On the regular leaf space of the cauliﬂower
Tomoki Kawahira ∗
Department of Mathematics
Graduate School of Science
Kyoto University
Email: kawahira@math.kyoto-u.ac.jp
June 4, 2003
Abstract
We construct a pinching semiconjugacy from a quadratic polynomialfc(z) = z2 + c with c ∈ (0, 1/4) to f1/4(z) = z2 + 1/4 on the sphere. Bylifting this semiconjugacy to their natural extensions, we investigate thestructure of the regular leaf space of f1/4 in detail.
1Introduction
As an analogy to hyperbolic 3-orbifolds associated with Kleinian groups, Lyu-bich and Minsky[2] introduced hyperbolic orbifold 3-laminations associated withrational maps. For a given rational map of degree ≥ 2, considering its naturalextension and regular leaf space is the ﬁrst step to the construction of such ahyperbolic orbifold 3-lamination.However, the global structures of the regular leaf spaces of rational maps arenot precisely known except only a few examples. Here is one of such examples.For fc(z) = z2 + c with c in the main cardioid of the Mandelbrot set, all regularleaf spaces of fc are topologically the same as that of f0(z) = z2, which is 2-dimensional extension of 2-adic solenoid[3, Example 2][2, §11].Now we may expect the simplest deformation of these regular leaf spacesas c tends to 1/4 along the real axis. Then the dynamics inside the Julia setsdegenerate from “hyperbolic” to “parabolic”, though the dynamics on and outsidethe Julia sets are still topologically the same as f0. In this paper, we describe thestructure of the regular leaf space of f1/4(z) = z2 + 1/4, whose Julia set is calledthe cauliﬂower, by using a pinching semiconjugacy from fc with c ∈ (0, 1/4) to
∗Research partially supported by JSPS Research Fellowships for Young Scientists
f1/4. We will see that the transversal structure of those regular leaf spaces arepreserved, however, the dynamics on the invariant leaves corresponding to theﬁxed points are change from “hyperbolic” to “parabolic”.In this section, we ﬁrst survey some basic notion of complex dynamics ofquadratic maps and their natural extensions. In §2, we construct tessellation(or tiling) of the interior of the ﬁlled Julia sets of fc with c ∈ (0, 1/4] in adynamically natural way. In §3, we construct a pinching semiconjugacy from fcwith c ∈ (0, 1/4) to f1/4 by gluing tile-to-tile homeomorphisms and the conjugacyon and outside the Julia sets. In §4, we lift such a semiconjugacy to their regularleaf spaces and describe their degeneration.
1.1Preliminaries
The Julia set.Let us set fc(z) = z2 + c (c ∈ C) and consider it as a rationalmap on the Riemann sphere ¯C = C ∪ {∞} with fc(∞) = ∞. The ﬁlled Julia setKc of fc is deﬁned by
Kc :=�z ∈ ¯C : {f nc (z)}∞n=0 is bounded�.
The Julia set Jc of fc is the boundary of Kc. One can easily check that those setsare forward and backward invariant under the action of fc.Now suppose that Kc is connected. (Thus so is Jc.) We denote the unit disk byD. For the outside of Kc, there exists a unique conformal map φc : ¯C−Kc → ¯C−¯Dsuch that
• φc(fc(z)) = φc(z)2; and
• φc(z)/z → 1 as z → ∞.
Moreover, if Jc is a Jordan curve, φ−1ccontinuously extends to ¯φ−1c: ¯C − D →¯C − K◦c .Now let us restrict our interest to the case of c ∈ (0, 1/4]. In our particularsituation, it is known that Jc are Jordan curves, and thus by looking through themap ¯φ−1c , the dynamics on and outside the Julia sets are topologically the sameas z �→ z2. For θ ∈ R/Z, set γc(θ) := ¯φ−1c (e2πiθ). Then points on Jc are bijectivelyparameterized by angles in R/Z. See [4, §18] for more details.We deﬁne some more notation:
• αc := (1 − √
1 − 4c)/2 which is the attracting (or parabolic iﬀ c = 1/4)ﬁxed point of fc with multiplier λc = 1 − √
1 − 4c.
• βc := (1 + √
1 − 4c)/2 which is the repelling (or parabolic iﬀ c = 1/4) ﬁxedpoint of fc with multiplier λ′c = 1 + √
1 − 4c.
The natural extension.Next we follow [2, §3]. For fc as above, let us considerthe set of all possible backward orbits
Nc :=�ˆz = (z0, z−1, . . .) : z0 ∈ ¯C, fc(z−n−1) = z−n�.
This set is called the natural extension of fc, and is equipped with a topologyfrom ¯C× ¯C×· · · . On this natural extension, the lift of fc and a natural projectionare deﬁned by
ˆfc(ˆz) := (fc(z0), z0, z−1, . . .) andπc(ˆz) := z0.
It is clear that ˆfc is a homeomorphism, and satisﬁes πc ◦ ˆfc = fc ◦ πc. For a ﬁxedpoint a ∈ ¯C of fc, set ˆa := (a, a, . . .) ∈ Nc.
The regular leaf space.An element ˆz = (z0, z−1, . . .) ∈ Nc is regular if thereexists a neighborhood U0 of z0 such that its pull-back U−n along the backwardorbit ˆz are eventually univalent. For example, ˆ∞ = (∞, ∞, . . .) is not regular forany c ∈ C.Let Rc denote the set of regular points in Nc. Rc is called the regular leafspace of fc. A leaf of Rc is a path connected component of Rc. By [2, Lemma3.1], leaves of Rc are Riemann surfaces:
Lemma 1.1 Leaves of Rc have the following properties:
• For each leaf L, we can introduce a complex structure such that πc : L → ¯Cis an analytic map.
• πc branches at ˆz = (z0, z−1, . . .) if and only if ˆz contains a critical point in{z−n}.
• ˆfc maps a leaf to a leaf isomorphically.
This lemma holds for any c ∈ C. In our case, we have:
Proposition 1.2 If c ∈ (0, 1/4], Rc has the following properties:
• Rc = Nc − { ˆ∞, ˆαc}.
• Each leaf of Rc is isomorphic to C.
This proposition is immediate from lemmas in [2, §3].
2Making tiles
For the ﬁrst step, we will decompose the interior K◦c of Kc with c ∈ (0, 1/4] intocountably many tiles and describe their relations.Before making tiles, we introduce some notation. For c ∈ (0, 1/4], let Ic(0)denote the interval [αc, βc] ⊂ R.In particular, I1/4(0) = {1/2}.Since Ic(0)contains no critical orbit, preimages of Ic(0) by fc are univalently spread outwith one of their endpoints on the Julia set.Set Ic := �
i≥0 f −ic (Ic(0)).Forθ ∈ Q/Z of the form k/2n, we denote the connected component of Ic containingγc(θ) by Ic(θ). As c → 1/4, Ic degenerates into I1/4 which is the grand orbit ofthe parabolic ﬁxed point 1/2.
2.1Tiles of K◦c with c ∈ (0, 1/4)
Suppose c ∈ (0, 1/4). Then αc is an attracting ﬁxed point and K◦c is its attractingbasin. On a neighborhood of αc, there exists a linearizing coordinate Φc whichanalytically conjugates the action of fc near αc to w �→ λcw near the origin.Moreover, we can extend this map to Φc : K◦c → C, and it is unique up tomultiplication by a constant[4].For the purpose of better understanding the diﬀerence between the cases of c ∈(0, 1/4) and c = 1/4, let us take an additional conjugation by w �→ w +1/(1−λc)on C. Then we can uniquely deﬁne Φc : K◦c → C such that
• Φc(fc(z)) = λcΦc(z)+1;
• Φc(αc) = 1/(1 − λc), Φc(0) = 0; and
• Φc is inﬁnitely branched covering whose branch points are �
i≥0 f −ic ({0}).
Now let us set ac := 1/(1 − λc), which tends to +∞ as c → 1/4.For m ∈ Z, set
Ac(m, +) :=�w ∈ C : λm+1cac ≤ |w − ac| ≤ λmc ac, Im w ≥ 0�
Ac(m, −) :=�w ∈ C : λm+1cac ≤ |w − ac| ≤ λmc ac, Im w ≤ 0�
and we call them the fundamental semi-annuli. Let Ac denote the collection ofall of the fundamental semi-annuli.Note the following two facts:
• The vertices of fundamental semi-annuli on the interval (−∞, ac) are theimages of the grand orbit of 0. In particular, all of the ramiﬁed points(critical values) of Φc are on the interval (−∞, 0].
• All components of Ic∩K◦c are mapped univalently onto the interval [ac, ∞).
For the boundary of Ac(m, ±), we call the edge on the interval (−∞, ac) (resp.[ac, ∞)) the critical edge (resp. degenerating edge). We call the edges shared byAc(m − 1, ±) or Ac(m + 1, ±) the circular edges.By the facts above, Φ−1c: C − (−∞, 0] → K◦c is a multi-valued functionwith univalent branches. Such a branch Ψc : C − (−∞, 0] → K◦c determines aunique angle θ ∈ Q/Z of the form k/2n such that Ψc([ac, ∞)) ∪ γc(θ) = Ic(θ).Thus for Ac(m, +) ∈ Ac, Ψc also determines a unique simply connected set T =Tc(θ, m, +) ⊂ K◦c such that T is the closure of Ψc(Ac(m, +) − (−∞, 0]). We callsuch a T a tile, and the triple (θ, m, +) the address of the tile T. We also deﬁneTc(θ, m, −) in the same way. We denote the collection of all possible T by Tc, andcall it the tessellation of K◦c . Now it is clear that Kc is the closure of � {T ∈ Tc}.For each T ∈ Tc, Φc maps T to an A ∈ Ac homeomorphically.For theboundary of T, critical (resp. degenerating, circular) edges are deﬁned by theboundary arcs corresponding to the critical (resp. degenerating, circular) edgesof A.Now we give relations among tiles:
Proposition 2.1 For T = Tc(θ, m, +) ∈ Tc, we have the following properties:
(1) fc(T) = Tc(2θ, m + 1, +). Moreover, f −1c (T) = Tc(θ/2, m − 1, +) ∪ Tc(θ/2 +1/2, m − 1, +).
(2) T shares the circular edges with Tc(θ, m − 1, +) and Tc(θ, m + 1, +).
(3) T shares the degenerating edge with Tc(θ, m, −).
(4) T shares the critical edge with Tc(θ + 2m, m, −).
The similar holds if we replace T by Tc(θ, m, −). In particular, Tc(θ, m, −) sharesthe critical edge with Tc(θ − 2m, m, +).
Proof.The ﬁrst three properties come from the deﬁnition of tiles. One caneasily check property (4) in the case of θ = 0. For general θ of the form k/2n ∈Q/Z, suppose that T = Tc(θ, m, +) shares the critical edge with T ′ ∈ Tc. ThenT ′ is the form Tc(θ′, m, −) since Φc maps T and T ′ to Ac(m, +) and Ac(m, −)respectively. Now f n sends T univalently to Tc(0, m + n, +) which shares thecritical edge with Tc(2m+n, m + n, −). Thus f n(T ′) = Tc(2m+n, m + n, −). Notethat f n(T ∪T ′) joins Ic(0) and Ic(2m+n). By pulling it back by a suitable univalentbranch of f n, T ∪ T ′ must join Ic(θ) and Ic(θ + 2m). This implies θ′ = θ + 2m. ■
2.2Tiles of K◦1/4
Next we make the tessellation T1/4 for K◦1/4 in the same way as above.Nowα1/4 = 1/2 is the parabolic ﬁxed point and K◦1/4 is its parabolic basin. On an
Figure 1: The tessellations of z2 + 0.2 and z2 + 1/4
attracting petal of α1/4, there exists a Fatou coordinate Φ1/4 which analyticallyconjugates the action of f1/4 to w �→ w + 1. Moreover, we can uniquely extendthis Fatou coordinate to Φ1/4 : K◦1/4 → C such that
• Φ1/4(f1/4(z)) = Φ1/4(z) + 1;
• Φ1/4(0) = 0; and
• Φ1/4 is inﬁnitely branched covering whose branch points are �
n≥0 f −n1/4({0}).
(See [4] again.)For m ∈ Z, set
A1/4(m, +) := {w ∈ C : m ≤ Re w ≤ m + 1, Im w ≥ 0}A1/4(m, −) := {w ∈ C : m ≤ Re w ≤ m + 1, Im w ≤ 0}
and we call them the fundamental semi-cylinders. Let A1/4 denote the collectionof all of the fundamental semi-cylinders.Note the following two facts, and compare with the case of c ∈ (0, 1/4):
• The vertices of fundamental semi-cylinders on the real axis (−∞, ∞) arethe images of the grand orbit of 0. In particular, all of the ramiﬁed points(critical values) of Φ1/4 are on the interval (−∞, 0].
• All components of I1/4 are outside of the domain of Φ1/4.
For the boundary of A1/4(m, ±), we call the edge on the real axis the critical edge.We also call the edges shared by A1/4(m − 1, ±) or A1/4(m + 1, ±) the circularedges. Note that A1/4(m, ±) has no edges corresponding to degenerating edges offundamental semi-annuli.
Now Φ−11/4 : C − (−∞, 0] → K◦1/4 is a multi-valued function with univalentbranches. Such a branch Ψ1/4 : C − (−∞, 0] → K◦1/4 determines a unique angleθ ∈ Q/Z of the form k/2n such that I1/4(θ) ∈ I1/4 is the limit of Ψ1/4(t) as t tendsto +∞ along the real axis. Thus for A1/4(m, +) ∈ A1/4, Ψ1/4 maps the interior ofA1/4(m, +) to a simply connected subset of K◦1/4. Since Ψ1/4 extends continuouslyto the boundary of A1/4(m, +), we denote its image by T = T1/4(θ, m, +) ⊂ K◦1/4.We also deﬁne T1/4(θ, m, −) in the same way. We denote the collection of allpossible T by T1/4, and call it the tessellation of K◦1/4. Now it is also clear thatK1/4 is the closure of � �T ∈ T1/4�.For each T ∈ T1/4, Φc maps T to an A ∈ A1/4 homeomorphically. For theboundary of T, critical (resp. circular) edges are deﬁned by the boundary edgescorresponding to the critical (resp. circular) edges of A. Note that T1/4(θ, m, ±)does not contain the point I1/4(θ) ∈ J1/4, and has no edges corresponding todegenerating edges.The relations among tiles are given in the same way as Proposition 2.1:
Proposition 2.2 For T = T1/4(θ, m, +) ∈ T1/4, we have the following properties:
(1) f1/4(T) = T1/4(2θ, m + 1, +).Moreover, f −11/4(T) = T1/4(θ/2, m − 1, +) ∪T1/4(θ/2 + 1/2, m − 1, +).
(2) T shares the circular edges with T1/4(θ, m ± 1, +).
(3) For any n ∈ Z,
T shares a point I1/4(θ) ∈ I1/4 with
T1/4(θ, n, +) and
T1/4(θ, n, −).
(4) T shares the critical edge with T1/4(θ + 2m, m, −).
The similar holds if we replace T by T1/4(θ, m, −).
3Construction of the semiconjugacy
Here we construct a semiconjugacy which pinches Ic to I1/4:
Theorem 3.1 For c ∈ (0, 1/4), there exists a semiconjugacy Hc : ¯C → ¯C fromfc to f1/4 such that
• Hc maps ¯C−Ic to ¯C−I1/4 homeomorphically and is a topological conjugacybetween fc|(¯C − Ic) and f1/4|(¯C − I1/4);
• For each θ = k/2n ∈ Q/Z, Hc maps an arc Ic(θ) onto a point I1/4(θ).
Proof.The rest of this section is devoted to the proof of this theorem. Let usﬁx c ∈ (0, 1/4).
Conjugation on tiles.First we make a homeomorphism between Ac(0, +) −[ac, ∞) (a tile minus its degenerating edge) and A1/4(0, +). Take w ∈ Ac(0, +) −[ac, ∞), and set reit := w − ac with (ac − 1 =)λcac ≤ r ≤ ac and 0 < t ≤ π. Nowsethc(w) := (ac − r) + i tan π − t
2∈ A1/4(0, +).
Then hc is a homeomorphism between Ac(0, +) − [ac, ∞) and A1/4(0, +) whichpreserves the action of Sc(w) = λcw +1 on the (outer) circular edge of Ac(0, +)−[ac, ∞) to that of S1/4(w) = w + 1 on the (left) circular edge of A1/4(0, +).
Figure 2: hc : Ac(0, +) → A1/4(0, +)
By using this property, for any m ∈ Z, we can extend hc to Ac(m, +)−[ac, ∞)by
hc : Ac(m, +) − [ac, ∞) → A1/4(m, +)w �→ hc(S−mc(w)) + m.
Then hc gives a homeomorphism between Ac(m, +) − [ac, ∞) and A1/4(m, +).Similarly, we deﬁne hc on Ac(m, −)−[ac, ∞). Then we obtain a homeomorphismhc : C − [ac, ∞) → C, moreover, hc is a topological conjugacy between Sc|(C −[ac, ∞)) and S1/4.Next, let us take θ of the form k/2n ∈ Q/Z, and take m ∈ Z.We de-ﬁne a map between Tc(θ, m, +) − Ic(θ) (a tile minus its degenerating edge) andT1/4(θ, m, +) as follows: There is a unique branch Ψθ1/4 of Φ−11/4 which maps theinterior of A1/4(m, +) to the interior of T1/4(θ, m, +). It is clear that Ψθ1/4 extendscontinuously to the boundary of A1/4(m, +). For z ∈ Tc(θ, m, +) − Ic(θ), set
Hc : Tc(θ, m, +) − Ic(θ) → T1/4(θ, m, +)
z �→ Ψθ1/4 ◦ hc ◦ Φc(z).
This deﬁnition gives a homeomorphism Hc : K◦c − Ic → K◦1/4, moreover, by thedeﬁnition and the combinatorics of tiles, Hc is a topological conjugacy betweenfc|(K◦c − Ic) and f1/4|K◦1/4.
Continuous extension.For θ ∈ Q/Z of the form k/2n, set Hc(Ic(θ)) :=I1/4(θ) = γ1/4(θ). Then Hc : K◦c ∪ Ic → K◦1/4 ∪ I1/4 is a semiconjugacy whichpinches the arcs in Ic to the points in I1/4. Now we claim that we can continu-ously extend this Hc to Hc : Kc → K1/4. Fix a point ζ ∈ Jc. Since Jc is a Jordancurve, there exists an angle θ ∈ R/Z such that ζ = γc(θ). Let zn ∈ K◦c ∪ Ic be asequence converging to ζ. We show that wn := Hc(zn) ∈ K◦1/4 ∪ I1/4 converges toγ1/4(θ).Take a small interval of angle [t, t′] containing θ, where t and t′ are of the forms(2k − 1)/2m and (2k + 1)/2m respectively with the same k and m ≫ 0. Thenγc(t) and γc(t′) bound a small piece of Jc, and the piece, say J′c, is a Jordan arccontaining ζ. Take an open arc C ⊂ K◦c such that C only passes though tiles ofangles in [t, t′] and ¯C ∩ Jc = {γc(t), γc(t′)}. (See Remark 2 below.) Let V denotethe small open set with ∂V = C ∪J′c. By the deﬁnition of Hc,
Hc(V )∩J1/4 =: J′1/4is a piece of J1/4 which is a small Jordan arc with endpoints γ1/4(t) and γ1/4(t′).Since zn ∈ V ∪ J′c for all n ≫ 0, wn ∈ Hc(V ) ∪ J′1/4 for all n ≫ 0. If thereexists a subsequence {ni} ⊂ {n} such that wni converges to a point in K◦1/4,then zni → ζ ∈ K◦c by the deﬁnition of Hc. This contradicts ζ ∈ Jc. Thus wnaccumulates on J′1/4. Since t and t′ are arbitrarily close to θ, wn must convergesto γ1/4(θ).
Extension to a global semiconjugacy.Finally we deﬁne Hc outside the Juliaset by
Hc : ¯C − Kc → ¯C − K1/4z �→ φ−11/4 ◦ φc(z),
which gives a topological conjugacy on the domain, and continuously extends tothe conjugacy Hc : ¯C − K◦c → ¯C − K◦1/4. Then Hc inside and outside Jc arecontinuously glued along Jc. Now Hc : ¯C → ¯C is a desired semiconjugacy. ■
Remark.There are other possible choices of the conjugacy hc : C−[ac, ∞) → Cwith better regularity. For example, for w = ac + reit with r > 0 and 0 < t < 2π,one may also use
hc(w) = log r − log ac
log λc+ i tan π − t
Remark 2.The arc C in the proof above exists. First, take tiles T = Tc(t, −m, +)and T ′ = Tc(t′, −m, −). By Proposition 2.1, T and T ′ share the critical edgeswith T1 = Tc(k/2m−1, −m, −) and T ′1 = Tc(k/2m−1, −m, +) respectively, and thenT1 and T ′1 share the degenerating edges. Now we can join γc(t) and γc(t′) by anarc via Ic(t), T, T1, T ′1, T ′, and Ic(t′).
4Degeneration of the regular leaf spaces
Finally we investigate the structure of R1/4, the regular leaf space of f1/4, andthe degeneration of the invariant leaf corresponding to ˆβc.We begin with some notation and preliminary remarks. For c ∈ (0, 1/4], letus set ˆIc := π−1c (Ic). Now we take a sequence of angles ˆθ := (θ0, θ−1, . . .) suchthat θ0 is of the form k/2m and θ−n = 2θ−n−1. Since Ic contains no critical orbit,ˆθ corresponds bijectively to a connected component of ˆIc which consists of thebackward orbits {z−n}∞n=0 satisfying z−n ∈ Ic(θ−n). We denote such a componentˆIc(ˆθ). Note that ˆIc(ˆθ) is an arc if c ∈ (0, 1/4), or a point if c = 1/4. Note alsothat for ˆ0 = (0, 0, . . .), ˆIc(ˆ0) is the component corresponding to the backwardorbits which are always in the interval Ic(0) = [αc, βc]. Thus ˆIc(ˆ0) contains ˆαc,one of the two irregular points of Nc.For c ∈ (0, 1/4], the set
Lc := {ˆz = (z0, z−1, . . .) ∈ Rc : z−n → βc}
is invariant under the action of ˆfc and is a leaf isomorphic to C. (We will constructthe isomorphism later.) Note that ˆI1/4(ˆ0) = ˆβ1/4 = ˆα1/4 does not belong to L1/4.On the other hand, for c ∈ (0, 1/4), ˆIc(ˆ0) − {ˆαc} is a subset of Lc.Now the main result is:
Theorem 4.1 For c ∈ (0, 1/4), there exists a semiconjugacy ˆHc : Nc → N1/4from ˆfc to ˆf1/4 with the following properties:
(1) ˆHc : Nc − ˆIc → N1/4 − ˆI1/4 is a topological conjugacy between ˆfc|(Nc − ˆIc)and ˆf1/4|(N1/4 − ˆI1/4).
(2) For ˆθ as above, ˆHc maps the arc ˆIc(ˆθ) to the point ˆI1/4(ˆθ).
(3) ˆH−1c (L1/4 ∪�ˆα1/4�) = Lc ∪ {ˆαc}.
(4) For a leaf L in Rc − Lc, ˆHc(L) is a leaf in R1/4 − L1/4.
(5) For a leaf L in R1/4 − L1/4, ˆH−1c (L) is a leaf in Rc − Lc.
Proof.For ˆz = (z0, z−1, . . .) ∈ Nc, set
ˆHc(ˆz) := (Hc(z0), Hc(z−1), . . .) ∈ N1/4.
Since Hc is a semiconjugacy from fc to f1/4, one can easily check that ˆHc issurjective, continuous, and satisﬁes ˆHc◦ ˆfc = ˆf1/4◦ ˆHc. Thus ˆHc is a semiconjugacyfrom ˆfc to ˆf1/4 on their respective natural extensions. In particular, since Hc :¯C − Ic → ¯C − I1/4 is a topological conjugacy, corresponding lift to the natural
extensions ˆHc : Nc − ˆIc → N1/4 − ˆIc is also a topological conjugacy. Thus weobtain property (1).Property (2) comes from the deﬁnition of ˆHc above and the one-to-one corre-spondenceˆIc(ˆθ) ←→ ˆθ ←→ ˆI1/4(ˆθ).
Let us show property (3). Take ˆz = {z−n} ∈ Lc. Then z−n → βc impliesHc(z−n) → Hc(βc) = β1/4 = α1/4 and thus ˆHc(Lc) ⊂ L1/4 ∪�ˆα1/4�.SinceˆHc(ˆαc) = ˆα1/4, we have ˆHc(Lc ∪ {ˆαc}) ⊂ L1/4 ∪�ˆα1/4�. On the other hand, weclaim that for ˆw = {w−n} ∈ L1/4 and ˆz = {z−n} ∈ ˆH−1c ( ˆw), w−n → β1/4 impliesz−n → βc, and thus ˆH−1c (L1/4) ⊂ Lc. Take a subsequence z−ni converging toa point ζ. Note that ζ must be in the Julia set, since otherwise one can showthat ˆz = ˆα or ˆ∞ and it contradicts to ˆHc(ˆz) = ˆw ∈ R1/4. By continuity of Hc,Hc(z−ni) = w−ni → Hc(ζ) and this implies ζ ∈ H−1c (β1/4) = Ic(0) = [αc, βc]. Sinceζ ∈ Jc, ζ = βc and we conclude the claim. Since ˆH−1c (ˆα1/4) = ˆH−1c (ˆI1/4(ˆ0)) =ˆIc(ˆ0) ⊂ Lc ∪ {ˆαc}, we have
ˆH−1c (L1/4 ∪�ˆα1/4�) = ˆH−1c (L1/4) ∪ ˆH−1c (ˆα1/4) ⊂ Lc ∪ {ˆαc}.
Now one can easily check property (3).To show properties (4) and (5), for ∗ ∈ {c, 1/4} and ˆθ as above, we deﬁneopen arcs η∗[ˆθ] : (1, 2) → R∗ − ˆI∗ by
η∗[ˆθ](r) := (φ−1∗ (re2πiθ0), φ−1∗ (r1/2e2πiθ−1), φ−1∗ (r1/4e2πiθ−2), . . .).
Note that the points
ˆz∗[ˆθ] = (γ∗(θ0), γ∗(θ−1), . . .) ∈ ˆI∗
are accessible by η∗[ˆθ](r) by letting r tend to 1. Thus for each ∗ ∈ {c, 1/4}, ˆz∗[ˆθ]and η∗[ˆθ] are in the same leaf of R∗, except when ∗ = 1/4 and ˆθ = ˆ0, that is,ˆz∗[ˆθ] = ˆα1/4.We show property (5) ﬁrst. Take a leaf L ̸= L1/4 from R1/4. By property(1), ˆH−1c (L − ˆI1/4) is homeomorphic to L − ˆI1/4, which is path connected. Nowany connected component of ˆH−1c (L ∩ ˆI1/4) ⊂ ˆIc is an arc and has an endpointof the form ˆzc[ˆθ], which is accessible by ηc[ˆθ]. Since ˆHc(ˆzc[ˆθ]) = ˆz1/4[ˆθ] ∈ L isaccessible by η1/4[ˆθ], we have η1/4[ˆθ] ⊂ L − ˆI1/4 and thus ηc[ˆθ] ⊂ ˆH−1c (L − ˆI1/4).Then we conclude that any component of ˆH−1c (L ∩ ˆI1/4) is attached to an arc inˆH−1c (L − ˆI1/4), and thus ˆH−1c (L) is path connected.Since ˆH−1c ( ˆ∞) = ˆ∞, ˆH−1c (ˆα1/4) = ˆIc(ˆ0), and L ⊂ R1/4 = N1/4 −�ˆ∞, ˆα1/4�
(Proposition 1.2), we have
ˆH−1c (L) ⊂ Nc − { ˆ∞} ∪ ˆIc(ˆ0) ⊂ Rc
and thus there is a leaf L′ of Rc which contains ˆH−1c (L). By property (3) and L ̸=L1/4, L′ is not Lc. Now L ⊂ ˆHc(L′) ⊂ R1/4 − L1/4 and ˆHc(L′) is path connectedby the continuity of ˆHc. Thus ˆHc(L′) is contained by a leaf in R1/4 − L1/4, whichmust be L. This implies ˆHc(L′) = L and we have
L′ ⊂ ˆH−1c ( ˆHc(L′)) = ˆH−1c (L) ⊂ L′.
Thus ˆH−1c (L) is L′, a leaf in Rc − Lc.Property (4) comes from property (5). Take a leaf L′ from Rc − Lc. SinceˆHc(L′) contains no irregular point and is path connected, there is a leaf L ̸= L1/4of R1/4 containing ˆHc(L′). Then L′ ⊂ ˆH−1c (L). By property (5), ˆH−1c (L) is a leaf̸= Lc, which must be L′. Thus L′ = ˆH−1c (L) and this implies ˆHc(L′) is L, a leafin R1/4 − L1/4. ■
Dynamics on the invariant leaves.Let us describe property (3) in furtherdetail. Now the semiconjugacy ˆHc maps Lc ∪ {ˆαc} onto L1/4 ∪�ˆα1/4�. WithinRc and R1/4, we observe this as following.For c ∈ (0, 1/4), Lc compactly contains all but one component of π−1c (ˆIc)∩Lc.The exception is ˆIc(ˆ0)−{ˆαc}. Since ˆIc(ˆ0) (resp. ˆα1/4) is invariant under the actionof ˆfc (resp. ˆf1/4) and ˆH−1c (ˆα1/4) = ˆIc(ˆ0), the map
ˆHc : Lc − ˆIc(ˆ0) → L1/4
is a semiconjugacy from ˆfc|(Lc − ˆIc(ˆ0)) to ˆf1/4|L1/4.Let us describe this semiconjugacy more precisely. For c ∈ (0, 1/4) (resp.c = 1/4), take a linearizing (resp. Fatou) coordinate Φ+c on a neighborhood (resp.repelling petal) Πc of βc such that the action of fc is conjugate to S+c (w) = λ′cw+1.(Recall that λ′1/4 = 1.) In particular, we may assume that Πc contains z = 1 andΦ+c (1) = 0. Then for any ˆz = (z0, z−1, . . .) ∈ Lc, there exists an N such thatz−n ∈ Πc for n ≥ N. By [2, §4], the isomorphism between Lc and C is given by:
ˆΦ+c (ˆz) := (S+c )N(Φ+c (z−N)).
One can easily check that ˆΦ+c (ˆz) does not depend on the choice of N. Then theisomorphism ˆΦ+c : Lc → C has the following properties:
• ˆΦ+c ( ˆfc(ˆz)) = λ′c ˆΦ+c (ˆz) + 1;
• (ˆΦ+c )−1(0) is the backward orbit of z = 1 along the interval (βc, 1];
• if c ∈ (0, 1/4), bc := ˆΦ+c (ˆβc) = 1/(1 − λ′c) tends to −∞ as c → 1/4; and
• if c ∈ (0, 1/4), ˆΦ+c (ˆIc(ˆ0) − {ˆαc}) = (−∞, bc].
Now let us consider the map
ˆΦ+1/4 ◦ ˆHc ◦ (ˆΦ+c )−1 : C − (−∞, bc] → C
for c ∈ (0, 1/4), which is a semiconjugacy from S+c |(C − (−∞, bc]) to S+1/4. Theslit (−∞, bc] is just like pinched and pushed away to “inﬁnity”. Topologically thesame thing happens on the invariant leaves. By ˆHc, a slit ˆIc(ˆ0)∩Lc is pinched, andpushed away to ˆβ1/4. As a result, π−11/4(J1/4) ∩ L1/4 is split into two components.(See Figure 3)
Figure 3: Lc for c = 0.2 and L1/4
Notes.
1. For c ∈ (0, 1/4], Rc has the structure of Riemann surface lamination. Moreprecisely, each point of Rc has a neighborhood homeomorphic to D×T, whereD is a topological disk and T is a Cantor set, and each t ∈ T, D × {t}corresponds to a topological disk on a leaf of Rc.(See [2, §2].)For c ∈(0, 1/4), ˆHc preserves the Cantor set direction of such neighborhoods, and theholonomies of ﬁbers of πc and π1/4.
2. The hyperbolic 3-lamination of fc is constructed by adding “height” to theleaves of Rc to obtain leaves isomorphic to H3. Though the actual constructionin [2] is very complicated, we may hope that the pinching ˆHc will naturallyextend to this hyperbolic 3-lamination and describe the degeneration as c tendsto 1/4.
3. For a quadratic polynomial with an attracting cycle, we can consider its degen-eration to a parabolic cycle with multiple petals. To investigate the associateddegeneration of the regular leaf spaces, the method developed in this paperwould be useful. Make a semiconjugation between the maps, and lift it to
their natural extensions. Then the lifted semiconjugation would give us es-sential information about the degeneration (or bifurcation) of the regular leafspaces. See part 2 of [1].
Acknowledgment.I would like to thank C. Cabrera, M. Lyubich, and M.Shishikura for stimulating discussions and their helpful advice, and the refereefor helpful suggestions and corrections. I also would like to thank people in SUNYat Stony Brook for their hospitality while this work was being prepared.
References
[1] T. Kawahira. Semiconjugacies in complex dynamics with parabolic cycles.Thesis, University of Tokyo, 2003.
[2] M. Lyubich and Y. Minsky. Laminations in holomorphic dynamics. J. Diﬀ.Geom. 47(1997) 17–94.
[3] D. Sullivan. Linking the universalities of Milnor-Thurston, Feigenbaum andAhlfors-Bers. Topological Methods in Modern Mathematics, L. Goldberg andPhillips, editor, Publish or Perish, 1993, 543–564.
[4] J. Milnor. Dynamics in One Complex Variable:Introductory Lectures.Vieweg, 1999.