\&\quad +R(\zeta ) = {\left( \frac{d}{\delta }\right) }^{n-1} \left( \beta - \alpha ^{d^{n-1}(d+1)}\right) \left( 1-e^{\frac{2\pi i}{d}}\right) \zeta ^{d^{n-1}(d+1)} \left[ 1 + {\tilde{R}}(\zeta )\right] , \end{aligned}$$

where $R(\zeta )=O\left( \zeta ^{d^n}\right)$ and

$$\begin{aligned} {\tilde{R}}(\zeta )=R(\zeta ){\left( {d}/{\delta }\right) }^{-n+1} {\left( \beta - \alpha ^{d^{n-1}(d+1)}\right) }^{-1}{\left( 1-e^{\frac{2\pi i}{d}}\right) }^{-1}\zeta ^{-d^{n-1}(d+1)}. \end{aligned}$$

We claim that $\alpha ^{d^{n-1}(d+1)}\rightarrow \beta$, as $n\rightarrow \infty$. If not, then there exists a subsequence $\{n_l\}_{l\ge 1}$ such that

$$\begin{aligned} \left|\alpha ^{d^{n_l-1}(d+1)}- \beta \right|>c>0, \end{aligned}$$

(4.6)

for all $l\ge 1$. Thus, if $|{d}/{\delta } |\ge 1$, then (note that R is a polynomial of degree at most $d^n$)

$$\begin{aligned} |{\tilde{R}}(\zeta )|\le \frac{4nK_1}{{|\zeta |}^{d^{n-1}}}, \end{aligned}$$

for some $K_1>1$. On the other hand, if $|{d}/{\delta } |<1$, then

$$\begin{aligned} |{\tilde{R}}(\zeta )|\le \frac{4nK_2}{{|\zeta |}^{d^{n-1}}}{\left( \frac{\delta }{d}\right) }^{n-1}, \end{aligned}$$

for some $K_2>1$. Therefore, since the modulus of $\Delta _H\left( \zeta ,e^{\frac{2\pi i}{d^n}}\zeta \right) -\Delta _F\left( \alpha \zeta , \alpha e^{\frac{2\pi i}{d^n}}\zeta \right)$ is uniformly bounded, for all $n\ge 1$, we get a contradiction if (4.6) holds. Thus

$$\begin{aligned} \alpha ^{d^{n}(d+1)} \rightarrow \beta , \end{aligned}$$

(4.7)

as $n \rightarrow \infty$. Also,

$$\begin{aligned} \alpha ^{d^{n+1}(d+1)} \rightarrow \beta \end{aligned}$$

(4.8)

as $n\rightarrow \infty$ and thus dividing (4.8) by (4.7) and taking the limit, we get

$$\begin{aligned} \alpha ^{(d+1)(d-1)d^{n}} \rightarrow 1, \end{aligned}$$

(4.9)

as $n\rightarrow \infty$. Therefore, we get

$$\begin{aligned} \beta ^{d-1}=1. \end{aligned}$$

On the other hand, it follows from (4.7) that ${\left( \alpha ^{d+1}\right) }^{d^n} \rightarrow \beta$ and also note that $\beta$ is a repelling fixed point for the map $z\mapsto z^d$. Therefore, (4.7) holds if and only if

$$\begin{aligned} \alpha ^{d+1}=\beta . \end{aligned}$$

Thus $\left( \alpha ^{d^{n}(d+1)} - \beta \right) =0$, for all $n\ge 1$, which in turn gives $\gamma (\zeta )=\gamma \left( e^{\frac{2k\pi i}{d^n}}\zeta \right)$, for all $k\ge 1$ and $n\ge 0$. Thus $\gamma \equiv \gamma _0$, for some $\gamma _0\in {\mathbb {C}}$.

Relation between $Q_H$ and $Q_F$: Note that it follows from Lemma 2.4 that next to the highest degree coefficients of $Q_H$ and $Q_F$ vanish. Let

$$\begin{aligned} Q_H(\zeta )=\zeta ^{d+1}+A^H_{d-1} \zeta ^{d-1}+ A_{d-2}^H \zeta ^{d-2} + \cdots + A_1^H \zeta +A_0^H \end{aligned}$$

(4.10)

and

$$\begin{aligned} Q_F(\zeta )=\zeta ^{d+1}+A^F_{d-1} \zeta ^{d-1}+ A_{d-2}^F \zeta ^{d-2} + \cdots + A_1^F \zeta +A_0^F. \end{aligned}$$

(4.11)

Since $\gamma$ is identically constant in the complex plane, taking $k,n=1$, it follows from (4.4) and (4.5) that

$$\begin{aligned} \beta \frac{d}{\delta } \left[ Q_H(\zeta )-Q_H\left( e^{\frac{2\pi i}{d}} \zeta \right) \right]= & {} \Delta _H(\zeta , \zeta ')=\Delta _F (\alpha \zeta , \alpha \zeta ') \nonumber \\= & {} \frac{d}{\delta } \left[ Q_F(\alpha \zeta )-Q_F\left( e^{\frac{2\pi i}{d}} \alpha \zeta \right) \right] . \end{aligned}$$

(4.12)

Comparing coefficients of both sides of (4.12), we get

$$\begin{aligned} \beta A_{d-k}^H=\alpha ^{d-k} A_{d-k}^F, \end{aligned}$$

for $1\le k \le (d-1)$ and since $\beta =\alpha ^{d+1}$, equivalently we get

$$\begin{aligned} A_{d-k}^H=\alpha ^{-(k+1)} A_{d-k}^F, \end{aligned}$$

(4.13)

for $1\le k \le (d-1)$. Note that the relation between the constant terms of $Q_H$ and $Q_F$, i.e., the relation between $A_0^H$ and $A_0^F$ cannot be extracted from (4.12). However, as we are going to see in the next section that the explicit relation between $A_{d-i}^H$ and $A_{d-i}^F$, for $1\le i\le (d-1)$, obtained in (4.13) is sufficient to track down the relation between $p_H$ and $p_F$.

5 Relation between $p_H$ and $p_F$

It follows from Lemma 2.4 that there exist

$$\begin{aligned} y_H(\zeta )\equiv y_H(0, \zeta ) =\zeta +\frac{D_1^H}{\zeta }+ \frac{D_2^H}{\zeta ^2} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}} +O\left( \frac{1}{\zeta ^d}\right) \end{aligned}$$

(5.1)

and

$$\begin{aligned} y_F(\zeta )\equiv y_F(0, \zeta )=\zeta + \frac{D_1^F}{\zeta }+ \frac{D_2^F}{\zeta ^2} +\cdots + \frac{D_{d-1}^F}{\zeta ^{d-1}} + O\left( \frac{1}{\zeta ^d}\right) , \end{aligned}$$

(5.2)

with $D_i^H, D_i^F\in {\mathbb {C}}$ for $i\ge 1$, such that $Q_H(\zeta )$ and $Q_F(\zeta )$ are polynomial parts of $\zeta ^d y_H(0,\zeta )$ and $\zeta ^d y_F(0,\zeta )$, respectively. Therefore, the precise relation obtained in (4.13) determines a similar relation between the first few corresponding coefficients of the power series expansion of $y_H(\zeta )$ and $y_F(\zeta )$, namely

$$\begin{aligned} D_k^H=\alpha ^{-(k+1)}D_k^F, \end{aligned}$$

(5.3)

for $1\le k \le (d-1)$.

Since we assume that the Jacobian determinants of H and F are the same, that is, $\delta _H=\delta _F=\delta$, understanding the relation between the polynomials $p_H$ and $p_F$ is sufficient to capture the relation between H and F. We show in this section that the relation between the coefficients of $p_H$ and $p_F$ can be extracted via the Böttcher coordinates of H and F, namely via the functions $\zeta _H$ and $\zeta _F$, respectively. Let

$$\begin{aligned} \zeta _H(y)\equiv \zeta _H(0,y)=y+\frac{L_1^H}{y}+ \frac{L_2^H}{y^2} +\cdots + \frac{L_{d-1}^H}{y^{d-1}} + O\left( \frac{1}{y^d}\right) \end{aligned}$$

(5.4)

and

$$\begin{aligned} \zeta _F(y)\equiv \zeta _F(0,y)=y+\frac{L_1^F}{y}+ \frac{L_2^F}{y^2} +\cdots + \frac{L_{d-1}^F}{y^{d-1}} + O\left( \frac{1}{y^d}\right) . \end{aligned}$$

(5.5)

Recall from Section 2 that $\zeta _H \circ y_H (0,\zeta )= \zeta$ and $\zeta _F \circ y_F (0,\zeta )= \zeta$, for all $\zeta \in {\mathbb {C}}$ with $|\zeta |>R$, where $R>0$ is sufficiently large. Thus implementing (5.1), (5.2), (5.4) and (5.5) together, we get

$$\begin{aligned}&\left[ \frac{D_1^H}{\zeta }+\frac{D_2^H}{\zeta ^2} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}} + O\left( \frac{1}{\zeta ^d}\right) \right] \nonumber \\&\quad +\frac{L_1^H}{\zeta } \left[ 1+\frac{D_1^H}{\zeta ^2}+\frac{D_2^H}{\zeta ^3} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}} \right. \nonumber \\&\quad \left. + O\left( \frac{1}{\zeta ^d}\right) \right] ^{-1} +\frac{L_2^H}{\zeta ^2} \left[ 1+\frac{D_1^H}{\zeta ^2}+\frac{D_2^H}{\zeta ^3} \right. \nonumber \\&\quad \left. +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}} + O\left( \frac{1}{\zeta ^d}\right) \right] ^{-2}+ \cdots =0 \end{aligned}$$

(5.6)

and

$$\begin{aligned}&\left[ \frac{D_1^F}{\zeta }+\frac{D_2^F}{\zeta ^2} +\cdots + \frac{D_{d}^F}{\zeta ^d} + O\left( \frac{1}{\zeta ^d}\right) \right] \nonumber \\&\quad +\frac{L_1^F}{\zeta } \left[ 1+\frac{D_1^F}{\zeta ^2}+\frac{D_2^F}{\zeta ^3} +\cdots + \frac{D_{d-1}^F}{\zeta ^{d-1}}\right. \nonumber \\&\quad \left. + O\left( \frac{1}{\zeta ^d}\right) \right] ^{-1} + \frac{L_2^F}{\zeta ^2} \left[ 1+\frac{D_1^F}{\zeta ^2}+\frac{D_2^F}{\zeta ^3} \right. \nonumber \\&\quad \left. +\cdots + \frac{D_{d-1}^F}{\zeta ^{d-1}} + O\left( \frac{1}{\zeta ^d}\right) \right] ^{-2}+ \cdots =0. \end{aligned}$$

(5.7)

Note that expanding (5.6) and (5.7), one can express $L_k^H$ and $L_k^F$, for any $k\ge 1$, in terms of $D_1^H, D_2^H,\ldots$ and $D_1^F, D_2^F,\ldots$, respectively. In fact, using the relation between $D_k^H$ and $D_k^F$, for $1\le k \le (d-1)$, obtained in (5.3), one can establish an explicit relation between $L_k^H$ and $L_k^F$ for $1\le k \le (d-1)$.

Claim ${\mathcal {L}}$: For $1\le k \le d-1$, $L_k^H=\alpha ^{-(k+1)}L_k^F$.

Note that implementing claim ${\mathcal {L}}$ to (5.4) and (5.5), we obtain

$$\begin{aligned} \alpha \zeta _H(y)-\zeta _F(\alpha y)=O\left( {1}/{y^{d}}\right) . \end{aligned}$$

(5.8)

As we shall see shortly that the above relation between $\zeta _H$ and $\zeta _F$ plays the key role in establishing the relation between $p_H$ and $p_F$. For now assuming claim ${\mathcal {L}}$ is true and thus assuming (5.8) holds, let us first determine the relation between $p_H$ and $p_F$. We see a proof of the claim ${\mathcal {L}}$ in the end of the present section.

Relation between $p_H$ and $p_F$: Recall from Section 2 that for $|y|>R$, with R large enough, one can write

$$\begin{aligned} \zeta _H(y)\equiv \zeta _H(0,y)=y.{\left( \frac{p_H(y)}{y^d}\right) }^{\frac{1}{d}}{\left( \frac{p_H(y_{1,H})}{y_{1,H}^d}\right) }^{\frac{1}{d^2}}\cdots \end{aligned}$$

(5.9)

and

$$\begin{aligned} \zeta _F(y)\equiv \zeta _F(0,y)=y.{\left( \frac{p_F(y)}{y^d}\right) }^{\frac{1}{d}}{\left( \frac{p_F(y_{1,F})}{y_{1,F}^d}\right) }^{\frac{1}{d^2}}\cdots , \end{aligned}$$

(5.10)

where $y_{1,H}={(H(x,y))_2}=p_H(y)-\delta x$ and $y_{1,F}={(F(x,y))_2}=p_F(y)-\delta x$. Thus $\zeta _H(y)-O({1}/{y^{d}})$ and $\zeta _F(y)-O({1}/{y^{d}})$ are determined by

$$\begin{aligned} y.{\left( \frac{p_H(y)}{y^d}\right) }^{\frac{1}{d}} \text { and } y.{\left( \frac{p_F(y)}{y^d}\right) }^{\frac{1}{d}}, \end{aligned}$$

respectively. The power series expansion of the holomorphic function ${(1+z)}^{{1}/{d}}$ (principal branch) is

$$\begin{aligned} 1+({1}/{d}) z+\left( {({1}/{d}-1)}/{2d}\right) z^2+\cdots , \end{aligned}$$

for $z\in {\mathbb {D}}=\left\{ z\in {\mathbb {C}}: |z|<1\right\}$. Now note that

$$\begin{aligned} y.{\left( \frac{p_H(y)}{y^d}\right) }^{\frac{1}{d}}= & {} y. {\left( \frac{y^d+a_{d-2}^H y^{d-2}+\cdots +a_1^H y+a_0^H}{y^d}\right) }^{\frac{1}{d}} \nonumber \\= & {} y \left[ 1+ \frac{1}{d} \left( \frac{a_{d-2}^H}{y^2}+\cdots +\frac{a_0^H}{y^d}\right) \right. \nonumber \\{} & {} \left. +\frac{1}{2d}\left( \frac{1}{d}-1\right) {\left( \frac{a_{d-2}^H}{y^2}+\cdots +\frac{a_0^H}{y^d}\right) }^2+\cdots \right] \end{aligned}$$

(5.11)

and

$$\begin{aligned} \alpha y.{\left( \frac{p_F(\alpha y)}{{(\alpha y)}^d}\right) }^{\frac{1}{d}}= & {} \alpha y. {\left( \frac{{(\alpha y)}^d+a_{d-2}^F {(\alpha y)}^{d-2}+\cdots +a_1^F (\alpha y)+a_0^F}{{(\alpha y)}^d}\right) }^{\frac{1}{d}} \nonumber \\= & {} \alpha y \left[ 1+ \frac{1}{d}\left( \frac{a_{d-2}^F}{\alpha ^2 y^2}+\cdots +\frac{a_0^F}{\alpha ^d y^d}\right) \right. \nonumber \\{} & {} \left. +\frac{1}{2d}\left( \frac{1}{d}-1\right) {\left( \frac{a_{d-2}^F}{\alpha ^2 y^2}+\cdots +\frac{a_0^F}{\alpha ^d y^d}\right) }^2+\cdots \right] , \end{aligned}$$

(5.12)

for all $y\in {\mathbb {C}}$ with $|y |$ large enough. Since (5.8) holds, comparing (5.11) and (5.12), we get

$$\begin{aligned} a_{d-2}^F=\alpha ^2 a_{d-2}^H. \end{aligned}$$

Let us assume that

$$\begin{aligned} a_{d-k}^F=\alpha ^k a_{d-k}^H, \end{aligned}$$

up to some k, where $2\le k \le (d-1)$. Now it follows from (5.11) that the coefficient of ${1}/{y^{k}}$ in the expansion of $\alpha \zeta _H(y)$ is

$$\begin{aligned} c_{k+1}^H=\alpha \left[ \left( a_{d-(k+1)}^H/{d}\right) + G\left( a_{d-2}^H, \cdots ,a_{d-k}^H \right) \right] , \end{aligned}$$

(5.13)

where G is defined as follows. The function G is the polynomial in $(k-1)$ complex variables determined by the coefficient of ${1}/{y^{k+1}}$ of the power series expansion of

$$\begin{aligned}{} & {} {\left( \frac{p_H(y)}{y^d}\right) }^{\frac{1}{d}}-\frac{1}{d} \left( \frac{a_{d-2}^H}{y^2}+\cdots +\frac{a_0^H}{y^d}\right) -1 \nonumber \\{} & {} \quad = \frac{1}{2d}\left( \frac{1}{d}-1\right) {\left( \frac{a_{d-2}^H}{ y^2}+\cdots +\frac{a_0^H}{ y^d}\right) }^2+\cdots \end{aligned}$$

(5.14)

(see 5.11). In other words, if the coefficient of ${1}/{y^{k+1}}$ in (5.14) is $\sum _{i=1}^n g_i \left( \Pi _{j=2}^k {(a_{d-j}^H)}^{i(j)}\right)$ with $g_i\in {\mathbb {C}}$, then

$$\begin{aligned} G(x_1,x_2,\ldots ,x_{k-1})=\sum _{i=1}^n g_i \left( \Pi _{j=2}^k {x_j}^{i(j)}\right) . \end{aligned}$$

Similarly, it follows from (5.12) that the coefficient of ${1}/{y^{k}}$ in the expansion of $\zeta _F(\alpha y)$ is

$$\begin{aligned} c_{k+1}^F=\frac{a_{d-(k+1)}^F}{\alpha ^{k} d}+\alpha G\left( \frac{a_{d-2}^F}{\alpha ^2}, \cdots ,\frac{a_{d-k}^F}{\alpha ^k} \right) =\frac{a_{d-(k+1)}^F}{\alpha ^{k} d}+ \alpha G(a_{d-2}^H, \cdots ,a_{d-k}^H ). \nonumber \\ \end{aligned}$$

(5.15)

Since $c_{k+1}^H=c_{k+1}^F$, comparing (5.13) and (5.15), we have

$$\begin{aligned} a_{d-(k+1)}^F=\alpha ^{k+1}a_{d-(k+1)}^H. \end{aligned}$$

Therefore, the following relation between the corresponding coefficients of $p_H$ and $p_F$ holds:

$$\begin{aligned} a_{d-k}^F=\alpha ^{k}a_{d-k}^H, \end{aligned}$$

(5.16)

for $2\le k \le d$.

Now since $\beta =\alpha ^{d+1}$ and (5.16) holds, we have

$$\begin{aligned} p_F(y)= & {} y^d+a_{d-2}^F y^{d-2}+\cdots +a_1^F y+a_0^F\\= & {} y^d+ \alpha ^2 a_{d-2}^H y^{d-2}+\alpha ^3 a_{d-3}^H y^{d-3}+\cdots +\alpha ^{d-1} a_{1}^H y+\alpha ^d a_0^H\\= & {} y^d+\beta \alpha ^{-(d-1)} a_{d-2}^H y^{d-2}+\beta \alpha ^{-(d-2)} a_{d-3}^H y^{d-3}+\cdots +\beta \alpha ^{-2} a_{1}^H y+\beta \alpha ^{-1} a_0^H\\= & {} \beta \alpha ^{-1} \left( {(\alpha ^{-1}y)}^d+ a_{d-2}^H {(\alpha ^{-1}y)}^{d-2}+ a_{d-3}^H {(\alpha ^{-1}y)}^{d-3}+\cdots + a_{1}^H {(\alpha ^{-1}y)}+a_0^H\right) \\= & {} (\beta \alpha ^{-1}) p_H(\alpha ^{-1}y). \end{aligned}$$

In other words, we obtain

$$\begin{aligned} \alpha p_F(\alpha y)=\beta p_H(y), \end{aligned}$$

(5.17)

for all $y\in {\mathbb {C}}$, with $\alpha ^{d+1}=\beta$ and $\beta ^{d-1}=1$.

Before giving a formal proof of the claim ${\mathcal {L}}$, let us establish the relation between $L_k^H$ and $L_k^F$ with bare hands when the common degree d of the Hénon maps H and F is small. To start with, first note that the coefficients of ${1}/{\zeta ^k}$ vanish in both (5.6) and (5.7), for all $k\ge 1$.

$\bullet$ Let $d=2$. Thus by (5.3), we have $D_1^H=\alpha ^{-2}D_1^F$. The coefficients of ${1}/{\zeta }$ in (5.6) and (5.7) are $D_1^H+ L_1^H$ and $D_1^F+ L_1^F$, respectively and both vanish. Thus

$$\begin{aligned} L_1^H=-D_1^H=-\alpha ^{-2} D_1^F=\alpha ^{-2} L_1^F. \end{aligned}$$

$\bullet$ Let $d=3$. By (5.3), we have $D_i^H=\alpha ^{-(i+1)}D_i^F$, for $i=1,2$. Since the coefficients of ${1}/{\zeta }$ vanish in (5.6) and (5.7), as before we can show that $L_1^H=\alpha ^{-2} L_1^F$. Using the fact that coefficients of ${1}/{\zeta ^2}$, which are

$$\begin{aligned} D_2^H+L_1^H. 0+L_2^H \text { and } D_2^F+L_1^F. 0+L_2^F, \end{aligned}$$

vanish, we have

$$\begin{aligned} L_2^H=-D_2^H=-\alpha ^{-3} D_2^F= \alpha ^{-3} L_2^F. \end{aligned}$$

$\bullet$ Let $d=4$. By (5.3), we have $D_i^H=\alpha ^{-(i+1)}D_i^F$, for $1\le i \le 3$. Now since the coefficients ${1}/{\zeta ^i}$ vanish in (5.6) and (5.7), as before we can show that $L_i^H=\alpha ^{-(i+1)} L_i^F$, for $1\le i \le 2$. Now the coefficients of ${1}/{\zeta ^3}$ in (5.6) and (5.7) are

$$\begin{aligned} D_3^H+L_1^H. \left( C_{1}D_1^H \right) +L_2^H. 0+ L_3^H \text { and } D_3^F+L_1^F. \left( C_{1}D_1^F \right) +L_2^F. 0+ L_3^F, \end{aligned}$$

respectively for some $C_{1}\in {\mathbb {C}}$ and they vanish. Therefore,

$$\begin{aligned}{} & {} L_3^H =-D_3^H-L_1^H. \left( C_{1}D_1^H\right) , \end{aligned}$$

which implies

$$\begin{aligned} L_3^H =-\alpha ^{-4} D_3^F-\alpha ^{-4} L_1^F. \left( C_{1} D_1^F\right) = \alpha ^{-4} L_3^F. \end{aligned}$$

$\bullet$ Let $d=5$. Thus we have $D_i^H=\alpha ^{-(i+1)}D_i^F$, for $1\le i \le 4$. Using the same arguments as before we can show $L_i^H=\alpha ^{-(i+1)} L_i^F$, for $1\le i \le 3$. Now the coefficients of ${1}/{\zeta ^4}$ in (5.6) and (5.7) are

$$\begin{aligned} D_4^H+ L_1^H. \left( C_{2} D_2^H\right) +L_2^H. \left( C_{3} D_1^H\right) + L_3^H. 0 + L_4^H \end{aligned}$$

and

$$\begin{aligned} D_4^F+ L_1^F. \left( C_{2} D_2^F\right) +L_2^F. \left( C_{3} D_1^F\right) + L_3^F. 0 + L_4^F, \end{aligned}$$

for some $C_2, C_3 \in {\mathbb {C}}$ and since they vanish, a simple calculation as above gives that

$$\begin{aligned} L_4^H= \alpha ^{-5} L_4^F. \end{aligned}$$

$\bullet$ Let $d=6$. Thus we have $D_i^H=\alpha ^{-(i+1)}D_i^F$, for $1\le i \le 5$. As before we can show $L_i^H=\alpha ^{-(i+1)} L_i^F$, for $1\le i \le 4$. The coefficients of ${1}/{\zeta ^5}$ in (5.6) and (5.7) are $M_5^X$. Here for $X=H, F$,

$$\begin{aligned} M_5^X= D_5^X + L_1^X. \left( C_4 D_3^X+C_5{(D_1^X)}^2\right) +L_2^X. \left( C_6 D_2^X\right) + L_3^X . (C_7 D_1^X)+L_4^X .0+ L_5^X, \end{aligned}$$

where $C_4, C_5$ and so on are from ${\mathbb {C}}$. Since $M_5^H$ and $M_5^F$ vanish, calculations as above give us

$$\begin{aligned} L_5^H= \alpha ^{-6} L_5^F. \end{aligned}$$

$\bullet$ Let $d=7$. Thus $D_i^H=\alpha ^{-(i+1)}D_i^F$, for $1\le i \le 6$. Hence we can show $L_i^H=\alpha ^{-(i+1)} L_i^F$, for $1\le i \le 5$ as before. The coefficients of ${1}/{\zeta ^6}$ in (5.6) and (5.7) are $M_6^X$. Here for $X=H, F$,

$$\begin{aligned} M_6^X&=D_6^X + L_1^X. \left( C_8 D_4^X+C_9 D_1^X D_2^X\right) \\&\quad +L_2^X.\left( C_{10} D_3^X+C_{11} {(D_1^X)}^2\right) + L_3^X. (C_{12}D_2^X)\\&\quad +L_4^X. (C_{13} D_1^X) + L_5^X . 0+L_6^X, \end{aligned}$$

where $C_8, C_9$ and so on are from ${\mathbb {C}}$. Since $M_6^H$ and $M_6^F$ vanish, we get

$$\begin{aligned} L_6^H= \alpha ^{-7} L_6^F. \end{aligned}$$

$\bullet$ Let $d=8$. So $D_i^H=\alpha ^{-(i+1)}D_i^F$, for $1\le i \le 7$. Hence we can show $L_i^H=\alpha ^{-(i+1)} L_i^F$, for $1\le i \le 6$ just as before. Now the the coefficients of ${1}/{\zeta ^7}$ in (5.6) and (5.7) are $M_7^X$. Here for $X=H, F$,

$$\begin{aligned} M_7^X= & {} D_7^X + L_1^X.\left( C_{14} D_5^X+C_{15} D_1^X D_3^X+ C_{16}{(D_2^X)}^2+ C_{17} {(D_1^X)}^3 \right) \\{} & {} \quad + L_2^X.\left( C_{18} D_4^X+C_{19} D_1^X D_2^X\right) \\{} & {} \quad + L_3^X.\left( C_{20} D_3^X +C_{21} {(D_1^X)}^2\right) \\{} & {} \quad +L_4^X.(C_{22} D_2^X)+ L_5^X.(C_{23} D_1^X)+ L_6^X.0+L_7^X, \end{aligned}$$

where $C_{14}, C_{15}$ and so on are from ${\mathbb {C}}$. A simple calculation as above gives us

$$\begin{aligned} L_7^H= \alpha ^{-8} L_7^F. \end{aligned}$$

As promised earlier, now we see a proof of the claim ${\mathcal {L}}$.

5.1 Proof of Claim ${\mathcal {L}}$

First we make a few observations, which follow immediately by chasing the expression of the coefficients of ${1}/{\zeta ^s}$ in (5.6) and (5.7), for $s\ge 1$ and then using the fact that all of them vanish.

(O1) For any $s\ge 1$, the coefficients of ${1}/{\zeta ^s}$ in (5.6) and (5.7) vanish. Further, they are of the form $D_s^H+R_s^H+ L_s^H$ and $D_s^F+R_s^F+ L_s^F$, respectively, where $R_s^H$ and $R_s^F$ are linear combinations of products of powers of $D_i^H$’s, $L_j^H$’s and $D_i^F$’s, $L_j^F$’s, respectively, with $1\le i,j<s$.

(O2) For $s \ge 1$, one can write

$$\begin{aligned} R_s^H= \sum _{i=1}^{s-1} L_i^H R_{i,s}^H \text { and } R_s^F=\sum _{i=1}^{s-1} L_i^F R_{i,s}^F. \end{aligned}$$

Let $2\le i \le s-1$. It follows from (5.6) that $R_{i,s}^H$ is the coefficient of ${1}/{\zeta ^{s-i}}$ of the power series expansion of

$$\begin{aligned} \left( 1+\frac{D_1^H}{\zeta ^2}+\frac{D_2^H}{\zeta ^3} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}} + O\left( \frac{1}{\zeta ^d}\right) \right) ^{-i}, \end{aligned}$$

and similarly, $R_{i-1,s-1}^H$ is the coefficient of ${1}/{\zeta ^{s-i}}$ of the power series expansion of

$$\begin{aligned} \left( 1+\frac{D_1^H}{\zeta ^2}+\frac{D_2^H}{\zeta ^3} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}}+ O\left( \frac{1}{\zeta ^d}\right) \right) ^{-(i-1)}. \end{aligned}$$

Therefore, for $2\le i \le s-1$, if

$$\begin{aligned} R_{i,s}^H=C(i,s,1) D^H(i,s,1) +C(i,s,2) D^H(i,s,2)+\cdots +C(i,s,i_s) D^H(i,s,i_s) \nonumber \\ \end{aligned}$$

(5.18)

and

$$\begin{aligned} R_{i,s}^F=C(i,s,1) D^F(i,s,1) +C(i,s,2) D^F(i,s,2)+\cdots +C(i,s,i_s) D^F(i,s,i_s),\nonumber \\ \end{aligned}$$

(5.19)

where for $1\le j \le i_s$, $D^H(i,s,j)$ and $D^F(i,s,j)$ are products of powers of $D_l^H$’s and $D_l^F$’s ($1\le l <s$), respectively, with $C(i,s,j)\in {\mathbb {C}}$, then

$$\begin{aligned} R_{i-1,s-1}^H= & {} {\tilde{C}}(i,s,1)D^H(i,s,1) + {\tilde{C}}(i,s,2)D^H(i,s,2)\nonumber \\{} & {} \quad + \cdots +{\tilde{C}}{(i,s,i_s)} D^H(i,s,i_s) \end{aligned}$$

(5.20)

and

$$\begin{aligned} R_{i-1,s-1}^F= & {} {\tilde{C}}(i,s,1)D^F(i,s,1) + {\tilde{C}}(i,s,2)D^F(i,s,2) \nonumber \\{} & {} \quad + \cdots +{\tilde{C}}{(i,s,i_s)} D^F(i,s,i_s), \end{aligned}$$

(5.21)

with ${\tilde{C}}(i,s,j)\in {\mathbb {C}}$, for $1\le j \le i_s$.

(O3) Let $s \ge 1$ and $i=1$. Then

$$\begin{aligned} R_{1,s}^H= & {} C(1,s,1) D^H(i,s,1) +C(1,s,2) D^H(i,s,2)\nonumber \\{} & {} \quad +\cdots +C(1,s,1_s) D^H(1,s,1_s), \end{aligned}$$

(5.22)

where for $1\le j \le 1_s$,

$$\begin{aligned} D^H(1,s,j)=D_{{s_j}_1}^H D_{{s_j}_2}^H \cdots D_{{s_j}_{m(s_j)}}^H, \end{aligned}$$

with ${s_j}_1,{s_j}_2,\ldots , {s_j}_{m(s_j)} \in \{1,2,\ldots ,s-1\}$ (${s_j}_1,{s_j}_2,\ldots , {s_j}_{m(s_j)}$ are not possibly all distinct), i.e., $D_{{s_j}_1}^H, D_{{s_j}_2}^H, \ldots , D_{{s_j}_{m(s_j)}}^H \in \left\{ D_1^H, \ldots , D_{s-1}^H\right\}$. Now $R_{1,s}^H$ is the coefficient of ${1}/{\zeta ^{s-1}}$ of the power series expansion of

$$\begin{aligned} \left( 1+\frac{D_1^H}{\zeta ^2}+\frac{D_2^H}{\zeta ^3} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}}+ O\left( \frac{1}{\zeta ^d}\right) \right) ^{-1}. \end{aligned}$$

Similarly, $R_{1,s+1}^H$ is the coefficient of ${1}/{\zeta ^{s}}$ of the power series expansion of

$$\begin{aligned} \left( 1+\frac{D_1^H}{\zeta ^2}+\frac{D_2^H}{\zeta ^3} +\cdots + \frac{D_{d-1}^H}{\zeta ^{d-1}}+ O\left( \frac{1}{\zeta ^d}\right) \right) ^{-1}. \end{aligned}$$

Therefore, if

$$\begin{aligned} R_{1,s}^H= \sum _{j=1}^{1_s} C(1,s,j) D_{{s_j}_1}^H D_{{s_j}_2}^H \cdots D_{{s_j}_{m(s_j)}}^H, \end{aligned}$$

(5.23)

then

$$\begin{aligned} R_{1,s+1}^H= & {} \sum _{j=1}^{1_s} \biggl [C^{(1)}{(1,s,j)} D_{{s_j}_1+1}^H D_{{s_j}_2} \cdots D_{{s_j}_{m(s_j)}}^H +C^{(2)}{(1,s,j)}D_{{s_j}_1}^H D_{{s_j}_2+1}^H \cdots D_{{s_j}_{m(s_j)}}^H\nonumber \\{} & {} +\cdots +C^{(m(s_j))}(1,s,j) D_{{s_j}_1}^H.D_{{s_j}_2}\cdots D_{{s_j}_{m(s_j)}+1}^H\biggr ]+C_s {(D_1^H)}^{\frac{s}{2}}, \end{aligned}$$

(5.24)

with $C_s, C^{(k)}(1,s,j)\in {\mathbb {C}}$, for $1\le j \le 1_s$ and $1\le k \le m(s_j)$. Further, $C_s=0$, if s is not divisible by 2 and $C_s\ne 0$, if s is divisible by 2. The same relation holds between $R_{1,s}^F$ and $R_{1,s+1}^F$.

Let $d\ge 4$. Note that we have already established claim ${\mathcal {L}}$ for $d=2,3$ just before starting the present proof. Let $1\le k \le d-2$. Then we prove that if for $1\le s \le k$,

• $(\bullet )$ $L_{s}^H=\alpha ^{-(s+1)}L_s^F$, and

• $(\bullet \bullet )$ for each fixed pair (i, s) satisfying $2 \le i\le s-1$, $D^H(i,s,j)= \alpha ^{-(s-i)}D^F(i,s,j),$ for $1\le j \le i_s$ and $R_{1,s}^H=\alpha ^{-(s-1)}R_{1,s}^F$ (which in turn gives $R_{i,s}^H=\alpha ^{-(s-i)}R_{i,s}^F$, for $1\le i \le s-1$),

then $L_{k+1}^H=\alpha ^{-(k+2)}L_{k+1}^F$.

Now observe that performing the same calculations, which we did immediately before starting this proof, ($\bullet$) and ($\bullet \bullet$) hold for a first few k’s with $1\le k \le d-2$. The coefficients of ${1}/{\zeta ^{k+1}}$ in (5.6) and (5.7) are

$$\begin{aligned} D_{k+1}^H + L_1^H R_{1,k+1}^H + L_2^H R_{2,k+1}^H + \cdots + L_{k}^H R_{k,k+1}^H+ L_{k+1}^H, \end{aligned}$$

(5.25)

and

$$\begin{aligned} D_{k+1}^F + L_1^F R_{1,k+1}^F + L_2^F R_{2,k+1}^F + \cdots + L_{k}^F R_{k,k+1}^F+ L_{k+1}^F, \end{aligned}$$

(5.26)

respectively (note that both vanish). Let $2\le i \le k$. Then by (5.18), it follows that

$$\begin{aligned} R_{i,k+1}^H= & {} C(i,k+1,1) D^H(i,k+1,1) +\cdots \nonumber \\{} & {} \quad +C(i,k+1,i_{k+1}) D^H(i,k+1,i_{k+1}), \end{aligned}$$

(5.27)

and thus by (5.20)

$$\begin{aligned} R_{i-1,k}^H= & {} {\tilde{C}}(i,k+1,1)D^H(i,k+1,1) \nonumber \\{} & {} \quad + \cdots +{\tilde{C}}{(i,k+1,i_{k+1})} D^H(i,k+1,i_{k+1}). \end{aligned}$$

(5.28)

Therefore, clearly for $1\le j \le i_{k+1}$, $D^H(i,k+1,j)=D^H(i-1,k,r_j)$, for some $1\le r_j \le {(i-1)}_{k}$. Similar conclusion holds for $D^F(i,k +1,j)$, i.e., for $1\le j \le i_{k+1}$, $D^F(i,k +1,j)=D^F(i-1,k,r_j)$. Therefore, using ($\bullet \bullet$) for $s=k$, we have

$$\begin{aligned} D^H(i,k+1,j)=D^H(i-1,k,r_j)= & {} \alpha ^{-(k-i+1)}D^F(i-1,k,r_j)\\{} & {} \quad =\alpha ^{-(k-i+1)} D^F(i,k+1,j), \end{aligned}$$

for $2\le i \le k$ and for $1\le j \le i_{k+1}$.

Therefore,

$$\begin{aligned} R_{i,k+1}^H= \alpha ^{-(k+1-i)}R_{i,k+1}^F, \end{aligned}$$

(5.29)

for $2\le i \le k$. By ($\bullet \bullet$), we have $R_{1,k}^H = \alpha ^{-(k-1)}R_{1,k}^F.$ Thus comparing (5.23) and (5.24) along with using the fact that $D_k^H=\alpha ^{-(k+1)}D_k^F$, for $1\le k \le d-1$ (see (5.3)), we obtain

$$\begin{aligned} R_{1,k+1}^H =\alpha ^{-k} R_{1,k+1}^F. \end{aligned}$$

(5.30)

Now as mentioned before (5.25) and (5.26) both are equal to zero. Therefore using (5.29), (5.30) and (5.3), we get

$$\begin{aligned} L_{k+1}^H= & {} -\left( D_{k+1}^H + L_1^H R_{1,k+1}^H + L_2^H R_{2,k+1}^H + \cdots + L_{k}^H R_{k,k+1}^H\right) \\= & {} -\alpha ^{-(k+2)}\left( D_{k+1}^F + L_1^F R_{1,k+1}^F + L_2^F R_{2,k+1}^F + \cdots + L_{k}^F R_{k,k+1}^F\right) \\= & {} \alpha ^{-(k+2)} L_{k+1}^F. \end{aligned}$$

Thus the claim follows.

6 Hénon maps with different Jacobians

Let $H(x,y)=\left( y, p_H(y)-\delta _H x\right)$ and $F(x,y)=(y, p_F(y)-\delta _F x)$ be such that $\delta _H\ne \delta _F$. As before, we assume that $\deg H=\deg F=d\ge 2$. It follows from [4, Thm. 2.7] that if $I_H^+$ and $I_F^+$ are biholomorphic, then

$$\begin{aligned} \delta _H^{d-1}=\delta _F^{d-1}=\delta . \end{aligned}$$

(6.1)

Now instead of working with H and F, we work with ${\tilde{H}}=H^{d-1}$ and ${\tilde{F}}=F^{d-1}$. Note that since (6.1) holds the Jacobians of ${\tilde{H}}$ and ${\tilde{F}}$ are the same. Further, $I_{H^{d-1}}^+=I_{F^{d-1}}^+$.

If ${\tilde{H}}: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}) \rightarrow {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ is the lift of $H:I_H^+ \rightarrow I_H^+$, then clearly ${{\tilde{H}}}^{d-1}: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}) \rightarrow {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}})$ is the lift of $H^{d-1}:I_H^+ \rightarrow I_H^+$. It follows from (2.6) that

$$\begin{aligned} {\tilde{H}}(z,\zeta )=\left( {(\delta _H}/{d}) z+Q_H(\zeta ), \zeta ^d \right) . \end{aligned}$$

A simple calculation gives

$$\begin{aligned} { {\tilde{H}}}^{d-1}(z,\zeta )=\left( {\left( {\delta _H}/{d}\right) }^{d-1}z+{\tilde{Q}}_H(\zeta ), \zeta ^{d^{d-1}}\right) , \end{aligned}$$

where

$$\begin{aligned} {\tilde{Q}}_H(\zeta )= {\left( {\delta _H}/{d}\right) }^{d-2} Q_H(\zeta )+{\left( {\delta _H}/{d}\right) }^{d-3} Q_H(\zeta ^d)+\cdots +Q_H\left( \zeta ^{d^{d-2}}\right) . \end{aligned}$$

Note that $\deg {\tilde{H}}=\deg {\tilde{F}}={\tilde{d}}=d^{d-1}$ and $\det J_H=\det J_H=\delta$, where $J_H$ and $J_F$ are the Jacobian matrices of H and F, respectively. Also it is easy to see that ${\mathbb {Z}}[{1}/{d^n}]={\mathbb {Z}}[{1}/{d}]$ for all $n\ge 1$, and in particular, ${\mathbb {Z}}[{1}/{{\tilde{d}}}]={\mathbb {Z}}[{1}/{d^{d-1}}]={\mathbb {Z}}[{1}/{d}]$. As in (4.1), one can show that for each element $\left[ {k}/{{\tilde{d}}^n}\right] \in {\mathbb {Z}}[{1}/{{\tilde{d}}}]/{\mathbb {Z}}$, there exists a unique deck transformation $\gamma _{{k}/{{\tilde{d}}^n}}$ of ${\mathbb {C}}\times ({\mathbb {C}}{{\setminus }}\bar{{\mathbb {D}}})$ such that

$$\begin{aligned} \gamma _{{k}/{{\tilde{d}}^n}} \begin{bmatrix} z\\ \zeta \end{bmatrix} =\begin{bmatrix} z+ \frac{{\tilde{d}}}{\delta } \sum _{l=0}^{n-1} {\left( \frac{{\tilde{d}}}{\delta }\right) }^l\left( {\tilde{Q}}_H(\zeta ^{{{\tilde{d}}}^l})-{\tilde{Q}}_H\left( {\left( e^{\frac{2 k\pi i}{{\tilde{d}}^n}}\zeta )\right) }^{{\tilde{d}}^l} \right) \right) \\ e^{\frac{2 k\pi i}{{\tilde{d}}^n}}\zeta \end{bmatrix}, \end{aligned}$$

(6.2)

for each $n\ge 0$, $k\ge 1$ and for all $(z,\zeta )\in {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }}\bar{{\mathbb {D}}})$. Thus, if ${\tilde{p}}=(z,\zeta )\in {\mathbb {C}}\times \left( {\mathbb {C}}{{\setminus }} \bar{{\mathbb {D}}}\right)$ is a point in the fiber of $p\in I_H^+=I_{{\tilde{H}}}^+$, then the other points in the fiber are precisely $\gamma _{{k}/{{\tilde{d}}^n}}(z,\zeta )$ with $n\ge 0$ and $k\ge 1$.

Note that the main hindrance to run a similar set of calculations as in Section 3 for any arbitrary pair of Hénon maps H and F of the same degree is that the Jacobians of H and F are possibly different. But if we work with $H^{d-1}$ and $F^{d-1}$, there is no such issues. Since (6.2) holds, a moment’s thought assures that exactly similar calculations as in Section 3 run smoothly for the maps $H^{d-1}$ and $F^{d-1}$. Therefore, if $a: I_{{\tilde{H}}}^+ \rightarrow I_{{\tilde{H}}}^+$ is a biholomorphism and if $A: {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }}\bar{{\mathbb {D}}}) \rightarrow {\mathbb {C}}\times ({\mathbb {C}}{{\setminus }}\bar{{\mathbb {D}}})$ is a lift, i.e, A is of the form $(z,\zeta )\mapsto (\beta (z)\zeta +\gamma (\zeta ), \alpha \zeta )$, then as in Section 3, we can show

• $\beta (\zeta ) \equiv \beta$ in ${\mathbb {C}}$;

• $\left( \alpha ^{d^{d-2}}\right) ^{(d+1)d^{n-1}}\rightarrow \beta$ as $n\rightarrow \infty$ and consequently, $\alpha ^{d^{d-2}(d+1)}=\beta$ with $\beta ^{d-1}=1$.

Thus as in (4.12) we obtain

$$\begin{aligned} \beta \frac{{\tilde{d}}}{\delta } \left[ {\tilde{Q}}_H(\zeta )-{\tilde{Q}}_H\left( e^{\frac{2\pi i}{{\tilde{d}}}} \zeta \right) \right] =\frac{{\tilde{d}}}{\delta } \left[ {\tilde{Q}}_F(\alpha \zeta )-{\tilde{Q}}_F\left( e^{\frac{2\pi i}{{\tilde{d}}}} \alpha \zeta \right) \right] , \end{aligned}$$

(6.3)

for all $\zeta \in {\mathbb {C}}{\setminus } \bar{{\mathbb {D}}}$, with $\alpha ^{d^{d-2}(d+1)}=\beta$ and $\beta ^{d-1}=1$. Expanding (6.3), we get

$$\begin{aligned} \beta \frac{\delta _H^{d-2}}{{{d}}^{d-2}} \left[ {Q}_H(\zeta )-{Q}_H \left( e^{\frac{2 \pi i}{{{\tilde{d}}}}} \zeta \right) \right] +\beta \frac{\delta _H^{d-3}}{{{d}}^{d-3}}\left[ {Q}_H(\zeta ^d)-{Q}_H\left( { \left( e^{\frac{2 \pi i}{{{\tilde{d}}}}} \zeta \right) }^d\right) \right] + \nonumber \\ \cdots +\beta \left[ {Q}_H\left( \zeta ^{d^{d-2}}\right) -{Q}_H \left( {\left( e^{\frac{2 \pi i}{{\tilde{d}}}} \zeta \right) }^{d^{d-2}} \right) \right] \nonumber \\ = \frac{\delta _F^{d-2}}{{{d}}^{d-2}} \left[ {Q}_F(\alpha \zeta )-{Q}_F \left( e^{\frac{2 \pi i}{{{\tilde{d}}}}}\alpha \zeta \right) \right] +\frac{\delta _F^{d-3}}{{{d}}^{d-3}}\left[ {Q}_F\left( {(\alpha \zeta )}^d\right) -{Q}_F\left( { \left( e^{\frac{2 \pi i}{{{\tilde{d}}}}} \alpha \zeta \right) }^d \right) \right] +\nonumber \\ \cdots +\left[ {Q}_F\left( {(\alpha \zeta )}^{d^{d-2}}\right) -{Q}_F\left( {\left( e^{\frac{2 \pi i}{{\tilde{d}}}} \alpha \zeta \right) }^{d^{d-2}} \right) \right] . \end{aligned}$$

(6.4)

While comparing the coefficients of the polynomials appearing in the right hand side and in the left hand side of (6.4), note that if $\zeta ^{d^r l}=\zeta ^{d^s m}$ for $0\le r,s \le d-2$ and $1\le l,m \le d+1$ with $r\ne s$ (without loss of generality $r>s$, say), then $m=d^{r-s} l$. Since $d\ge 2$ and $0\le r,s \le d-2$ and $1\le l,m \le d+1$, clearly $r=s+1$ and $l=1$. Thus since next to the highest coefficients of both $Q_H$ and $Q_F$ vanish ($Q_H$ and $Q_F$ are of the form (4.10) and (4.11), respectively), we obtain

$$\begin{aligned}{} & {} \beta \left[ {Q}_H\left( \zeta ^{d^{d-2}}\right) -{Q}_H\left( {\left( e^{\frac{2 \pi i}{{\tilde{d}}}} \zeta \right) }^{d^{d-2}}\right) \right] \nonumber \\{} & {} \quad = {Q}_F\left( {(\alpha \zeta )}^{d^{d-2}}\right) -{Q}_F\left( {\left( e^{\frac{2 \pi i}{{\tilde{d}}}} \alpha \zeta \right) }^{d^{d-2}} \right) . \end{aligned}$$

(6.5)

Comparing both sides of (6.5), we get

$$\begin{aligned} \beta A_{d-k}^H={\tilde{\alpha }}^{d-k} A_{d-k}^F, \end{aligned}$$

for $1\le k \le (d-1)$, where ${\tilde{\alpha }}=\alpha ^{d^{d-2}}$. Equivalently, we get

$$\begin{aligned} A_{d-k}^H={\tilde{\alpha }}^{-(k+1)}A_{d-k}^F \end{aligned}$$

(6.6)

for $1\le k \le (d-1)$, with ${{\tilde{\alpha }}}^{d+1}=1$. Note that (6.6) is an analogue of (4.13) and therefore, using the same set of arguments as in Section 4, we get

$$\begin{aligned} \beta p_H(y)={\tilde{\alpha }}p_F({\tilde{\alpha }}y), \end{aligned}$$

(6.7)

for all $y\in {\mathbb {C}}$, with ${{\tilde{\alpha }}}^{d+1}=\beta$ and $\beta ^{d-1}=1$.

7 Proofs of Theorem 1.1 and Theorem 1.2

Proof of Theorem 1.1: It follows from (5.17), (6.7) and (6.1) that

$$\begin{aligned} \alpha p_F(\alpha y)=\beta p_H(y), \end{aligned}$$

with $\alpha ^{d+1}=1$ and

$$\begin{aligned} \delta _F=\gamma \delta _H, \end{aligned}$$

with $\alpha ^{d^2-1}=1$ and $\beta ^{d-1}=\gamma ^{d-1}=1$. Thus if

$$\begin{aligned} A(x,y)=(\alpha x, \beta \alpha ^{-1} y) \text { and } B(x,y)=(\gamma \alpha \beta ^{-1}x, \alpha ^{-1}y), \end{aligned}$$

(7.1)

then a simple calculation gives

$$\begin{aligned} F\equiv A\circ H \circ B \end{aligned}$$

in ${\mathbb {C}}^2$. Now note that if we write $A=(A_1,A_2)$ and $B=(B_1,B_2)$, then

$$\begin{aligned} {\left( A_1(x,y)\right) }^{d-1}={(B_1(x,y))}^{d-1} \text { and }{\left( A_2(x,y)\right) }^{d-1}={(B_2(x,y))}^{d-1}. \end{aligned}$$

Thus,

$$\begin{aligned} A_1(x,y)=\mu _1 B_1(x,y) \text { and }A_2(x,y)=\mu _2 B_2(x,y), \end{aligned}$$

(7.2)

for some $\mu _1, \mu _2 \in {\mathbb {C}}$, with $\mu _1^{d-1}=\mu _2^{d-1}=1$. Now comparing (7.1) and (7.2), we obtain $\mu _1=\gamma ^{-1}\beta$ and $\mu _2=\beta$. Therefore,

$$\begin{aligned} A(x,y)=L_\mu \circ B(x,y), \end{aligned}$$

for all $(x,y)\in {\mathbb {C}}^2$, where $L_\mu (x,y)=(\mu _1 x, \mu _2 y)$. Thus,

$$\begin{aligned} F\equiv L_\mu \circ B \circ H \circ B. \end{aligned}$$

Proof of Theorem 1.2: Implementing the idea discussed just after stating the Theorem 1.1 in the Introduction, proof of Theorem 1.2 follows immediately once we prove Theorem 1.1.

It would be interesting to investigate the converse of Theorem 1.2.

Question: If H and F are two Hénon maps such that

$$\begin{aligned} F=A_1\circ H \circ A_2 \end{aligned}$$

in ${\mathbb {C}}^2$, where $A_1$ and $A_2$ are affine automorphisms in ${\mathbb {C}}^2$, then how are the escaping sets of H and F related?