Hénon mappings with biholomorphic escaping sets
 Sylvain Bonnot^{1}Email authorView ORCID ID profile,
 Remus Radu^{2} and
 Raluca Tanase^{2}
https://doi.org/10.1186/s4062701700109
© The Author(s) 2017
Received: 27 July 2017
Accepted: 20 October 2017
Published: 10 November 2017
Abstract
For any complex Hénon map \(H_{P,\,a}:\left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}P(x)ay\\ x\end{array}}\right),\) the universal cover of the forward escaping set \(U^{+}\) is biholomorphic to \({\mathbb D}\times \mathbb C\), where \({\mathbb D}\) is the unit disk. The vertical foliation by copies of \(\mathbb C\) descends to the escaping set itself and makes it a rather rigid object. In this note, we give evidence of this rigidity by showing that the analytic structure of the escaping set essentially characterizes the Hénon map, up to some ambiguity which increases with the degree of the polynomial P.
Keywords
1 Introduction
Those maps were first studied in [5–7]. The behaviour at infinity has been addressed in [8] and the structure of the associated Julia sets has been considered in a long series of articles by Bedford and Smillie, culminating in [1, 2]. We define the escaping set \(U^{+}\) to be the set of all points of \(\mathbb C^{2}\) with unbounded forward orbit.
The whole purpose of this article can be summarized as follows: can one recover a Hénon map from the knowledge of the analytic structure of its escaping set? This question was first formulated by Hubbard, based on the explicit formulas of [6]. Our study is built on the methods first developed by Bousch in his unpublished manuscript [3].
1.1 Results
Let us consider two complex Hénon mappings \(H_{i}:\left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}P_{i}(x)a_{i}y\\ x\end{array}}\right)\), \(i \in \{1,\,2\}\) together with their respective forward escaping sets \(U^{+}_{1}, \,U^{+}_{2}\).
Theorem 1.1
 (a)
if one of the degrees \(d_1,\, d_2\) is prime, then necessarily \(d_1 = d_2;\)
 (b)
if \(d_1 = d_2\) (not necessarily prime), then the quotient \(\left( \dfrac{a_1}{a_2}\right)\) is a \((d1)\) th root of unity.
Quadratic case
Observe that the theorem above already indicates that in degree 2 the Jacobian is prescribed. Actually, a more precise result can be obtained in this case:
Theorem 1.2
 (a)
the forward escaping sets are biholomorphic;
 (b)
\(H_{1}=H_{2}\).
Ambiguity in higher degrees
Theorem 1.3
 1.
\(H_{1}\) and \(H_{2}\) are conjugate by \(\left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}\,x\\ \,y\end{array}}\right)\) (equivalently \((a_2,\,A_2,\,B_2)=(a_1,\,A_1,\,\,B_1));\)
 2.
\((a_1,\,A_1)=(a_2,\,\;A_2)\) and \(B_1=B_2=0,\) and in this case the two applications are not conjugate on \(\mathbb C^{2},\) but their squares are conjugate by \(g: \left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}ix\\ \,iy\end{array}}\right).\) Moreover, g is an explicit isomorphism between the two escaping sets.
We notice that the Jacobian is still prescribed in the cubic case. Surprisingly, this is not necessarily true in higher degrees:
Theorem 1.4
In the last theorem, the map \(\theta _{1}\) is actually a conjugacy on \(\mathbb C^{2}\) between \(H_{1}^{\circ 3}\) and \(H_{2}^{\circ 3}\). Thus one can ask:
Conjecture
Two complex Hénon mappings of degree d have biholomorphic escaping sets if and only if their \((d1)\) th powers are conjugate.
1.2 Idea of the proof
The first ingredient is an explicit trivialization of the universal cover of the escaping set. Actually, we prefer to work in an intermediate cover described as follows:
Theorem 1.5
 (a)
\(Q(X)=X^{3}\frac{a_{0}}{2}X\) if \(d=2,\)
 (b)
\(Q(X)=X^{d+1}\frac{a_{d2}}{d}X^{d1}\frac{a_{d3}}{d}X^{d2}(\frac{a_{d4}}{d} \frac{a_{d2}^{2}}{d^{2}})X^{d3} + \cdots ,\) if \(d \ge 3.\)
As such, this theorem is a rather straightforward generalization to any degree d of a theorem due to Hubbard and ObersteVorth (compare [6] for the quadratic case). As these authors notice, we can obtain an explicit representation of the fundamental group as a group of biholomorphisms of \((\mathbb C\overline{\mathbb {D}}) \times \mathbb C{:}\)
Theorem 1.6
 (1)
the fundamental group of \(U^{+}\) is isomorphic to \({\mathbb Z }\left[\frac{1}{d}\right];\)
 (2)the element \(\left[ {\frac{k}{{d^{n} }}} \right]\) of \(\pi _{1}(U^{+})\) is represented by the biholomorphism$$\begin{aligned} \gamma _{\frac{k}{d^{n}}}: \left( {\begin{array}{c}\zeta \\ z\end{array}}\right) \mapsto \left( {\begin{array}{c}e^{i 2\pi \frac{k}{d^{n}}}\cdot \zeta \\ z + \frac{d}{a}\sum _{l=0}^{\infty }\left(\frac{d}{a}\right)^{l}\left(P\left(\zeta ^{d^{l}}\right)P\left(\left(e^{i 2\pi \frac{k}{d^{n}}}\zeta\right )^{d^{l}}\right)\right)\end{array}}\right) . \end{aligned}$$
The second ingredient of our proof is a construction used by Bousch [3]. Any biholomorphism between escaping sets that induces the identity on the fundamental groups can be lifted to an isomorphism \((\mathbb C \overline{\mathbb {D}})\times \mathbb C.\) But these have a very simple form, and due to the very particular form of the fibers of \(\pi :\widetilde{U^{+}}\rightarrow U^{+},\) it turns out that very few of these isomorphisms can satisfy the additional requirement of sending the fibers of \(\pi _{1}\) to the fibers of \(\pi _{2}:\widetilde{U_{2}^{+}} \rightarrow U_{2}^{+}.\) The rigidity derives exactly from this fact.
2 Equality of degrees
Our objective is the following:
Theorem 2.1
(Conditions on degrees) Let \(U^{+}_{1}\) be the escaping set of a complex Hénon mapping of degree d, with d being a prime integer. Assume that the escaping set \(U^{+}_{2}\) of some other Hénon mapping is biholomorphic to \(U^{+}_{1};\) then necessarily the polynomials \(P_{1},\, P_{2}\) have the same degree d.
Let us prove first an easy lemma:
Lemma 2.2
The degree \(d'\) of \(H_{2}\) is a power of d.
Proof
Let \(\theta :U^{+}_{1}\rightarrow U^{+}_{2}\) be the biholomorphism. The map \(\psi =\theta ^{1}\circ H_{2} \circ \theta : U^{+}_{1} \rightarrow U^{+}_{1}\) induces an automorphism on the fundamental group of \(U^{+}_{1}={\mathbb Z }\left[\frac{1}{d}\right]\). These automorphisms are the multiplications by an invertible element of \({\mathbb Z } \left[\frac{1}{d} \right]\). Therefore, \(\psi\) induces the multiplication by a power of d. Since \(H_{2}\) induces the multiplication by \(d'\) and all the induced maps \(\theta _{\star }, \,\theta ^{1}_{\star }\) commute (being multiplications by some integer), we deduce that \(d'=d^{n}\) for some integer n. \(\square\)
Lemma 2.3
There exists a biholomorphism between the escaping sets that induces the identity on the fundamental groups.
Proof
Since \(d'=d^{n}\), we deduce that \({\mathbb Z }[1/d']={\mathbb Z }[1/d]\). Therefore, the biholomorphism \(\theta\) must induce an automorphism of \({\mathbb Z }[1/d]\). These are made of multiplication by a power of d. By precomposing \(\theta\) with an iterate of \(H_{1}\), we can assume that \(\theta\) induces \(\pm \,Id\) on \({\mathbb Z }[1/d]\). Now we notice that the projection \(\pi :(\mathbb C \overline{\mathbb {D}}) \times \mathbb C\rightarrow U^{+}\) induces on the fundamental groups the map \(\pi _{\star }:{\mathbb Z }\mapsto {\mathbb Z }\subset {\mathbb Z }[1/d].\) Therefore, by standard covering theory \(\theta\) can be lifted to a selfmap of \((\mathbb C \overline{\mathbb {D}}) \times \mathbb C.\) If needed, one can postcompose with a deck transformation and assume that the lifted map \(\tilde{\theta }\) is actually an isomorphism of \((\mathbb C \overline{\mathbb {D}}) \times \mathbb C.\)
Lemma 2.4
The isomorphisms of \((\mathbb C \overline{\mathbb {D}}) \times \mathbb C\) are given by \(\phi : \left( {\begin{array}{c}\zeta \\ z\end{array}}\right) \mapsto \left( {\begin{array}{c}\alpha \zeta \\ \alpha (\zeta )z + \beta (\zeta )\end{array}}\right),\) where \(\alpha \in \mathbb C\) and \(\alpha (\zeta ),\, \beta (\zeta )\) are two holomorphic functions.
Proof
Left to the reader. See Bousch’s thesis [3] for details. \(\square\)
This implies, as Bousch notices, that \(\widetilde{\theta }\) actually induces \(+\,Id,\) because a loop \(\left( {\begin{array}{c}R\cdot {\rm e}^{i 2\pi t}\\ 0\end{array}}\right)\) in \((\mathbb C \overline{\mathbb {D}}) \times \mathbb C\) descends to the generator \(+\,1\) of \(\pi _{1}(U^{+})\) in \(U^{+}_{1}\) and is sent to the loop \(\left( {\begin{array}{c}\alpha R\cdot {\rm e}^{i 2\pi t}\\ \beta (R\cdot {\rm e}^{i 2\pi t})\end{array}}\right)\) which descends to the same generator in \(U^{+}.\)
2.1 Lifting of biholomorphisms
2.2 Fibers of \(\pi :\widetilde{U}^{+}\rightarrow U^{+}\) are mapped to fibers
Theorem 2.5
Proof

for \(n=1,\) we get the identity;

then we notice that \(\widetilde{H} \circ \gamma _{\frac{1}{d^{n+1}}}= \gamma _{\frac{1}{d^{n}}} \circ \widetilde{H}.\) Henceand also$$\begin{aligned} z'' = \frac{a}{d}z' + Q\left({\rm e}^{i \frac{2 \pi }{d^{n+1}}}\cdot \zeta\right ) \end{aligned}$$$$\begin{aligned} z'' = \frac{a}{d}z + Q(\zeta ) + \frac{d}{a}\sum _{l=0}^{\infty }\left(\frac{d}{a}\right)^{l}\left(Q\left(\zeta ^{d^{l}}\right)Q\left(\left({\rm e}^{i 2\pi \frac{1}{d^{n}}}\zeta \right)^{d^{l}}\right)\right) \end{aligned}$$$$\begin{aligned} \frac{a}{d}z' + Q\left({\rm e}^{i \frac{2 \pi }{d^{n+1}}}\cdot \zeta \right) = \frac{a}{d}z + Q\left(\zeta\right ) + \frac{d}{a}\sum _{l=0}^{\infty }\left(\frac{d}{a}\right)^{l}\left(Q\left(\zeta ^{d\cdot d^{l}}\right)Q\left(\left({\rm e}^{i 2\pi \frac{1}{d^{n}}}\zeta ^{d}\right)^{d^{l}}\right)\right). \end{aligned}$$
Conservation of fibers This whole section is an application of Bousch’s procedure.
Lemma 2.6
The degree \(d_{2}=d'=d^{t}\) is equal to d.
Proof
Fix \(z=0\) and take a fixed \(\zeta.\) Fix also \(k=1\) and see what happens for the particular choice of integers \(n=t\cdot b,\) with b going to infinity. It turns out that all the terms \(\alpha \left({\rm e}^{i\cdot \frac{2\pi }{d_{1}^{n}}}\cdot \zeta \right), \,\beta \left({\rm e}^{i\cdot \frac{2\pi }{d_{1}^{n}}}\cdot \zeta \right)\) remain bounded, whereas the rest grows like the leading terms in the sum which are \(\zeta ^{(d_{2}+1).d_{2}^{b1}}=\zeta ^{(d^{t}+1).d^{t.(b1)}}\) in one case, and \(\zeta ^{(d+1).d^{bt1}}\) in the other. This implies that t must be equal to one, and therefore \(d_{2}=d^{t}=d_{1},\) which is what we wanted. \(\square\)
Summary At this point, we have proved that if an escaping set of a Hénon map with prime degree d is biholomorphic to another escaping set, then necessarily the other Hénon map has degree d as well.
2.3 Condition on the Jacobians
Now that we have an explicit description of the fibers, one can write down necessary condition for them to coincide.
Theorem 2.7
(Condition on the Jacobians) If \(U^{+}_{1}\) and \(U^{+}_{2}\) are biholomorphic, then \(\left( \frac{a_{1}}{a_{2}}\right)\) is a \((d1){\text {th}}\) root of the unity, where d is the common degree of both maps.
Proof
We fix \(k=1\) and let n vary: one can first deduce that \(\alpha \left({\rm e}^{i \frac{2k \pi }{d^{n}}} \zeta\right )=\alpha (\zeta ),\) for any n, any \(\zeta\). Therefore, the function \(\alpha\) is a constant K (a nonzero constant because \(\widetilde{\theta }\) is an isomorphism): \(\forall \zeta \in (\mathbb C\overline{\mathbb {D}}), \,\alpha (\zeta )=K.\) Now fix \(\zeta\) and let n vary. The leading terms of the two expressions are as follows: \(K.\frac{d}{a_{1}}\cdot \left( \frac{d}{a_{1}}\right) ^{n1}\cdot \left( \zeta ^{(d+1)\cdot d^{n1}} {\rm e}^{i \frac{2 \pi }{d^{n}}\cdot d^{n1}\cdot (d+1)}\cdot \zeta ^{(d+1)\cdot d^{n1}}\right) ,\) for the first one, and \(\frac{d}{a_{2}}\cdot \left( \frac{d}{a_{2}}\right) ^{n1}\cdot \alpha ^{(d+1)\cdot d^{n1}}\cdot \zeta ^{(d+1)\cdot d^{n1}}\left( 1 {\rm e}^{i \frac{2 \pi }{d}}\right),\) for the second one.
Notice that the terms \(\beta (\zeta )\) and \(\beta ({\rm e}^{i \frac{2 \pi }{d^{n}}}\zeta )\) are bounded. After comparing these leading terms, we get for n large enough \(\lim _{n} \left( \left( \frac{a_{1}}{a_{2}}\right) ^{n}\alpha ^{(d+1)\cdot d^{n1}}\right) =K.\)
Similarly, we obtain \(\lim _{n} \left( \left( \frac{a_{1}}{a_{2}}\right) ^{n+1}\alpha ^{(d+1)\cdot d^{n}}\right) =K.\) Hence, by quotienting \(\lim _{n} \left( \left( \frac{a_{1}}{a_{2}}\right) \alpha ^{(d+1)\cdot (d1)\cdot d^{n1}}\right) =1.\)
By quotienting again one finally gets \(\lim _{n}\left( \alpha ^{(d+1)\cdot (d1)^{2}\cdot d^{n1}}\right) =1.\)
But this implies \(\left( \frac{a_{1}}{a_{2}}\right) ^{d1}=1.\) \(\square\)
2.4 Quadratic case
Theorem 2.8
Let \(U^{+}_{1}\) and \(U^{+}_{2}\) be the escaping sets of two quadratic complex Hénon mappings written in normalized form, \(H_{a,\,c}\) and \(H_{a',\,c'},\) respectively. If \(U^{+}_{1}\) and \(U^{+}_{2}\) are biholomorphic then \(a=a'\) and \(c=c'.\)
Proof
Since we know from the preceding theorem that the Jacobians are equal, we can pursue the comparison of the fibers by looking at the next leading term.
Therefore \(c \sim c'\cdot \alpha ^{2^{n}}\) which implies \(c=c'\). \(\square\)
2.5 Cubic case
This case is interesting because the degree is low enough to enable us to compute everything precisely, and at the same time it is the first case for which the analytic structure of the escaping set does not completely determine the Hénon map that created it.
Theorem 2.9
 1.
\(H_{1}\) and \(H_{2}\) are conjugate by \(\left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}\,x\\ \,y\end{array}}\right)\) (equivalently \((a',\,A',\,B')=(a,\,A,\,\,B));\)
 2.
\((a,\,A)=(a',\,\,A')\) and \(B=B'=0,\) and in this case the two applications are not conjugate on \(\mathbb C^{2},\) but their squares are conjugate by \(\left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}ix\\ \,iy\end{array}}\right).\) Moreover, this map is an explicit isomorphism between the two escaping sets.
Remark 1
A necessary condition for the escaping sets to be biholomorphic is that the Jacobians are equal. We will see that this is not true in degrees higher than 4.
Proof
Again we compare the successive leading terms:

A=A′ We already know that \(B \sim B' \alpha ^{3^{n}}\) and also that \(\alpha ^{4.3^{n}}\sim 1\) for n large enough. Therefore, either \(B=B'=0,\) or B and \(B'\) are nonzero and satisfy \(B= \pm \,B'.\)

A = − A′ Similarly, either \(B=B'=0,\) or B and \(B'\) are both nonzero, but then they must satisfy \(\pm \,i= \frac{B}{B'}=\alpha ^{3^{n}},\) which is impossible because \(\pm \,i^{3^{n}}\) is not constant.
2.6 In degree 3, \(U^{+}\) does not determine H
In this section we exhibit explicit examples of pairs of different maps having biholomorphic escaping sets.
Theorem 2.10
Let us consider \(H_{1}:\left( {\begin{array}{c}X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}X^{3}+CXaY\\ X\end{array}}\right)\) and \(H_{2}:\left( {\begin{array}{c}X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}X^{3}CXaY\\ X\end{array}}\right)\). Then the map \(\theta : \left( {\begin{array}{c}X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}iX\\ \,iY\end{array}}\right)\) is an isomorphism of \(\mathbb {C}^{2}\) into itself that restricts to an isomorphism from \(U_{2}^{+}\) into \(U_{1}^{+}.\) Moreover, \(\theta\) conjugates the squares of the two maps.
Proof

First notice that \(\,Id\) commutes with \(H_{1}\) (and also \(H_{2}\)):and that \(\theta ^{\circ 2}=Id\)$$\begin{aligned} \left( {\begin{array}{c}X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}\,X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}\,X^{3}CX+aY\\ \,X\end{array}}\right) \mapsto \left( {\begin{array}{c}X^{3}+CXaY\\ X\end{array}}\right) . \end{aligned}$$

Then one verifiesindeed$$\begin{aligned} \theta ^{1} \circ H_{1}= H_{2} \circ \theta ; \end{aligned}$$and$$\begin{aligned} \theta ^{1} \circ H_{1}: \left( {\begin{array}{c}X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}X^{3}+CXaY\\ X\end{array}}\right) \mapsto \left( {\begin{array}{c}iX^{3}iCX+aiY\\ iX\end{array}}\right) \end{aligned}$$$$\begin{aligned} H_{2} \circ \theta : \left( {\begin{array}{c}X\\ Y\end{array}}\right) \mapsto \left( {\begin{array}{c}iX\\ iY\end{array}}\right) \mapsto \left( {\begin{array}{c}iX^{3}iCX+aiY\\ iX\end{array}}\right) . \end{aligned}$$
To conclude, let \(p \in U^{+}_{2};\) then \(\mid \mid (\,Id)^{n} \circ H_{2}^{\circ n}(p) \mid \mid\) is unbounded but then \(\theta ^{1} \circ H_{1}^{\circ n} ( \theta (p))\) is unbounded and then \(\theta (p) \in U^{+}_{1}.\) It remains to show that \(\theta\) conjugates the squares of the maps: start from \(H_{1}=\theta \circ H_{2} \circ \theta\) and deduce \(H_{1}^{2}=\theta \circ H_{2} \circ (\,Id) \circ H_{2} \circ \theta,\) or if one prefers, by remembering that \(\,Id\) commutes with \(H_{i}\) and that \(Id \circ \theta =\theta ^{1},\) one obtains \(H_{1}^{2}=\theta ^{1} \circ H_{2}^{2} \circ \theta .\) \(\square\)
3 Jacobian not prescribed
Let us show with an explicit example that in degree 4 the Jacobian is not necessarily prescribed by the analytic structure of the escaping set.
Theorem 3.1

\(H_{1}: \left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}x^{4}y\\ x\end{array}}\right),\)

\(H_{2}: \left( {\begin{array}{c}x\\ y\end{array}}\right) \mapsto \left( {\begin{array}{c}x^{4} jy\\ x\end{array}}\right) ,\quad where \, j=e^{i\frac{2\pi }{3}},\)
Proof
If one recalls that L commutes with both \(H_{1}\), \(H_{2}\), one deduces that \(H_{2}^{\circ n} \circ \theta _{1}^{1}=\theta _{2} \circ L^{n} \circ H_{1}^{\circ n}\).
Precomposing by L or \(\theta _{i}\) does not change the norm of a vector (each component being multiplied by a complex number on the unit circle). Therefore, if \(p \in U^{+}_{1}\) then \(\Vert H_{1}^{\circ n }(p) \Vert\) is unbounded and also \(\Vert H_{2}^{\circ n }(\theta _{1}^{1}(p)\Vert\), but this shows that \(\theta _{1}^{1}(p)\) belongs to \(U^{+}_{2}\). \(\square\)
Declarations
Authors' contributions
SB wrote the first version, and all the authors discussed the topic and obtained the results. All authors read and approved the final manuscript.
Acknowledgements
The first author wishes to acknowledge support from the FAPESP Grants 2013/236434 and 2011/162658.
Competing interests
The authors declare that they have no competing interests.
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