Combinatorial rigidity of unicritical mapsCitation: Peng W J, Tan L. Combinatorial rigidity of unicritical maps. Sci China Math, 2010, 53(3): 831–848, doi: 10.1007/s11425-010-0022-x

SCIENCE CHINAMathematics

. ARTICLES . March 2010 Vol. 53 No. 3: 831–848

doi: 10.1007/s11425-010-0022-x

c© Science China Press and Springer-Verlag Berlin Heidelberg 2010 math.scichina.com www.springerlink.com

Combinatorial rigidity of unicritical mapsDedicated to Professor Yang Lo on the Occasion of his 70th Birthday

PENG WenJuan1,∗ & TAN Lei2

1School of Mathematical Sciences, Peking University, Beijing 100871, China;2Departement de Mathematiques, Universite d’Angers, Angers 49045, France

Email: [email protected], [email protected]

Received September 21, 2009; accepted November 6, 2009

Abstract In this paper, we combine the KSS nest constructed by Kozlovski, Shen and van Strien, and the

analytic method proposed by Avila, Kahn, Lyubich and Shen to prove the combinatorial rigidity of unicritical

maps.

Keywords combinatorial rigidity, unicritical maps, KSS nest

MSC(2000): 37F10, 37F20

Citation: Peng W J, Tan L. Combinatorial rigidity of unicritical maps. Sci China Math, 2010, 53(3): 831–848, doi:

10.1007/s11425-010-0022-x

1 Introduction

Rigidity is one of the fundamental and remarkable phenomena in holomorphic dynamics. The generalrigidity problem can be posed as

Rigidity problem [11]. Any two combinatorially equivalent rational maps are quasi-conformallyequivalent. Except for the Lattes examples, the quasi-conformal deformations come from the dynamicsof the Fatou set.

In the quadratic case, the rigidity problem is equivalent to the famous hyperbolic conjecture. TheMLC conjecture asserting that the Mandelbrot set is locally connected is stronger than the hyperbolicconjecture (cf. [4]). In 1990, Yoccoz [7] proved MLC for all parameter values which are at most finitelyrenormalizable. Lyubich [11] proved MLC for infinitely renormalizable quadratic polynomials of boundedtype. In [9], Kozlovski, Shen and van Strien gave the proof of the rigidity for real polynomials with allcritical points real. In [1], Avila, et al. proved that any unicritical polynomial fc : z �→ zd + c which is atmost finitely renormalizable and has only repelling periodic points is combinatorially rigid which impliesthat the connectedness locus (the Multibrot set) is locally connected at the corresponding parametervalues. The rigidity problem for the rational maps with Cantor Julia sets is totally solved (cf. [18, 19]).In [19], Zhai took advantage of a length-area method introduced by Kozlovski et al. (cf. [9]) to prove thequasi-conformal rigidity for rational maps with Cantor Julia sets. Kozlovski and van Strien proved thattopologically conjugate non-renormalizable polynomials are quasi-conformally conjugate (cf. [10]).

In the following, we list some other cases in which the rigidity problem is researched (see also [19]).(i) Robust infinitely renormalizable quadratic polynomials [12];

∗Corresponding author

832 PENG WenJuan et al. Sci China Math March 2010 Vol. 53 No. 3

(ii) Summable rational maps with small exponents [5];(iii) Holomorphic Collet-Eckmann repellers [10];(iv) Uniformly weakly hyperbolic rational maps [6].In this paper, we will give an alternative proof of the combinatorial rigidity of unicritical maps (see

the definition in Subsection 1.1). This is the first step to prove the combinatorial rigidity of multi-criticalmaps.

In the proof, we will exploit the powerful combinatorial tool called “puzzle” and a sophisticated choiceof puzzle pieces called the KSS nest constructed in [9]. We use the estimate obtained in [15] as a startingpoint of the proof. To get the quasi-conformal conjugation, we adapt the analytic method in [1] (seeLemma 3.2) to the KSS nest.

The paper is organized as follows. In Subsection 1.1, we present the main result of this paper, The-orem 1.1. In Subsection 1.2, we introduce the definitions of the puzzle and the tableau. In Subsection1.3, we propose Theorem 1.2. We summarize the construction of the KSS nest and prove Theorem 2.1in Section 2. The proof of Theorem 1.2 is given in Section 3. We deduce Theorem 1.2 to Theorem 1.1 inthe appendix.

1.1 Set up

V = �i∈IVi is the disjoint union of finitely many Jordan domains with disjoint

and quasi-circle boundaries,U is compactly contained in V, and is the union of finitely many

Jordan domains with disjoint closures; f : U → V is a proper holomorphic map with a unique critical point

c, with deg f |c = δ, and with c contained in Kf := {z ∈ U, fn(z) ∈ U, ∀n}.

Denote by Pm := {f j(c), j = 1, . . . ,m} and by P :=⋃

m�1 Pm the closure of the orbit of c (thepostcritical set).

For another such map f : U→ V, we mark the same objects with a tilde.Two such maps (f : U→ V), (f : U→ V) are said to be c-equivalent (combinatorially equivalent), if

there is a pair of orientation preserving homeomorphisms h0, h1 : V→ V such that⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

h1(U) = U and h1(P) = P,

h1 is isotopic to h0 rel ∂V ∪P,

h0 ◦ f ◦ h−11 |U = f ,

h1|V\U is C0-qc (qc is an abbreviation of quasi-conformal) for some C0 � 1,

(1.1)

in particular

V ⊃ U h1−→ U ⊂ Vf⏐�

⏐�f

V −→h0

Vcommutes.

This definition is to be compared with the notion of combinatorial equivalence introduced by McMullenin [13]. Notice that this definition is slightly different from the definitions of combinatorial eguivalencein [1] and [10], since we define it without using the external rays and angles.

We say that f and f are qc-conjugate off Kf if there is a qc map H : V\Kf → V\Kf so thatH ◦ f = f ◦H on U\Kf ,

i.e.

U\KfH−→ U\Kf

f⏐�

⏐�f

V\Kf −→H

V\Kf

commutes.

We say that f and f are qc-conjugate if there is a qc map H ′ : V→ V so that H ′ ◦ f = f ◦H ′ on U,

i.e.V ⊃ U H′

−→ U ⊂ Vf⏐�

⏐�f

V −→H′

Vcommutes.

PENG WenJuan et al. Sci China Math March 2010 Vol. 53 No. 3 833

Theorem 1.1. Assume that f is a map in the set up with the component of Kf containing c non-periodic, then Kf is totally disconnected. Assume that f is another map in the set up. Then the followingstatements are equivalent to each other:

A. f and f are c-equivalent; B. f and f are qc-conjugate off Kf ; C. f and f are qc-conjugate.

1.2 Puzzle and tableau

The connected components of f−m(V) are called puzzle pieces of depth m for every non-negative integerm.

For any x ∈ Kf , the tableau T (x), following Branner-Hubbard [2], is the graph embedded in {(u, v), u ∈R

−, v ∈ R} with the axis of u pointing upwards and the axis of v pointing rightwards (this is the standardR

2 with reversed orientation), with vertices indexed by −N × N, where N = {0, 1, . . .}, with the vertexat (−m, 0) being Pm(x), the puzzle piece of depth m containing x, and with f j(Pm(x)) occupying the(−m + j, j)-th entry of T (x). Therefore a given puzzle piece Q may appear at different entries of T (x)for different x’s, but will always be on the same row, denoted by row(Q). There are three types of edges:vertical, horizontal and diagonal. It is also equipped with the graph metric so that each edge is isometricto the unit interval [0, 1].

For a vertex Q with index (−m,n), we say that m is the depth, or the row number of Q.A vertex in a tableau is said to be critical if it represents a critical puzzle piece. A vertex in a tableau

is marked by ◦ if it is non-critical, by a solid • if it is a critical puzzle piece.

A vertical segment, say bounded by two vertices E and F , is assigned a length, denoted by∣∣∣E|F

∣∣∣, which

is simply the depth difference between F and E. It is also assigned a modulus, equaling to mod(E\F ).Recall that modulus of an annulus A is a conformal invariant, and is defined to be 1

2π logR if A is mappedconformally onto {z ∈ C | 1 < |z| < R} (see e.g. [14]).

The map f induces a (partial) dynamical system, indicated by the diagonal edges, mapping T (x)\{0-th row} onto T (x)\{0-th column} = T (f(x)). See Figure 1.

All tableaus satisfy the following three rules (see Figure 1).

Figure 1 Tableau

Rule 1 (vertical segment rule): in each column either there is no critical vertex

or the critical vertices form a single vertical segment with one end

on the top of the column.

An upper triangle T in a tableau is by the definition a sides-included filled triangle bounded by avertical segment on the left, a horizontal segment on the top and a diagonal segment as the third edge.Its size, denoted by |T |, is simply the length of any of its edges, and its depth is the depth of its lowestvertex.


Figure 2 Tableau rules, upper triangles and parallelograms, with D critically full

Rule 2 (double triangle rule): Two upper triangles in T (x) and T (x′)with identical size and lower vertex are identical.

A vertical parallelogram in a tableau is a sides-included filled parallelogram bounded by two verticalsegments and two diagonal segments.

A vertical parallelogram D is said critically full if its two vertical edges are entirely critical, and ifevery vertical critical segment that touches the top diagonal edge of D (if any) remains critical at leastuntil it reaches the lower diagonal edge of D.

Rule 3a (double parallelogram rule): Let D ⊂ T (x), D′ ⊂ T (x′) be two vertical

parallelograms with identical size and upper left vertex, such that D is critically full.

Then D′ is either identical to D, or has its lower diagonal edge completely non critical.

A vertical segmentI

|J

is called vacuous if (I\J)∩ (P ∪ {c}) = ∅. In other words, in any column of any

critical tableau T (c), whenever I appears the full vertical segment from I to J appears (otherwise theorbit of c would visit I\J).

Rule 3b: A vertical parallelogram with right edge vacuous has

every of its vertical segment vacuous,

and furthermore is critically full if its two top vertices are both critical.

For two puzzle pieces Q and I such that fk(Q) = I for some k > 0, we use [Q�I] to denote the

diagonal segment from Q to I. Thus the consecutive vertices on this segment are Q, f(Q), f2(Q), . . . ,fk(Q) = I. We use also [Q�

I[ if we exclude I. Define ]Q�I] and ]Q�

I[ accordingly. This diagonalsegment (whether it is closed, half open or open) is assigned a length, denoted by |Q�

I |, which is simplythe depth difference between Q and I. It is also assigned a degree, equal to deg(fk : Q→ I).

For the critical point c, the nest (Pm(c))m is called a critical nest.A child of a critical puzzle piece I is a critical puzzle piece J that is both a sub-piece of I and a

pullback of I, such that ]J�I [ does not meet any critical puzzle piece. See Figure 3.


Figure 3 Child

1.3 Statement

Hypothesis of recurrence. Let f be as in the set up. In this paper we assume• marching horizontally to the right of any vertex in T (c), one will meet a critical vertex; and T (c)

does not contain a full column of critical vertices (except of course on the 0-th column);

• a critical piece of any depth has at most finitely many children.(This is the commonly called persistently recurrent condition).The objective of this note is to present a proof of the following.

Theorem 1.2. Let f, f be two maps satisfying the hypothesis of recurrence and that Kf has no interior.Then the following statements are equivalent to each other:

A. f and f are c-equivalent; B. f and f are qc-conjugate off Kf ; C. f and f are qc-conjugate.

We will then show in the appendix how to deduce Theorem 1.1 from Theorem 1.2.

2 The KSS nest

In this section we will prove

Theorem 2.1. Assume that f is a map satisfying the hypothesis of recurrence. Then there are twoconstants CK (in fact CK = δ3) and Δ > 0, depending on δ and μ (see below), and a nested sequence ofcritical puzzle pieces Kn ⊂⊂ Kn−1, n � 1, with K0 being the critical puzzle piece of depth 0, satisfying

(i) each Kn, n � 1, is a pullback of Kn−1 and deg(Kn�

Kn−1)

� CK ;(ii) each Kn, n � 1, contains a sub-critical piece K−

n such that mod(Kn\K−n ) � Δ and (Kn\K−

n ) ∩P= ∅.

Hereμ = min{mod(P0(c)\W ) | W a component of U contained in P0(c)}. (2.1)

2.1 First hits have bounded degree

x

× first hit−−�1

I (critical)

| �

Lx(I)

,

x

× first hit−−�0

I (critical)

| �

Lx(I)

.

For any pair (I, x) such that I is a critical puzzle piece, and x ∈ Kf , we produce a puzzle piece Lx(I)(called the pullback of the first � 1 hit of x to I) as follows: start from the 0-th column of T (x) at row(I),march right k � 1 steps until the first hit of a critical spot1) (if any, this exists always if x ∈ P). Thenthat spot represents I and Lx(I) is the pulled-back piece by fk of I containing x. It is the lower vertexof an upper triangle whose left edge is on the 0-th column of T (x), whose right vertex is I and whose topedge does not contain critical spots (except at the ends).

1) We will frequently use the word ‘spot’ for a vertex in a tableau.


Similarly we define the pullback of the first � 0 hit of x to I to be a puzzle piece Lx(I) containing xwhich is equal to I if x ∈ I, otherwise is equal to Lx(I).Lemma 2.2. For I a critical piece and x ∈ Kf , the open diagonal ]Lx(I)�

I [ does not contain anycritical spot. And deg

(Lx(I)�

I)

= δ or = 1 depending whether Lx(I) is critical or not.

Proof. Otherwise we get two upper triangles of different sizes, both with I as the right vertex, andboth have a vertical left edge entirely critical. Now moving the smaller triangle to the left and applyingRule 2 would imply an earlier hit of critical spots on the top edge of the larger triangle.

2.2 The last child operator Γ

Lemma 2.3. (A consequence of the hypothesis of recurrence): Let J be a critical puzzle piece. Then2 � �{children of J} < +∞ .

The proof can be found in [17]. Denote by Γ(J) the last child of J , i.e. the child with the greatestdepth.

Definition. For J a critical puzzle piece, denote by r(J) (resp. R(J)) the minimal (resp. maximal)length of a horizontal segment linking two consecutive J vertices in T (c).

We might have R(J) = +∞ a priori, but the following corollary of Lemma 2.3 will exclude thispossibility. The estimates here play an essential role in the sequel.

Corollary 2.4. Let J be a critical puzzle piece.(i) If S is a critical sub-piece of J then R(S) � R(J) and r(S) � r(J);(ii) If J ′ (�= J) is a critical sub-piece of J and is also a pullback of J , then for k the number of critical

spots on the half open diagonal [J′�J ], we have, r(J) � |J′�J | � k · r(J ′);

(iii)R(J) <

2 · r(J) �

{

|Γ(J)�J | � r(Γ(J)), in particular r(Γ(J)) � 2r(J).

Proof. (i) Obvious, by Rule 1.(ii) Denote by T the upper triangle with vertices J ′, J, J . Then |J′�J | is also the length of the top

edge of T . Consequently |J′�J | � r(J).Let E,F be two consecutive critical spots on row(J ′) of T (c) with F on the right of E. By Rule 2 the

triangle T appears from column(E) with E as the lower vertex.Assume at first k = 1, i.e. [J′�J ] contains no critical spots. Applying Rule 1 one sees that the length

of the horizontal segment E–F must be at least |J′�J |. Therefore |J′�J | � r(J ′).In the general case, let (J ′ =)Jk, Jk−1, . . . , J1, J0 (= J) be the consecutive critical spots on the closed

diagonal [J′�J ]. Applying the above argument on each [Jl+1�Jl [ we get

|J′�J | � r(Jk) + r(Jk−1) + · · ·+ r(J1)(i)

� k · r(Jk) = k · r(J ′).

(iii) In T (c), let J—J be a horizontal segment bounded by two consecutive critical spots on row(J).Form an upper triangle T with this segment as the upper edge. Denote its lower vertex by W , and itslength by l.

If W is critical, we use Rule 2 to compare T with the upper triangle in T (c) with left edgeJ

|Lc(J)

and

(automatically) right vertex J , one concludes that W = Lc(J) and W is the first child of J . But thereare at least two children, so l < |Γ(J)�

J |.If W is not critical, follow its left-down diagonal in T (c) until the first critical vertex W ′ (such W ′

exists since the 0-th column vertex on that diagonal is critical). Then l < |W ′�J | � |Γ(J)�J |.

This proves R(J) < |Γ(J)�J |.

By the definition of a child, the half open diagonal [Γ(J)�J [ contains only one critical spot: Γ(J). So

|Γ(J)�J | � r(Γ(J)) by (ii). Denote by TΓ the upper triangle with vertices Γ(J), J, J . As J has at least

two children (Lemma 2.3) and Γ(J) is the last child, the top edge of TΓ contains at least three criticalspots (counting the ends). Therefore 2 · r(J) � |TΓ| = |Γ(J)�

J |.


2.3 The operators A and B

Definition. Given any critical puzzle piece I, define B(I) and A(I) as follows:

T (c) f t(c)

I Ifirst hit−−

�1I

∣∣∣ � �

B(I) = Γ(I) W∣∣∣ �

A(I)

(2.2)

Lemma 2.5. For any critical puzzle piece I, the vertical segmentB(I)

|A(I)

is vacuous; and

⎧⎨

⎩

deg(B(I)�

I)

= δ =: CB;

�({critical spots} ∩ [B(I)�

I [)

= 1;

⎧⎨

⎩

deg(A(I)�

I)

= δ2 =: CA;

�({critical spots} ∩ [A(I)�

I [)

= 2.

Proof. Let t be the length ofI|

Γ(I). Let W = Lft(c)(I).

At first W has to be critical (for otherwise A(I) is a child of I deeper than Γ(I)).Using then Rule 2 to compare the two triangles, one with vertices W, I, I the other with vertices

Lc(I), I, I, one sees that W = Lc(I).Now assume by contradiction that fs(c) ∈ B(I)\A(I) for some s > 0. This is seen in T (c) as below,

with E non critical:

fs(c)

Ifirst hit−−−−

�1I

� | ↙B(I) W ′

| � ↙E

... ↙Q

row(E) = row(A(I)), row(W ′) = row(W ) = row(Lc(I)) .

Consider the two parallelograms, the first with vertical edgesB(I)|

A(I)and

I|W

, the second with vertical

edgesB(I)|E

andI|W ′

. Applying Rule 3 to them one sees that W ′ is not critical. Now start from the

upper-right vertex of the second parallelogram and march right until the first critical vertex, and fromthere march diagonally left-downwards until the first critical vertex Q. Now Q cannot be on column(W ′)(for otherwise Q has to be aboveW ′ and one gets a smaller triangle than the one with vertices I, I,Lc(I)).Then Q must be a child of I deeper than B(I) = Γ(I). This is not possible.

ThereforeB(I)|

A(I)is vacuous. The rest is trivial.

2.4 The KSS nest

We now construct inductively the KSS nest (K ′n,Kn)n from a critical piece K0 by K ′

n = BΓ(Kn−1) =Γ2(Kn−1) and Kn = AΓ(Kn−1). See Figure 4.

By Corollary 2.4 (iii), we have 2 r(J) � |B(J)�J | = |Γ(J)�

J | � r(B(J)), therefore 3 r(J) � |A(J)�J | �

2 r(A(J)), by the construction of A and Corollary 2.4.(ii).


degree= δ2

one critical spot besides ends

degree = δno critical spots except at ends

Kn−1 Kn−1

In = B(Kn−1)In

vacuous

Kn

vacuous

(at least three times as long as the previous step)

In

B2(Kn) = K ′n+1

AB(Kn) = Kn+1

B(In) = K ′n

A(In) = Kn

Figure 4 The operators A, B, Γ and the KSS nest

Set In = Γ(Kn−1), un = |Kn=A(In)�In |, pn = |Kn�

Kn−1|, qn = |In+1=Γ(Kn)�Kn | .

We get successively the following estimates:

3 r(In) � un � 2 r(Kn) � qn � r(In+1); (2.3)

r(In) � 13un � 2

3r(Kn), pn = qn−1 + un � r(In) + un � 4

3un � 8

3r(Kn);

pn+1 = qn + un+1 � r(Kn) + 3 r(In+1) � r(Kn) + 3 · 2 r(Kn) = 8 r(Kn) � 3 pn;

|Kn�K0 | = pn + · · ·+ p1 <

(

1 +13

+132

+ · · ·)

pn =32pn.

Proposition 2.6. Starting from any critical piece K0, the sequence of pairs of critical pieces (K ′n,

Kn)n�1 satisfies ⎧⎪⎪⎨

⎪⎪⎩

K0 ⊃⊃ K ′1 ⊃⊃ K1 ⊃⊃ K ′

2 ⊃⊃ K2 ⊃⊃ · · · ;∀ n � 1, both K ′

n and Kn are pullbacks of Kn−1;

∀ n � 1, (K ′n\Kn) ∩P = ∅.

(2.4)

And, for any m � 1, for the maps g, f ξ constructed below, and for CK = δ3, d = δ4, β = δ5 such that:(a) deg(Km�

Km−1) � CK , deg(K′m

�Km−1) � CK ,

(b) deg(B′�K′m) = 1,

(c) deg(A′�B′) � d,

(d) deg(K′m+2

�U ) = deg(K′

m+2�Km) � β,

(e) f ξ(Km+2) ⊂ A.For any m � 1, we construct a proper map g : (U,A′, A) → (V,B′, B) as follows (see Figure 5). We

will look at T (c), rows of K0,Km,K′m+2 and Km+2.

Set ξ = |K′m+2

�Km |, M = |Km�

K0 |, x = f ξ(c) and y = fM (x). Set the triple U, V, g as:

V := K0 − V

| ↗g:=fM

U := Km


Figure 5 Construction of KL-maps from a KSS nest

By the hypothesis of recurrence, there is a minimal l � 0 such that from T (c) column of y, row(Km),marching horizontally to the right l steps one will meet again Km. If l = 0, i.e. y ∈ Km, set B = Km

and B′ = K ′m. If l > 0, let B′ ⊃⊃ B be the pair of puzzle pieces on column(y) forming a parallelogram

with K ′m, Km (i.e. y ∈ B and f l(B) = Km). In both cases let A′ ⊃⊃ A be the pair of puzzle pieces on

column(x) forming a parallelogram with B′, B.

Proof. Points (a) and (d) follow from

deg(Kn�

Kn−1)

= CA · CB = δ3 = CK ; deg(K′

n�Kn−1

)= C2

B < CK ; (2.5)

(K ′n\Kn) ∩P = ∅; (2.6)

deg(K′n+1

�Kn−1) = deg(K′

n+1�Kn) · deg(Kn�

Kn−1) = δ2 · δ3 = β . (2.7)

Point (b) is obvious.Starting from T (x), we define inductively the sequence of column numbers κj as follows: set κ0 = 0.

From κj-th column, row(Km), march at first um steps to the right, then march to the right until thefirst � 0 hit of a critical spot, and set κj+1 to be the corresponding column number. Let now L be theminimal integer so that κL > M := |Km�

K0 |.Therefore κL−1 � M . This implies that (L− 1)um � M < 3

2pm � 32 ·

43um = 2 um. So L � 2.

Now deg(A�

B)

� (deg(Km�

Im))L = (deg

(A(Im)�

Im))L � (CA)2 = δ4 = d.

But deg(A�B) = deg(A′�B′

) (as K ′m\Km contains no postcritical points), this gives Point (c).

Point (e). For this we will estimate the number of critical spots on the top edge of the upper triangle

T ′ (resp. T ), defined so to have a left edgeKm

|fξ(Km+2)

(resp.Km

|A

) on column(x).

At first, recall that B = Ly(Km), so κL is actually greater or equal to the column number (in T (x))of the right vertex of T . Also, by Rule 1, the number of critical spots on row(Km) from the κj-th


column (excluded) to the κj+1-th column (included) is at most equal to the number of critical spots on[Km=A(Im)�

Im], which is equal to 2 by Lemma 2.5. Therefore, denoting by top(T ) the top edge of in T

including the end vertices,

�{critical spots ∈ top(T )} − 1def. of L

� L · 2 � 4,

|Γ(Km)�Km | · (�{critical spots ∈ top(T ′)} − 1)

Cor.2.4 (iii)> R(Km) · (�{critical spots ∈ top(T ′)} − 1)

= R(Km) · (�{segments on top(T ′) linking two consecutive critical spots})∗� sum of the lengths of these segments = |T ′|∗∗� r(Im+2) =

r(Im+2)r(Im+1)

r(Im+1)(2.3)

� 3 · |Γ(Km)�Km |,

the one marked by ∗ is due to the definition of R(Km) as the maximal possible length between twoconsecutive critical spots on row(Km); the one marked by ∗∗ is due to the fact that the left edge of T ′

forms a parallelogram withK′

m+2|

Km+2

(see Figure 2.4), therefore |T ′| =∣∣∣∣

K′m+2|

Km+2

∣∣∣∣ =

∣∣∣∣

B(Im+2)|

A(Im+2)

∣∣∣∣, which in turn is

greater than or equal to r(Im+2) by (2.2).It follows then �{critical spots ∈ top(T ′)} − 1 � 4 � �{critical spots ∈ top(T )} − 1. Point (e) follows.

2.5 Further estimates

For any n � 1, let K−n be the critical puzzle piece so that

Kn

|K−

n

andB(Kn−1)

|A(Kn−1)

form a parallelogram. See

Figure 6.

Corollary 2.7. For every n � 1, the segmentKn

|K−

n

is vacuous,

∣∣∣∣∣∣

Kn+1

|K−

n+1

∣∣∣∣∣∣>

∣∣∣∣∣∣

Kn

|K−

n

∣∣∣∣∣∣;

∣∣∣∣∣∣

K′n+1

|Kn+1

∣∣∣∣∣∣>

∣∣∣∣∣∣

K′n

|Kn

∣∣∣∣∣∣, (2.8)

and

mod(Kn+1\K−n+1) >

mod(K ′n−1\Kn−1)C3

ACB. (2.9)

Proof. The vacuousness follows from the property of A and B (Lemma 2.5). Now

r(In) = r(Γ(Kn−1))Cor.2.4.(iii)

> R(Kn−1) �

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

∣∣∣∣∣∣

B(Kn−1)

|A(Kn−1)

∣∣∣∣∣∣=

∣∣∣∣∣∣

Kn

|K−

n

∣∣∣∣∣∣

R(In−1) �

∣∣∣∣∣∣

K′n−1

|Kn−1

∣∣∣∣∣∣

. (2.10)

Hence

∣∣∣∣∣∣

Kn+1

|K−

n+1

∣∣∣∣∣∣=

∣∣∣∣∣∣

B(Kn)

|A(Kn)

∣∣∣∣∣∣� r(Kn) � r(In)

(2.10)>

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

∣∣∣∣∣∣

Kn

|K−

n

∣∣∣∣∣∣

∣∣∣∣∣∣

K′n−1

|Kn−1

∣∣∣∣∣∣

,

∣∣∣∣∣∣

K′n+1

|Kn+1

∣∣∣∣∣∣� r(In+1)

(2.10)>

∣∣∣∣∣∣

K′n

|Kn

∣∣∣∣∣∣.


Figure 6 Control of moduli

Let J be the puzzle piece so thatJ

|K−

n+1

andK′

n−1

|Kn−1

form a parallelogram. See Figure 6. As

∣∣∣∣∣∣∣∣

J

|K−

n+1

∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣

K′n−1

|Kn−1

∣∣∣∣∣∣∣∣

<

∣∣∣∣∣∣∣∣

Kn+1

|K−

n+1

∣∣∣∣∣∣∣∣

, we know that J ⊂ Kn+1. Therefore

mod(Kn+1\K−n+1) � mod(J\K−

n+1)vacuous=

mod(K ′n−1\Kn−1)

deg(K−n+1

�Kn−1)� mod(K ′

n−1\Kn−1)C3

ACB.

2.6 Proof of Theorem 2.1

In the proof, we will use the following lemma:

Lemma 2.8 (Kahn-Lyubich Covering Lemma [18]). Fix η > 0 and D ∈ N. There is ε(η,D) > 0 suchthat: given any g : U → V ,

A ⊂⊂ A′ ⊂⊂ U, B ⊂⊂ B′ ⊂⊂ V (all discs),

g : U → V, A′ → B′, A→ B are proper holomorphic maps,

⎧⎪⎪⎨

⎪⎪⎩

deg(g|U ) � D,

deg(g|A′) � d,

mod(B′\B) � η ·mod(U\A),

=⇒ mod(U\A)either> ε(η,D) > 0;or>

η

2d2mod(V \B).

Now we prove Theorem 2.1.The fact (i) on the uniform bound of degree from Kn to Kn−1 is given by Proposition 2.6. The part

in (ii) that Kn\K−n is vacuous is proved in Corollary 2.7.

We will prove that∃ C = C(δ, μ) > 0, ∀n, mod(K ′

n\Kn) � C. (2.11)

It follows then from Corollary 2.7 that mod(Kn\K−n ) for any n have also a positive lower bound, denoted

by Δ, depending only on δ and μ. This proves the remaining part of (ii).

Proof of (2.11). Set Z = 2d3β2 + 1 = 2δ22 + 1. Set μn = mod(K ′n\Kn), n � 1.


Case 1. Assume there is an increasing sequence kn →∞ such that μm′∀m′<kn

� μkn .There is n0 so that any n � n0 satisfies kn − 2 � Z.Fix n � n0 and set m = kn − 2. Then μm′ � μm+2 for any m′ � m + 2. For this m, define as in

Proposition 2.6 the collection of objects

(ξ,M,U, V, g, B,B′, A,A′).

Set then V = Km−Z ; g = fM : U → V where M is chosen so that fM (U) = V ; B′ = g(A′) andB = g(A).

We then apply the estimates in Proposition 2.6. Note that, by (a) of Proposition 2.6, deg(g) � CZK =: Dwith D independent of m. Note also

deg(A′�B′), deg(B′�

B′) � deg(A′�B′

)(c)

� d, (2.12)

deg(B′�K′

m) = deg(B′�B′

) deg(B′�K′m)

(b)

� d . (2.13)

We have (see Figure 2.4)

μm+2 � mod(U\f ξ(Km+2))deg(K′

m+2�U )

(e)

�(d)

mod(U\A)β

; (2.14)

mod(B′\B) noP=μm

deg(B′�

K′m

)choice of kn

�(2.13)

μm+2

d

(2.14)

� mod(U\A)dβ

=: η ·mod(U\A),

with η := 1dβ independent of m.

Therefore for any n � n0 and m = kn − 2, we may then apply Kahn-Lyubich Covering Lemma tog : (U,A′, A)→ (V , B′, B), to conclude that,

β · μm+2

(2.14)

� mod(U\A)

either> ε(η,D) > 0;or> η·mod(V \B)

2 deg(A′�B′)2

(∗)� η

2d2 (μm + · · ·+ μm−Z+1)choice of kn

� ηZ2d2μm+2,

where the inequality (∗) is proved as follows: For j = m,m−1, . . . ,m−Z+1, and for lj � 0 minimal so thatf lj (g(x)) ∈ Kj , denote by Bj , resp. B′

j , the connected component of f−lj (Kj), resp. f−lj (K ′j), containing

g(x). Then {B′j\Bj}j are pairwise disjoint essential annuli in V \B with mod(B′

j\Bj) = μj

deg(Bj→Kj)= μj

(due to (K ′j\Kj) ∩P = ∅ and the fact that deg(Bj → Kj) = 1).

By our choice of Z we have Z > 2d3β2 = 2d2βη . Hence the second line above is impossible. So

μkn = μm+2 >1β · ε(η,D) > 0. Therefore ∀ l ∈ N, μl � limn→∞ μkn � 1

β · ε(η,D).Case 2. Assume there is k0 such that μk � μk0 for all k � 1.Case 2.1. k0 − 2 � Z. Then repeating the same argument as above we know that μk0 � 1

β · ε(η,D).Case 2.2. k0 − 2 < Z. Then k0 � Z + 1 = 2δ22 + 2. Notice that K ′

k0is mapped by some iterate of

f onto K0 = P0(c) with degree bounded by CZ+1K , due to (a) of Proposition 2.6. The same iterate of f

maps Kk0 into a component W of U contained in P0(c). So μk0 � 1CZ+1

K

mod(P0(c)\W ) � 1CZ+1

K

μ.

We have now proved (2.11) for C = min{ 1β · ε(η,D), 1

CZ+1K

μ}.

3 Proof of Theorem 1.2

Now we come to the proof of Theorem 1.2. Clearly if f and f are qc-conjugate then they are c-equivalentand qc-conjugate off Kf .


3.1 C-equivalence implies qc-conjugacy off Kf

We just repeat the standard argument (see for example Appendix in [13]).Assume that f, f are c-equivalent. Set U = U1, and Un = f−n(V). The same objects gain a tilde for

f . For t ∈ [0, 1], let ht : V→ V be an isotopy path linking h0 to h1.Then there is a unique continuous extension (t, z) �→ h(t, z), [0,∞[×V→ V such that0) each ht : z → h(t, z) is a homeomorphism;1) ht|∂V∪P = h0|∂V∪P, ∀t ∈ [0,+∞[;2) for n � 1, t > n and x ∈ V\Un we have ht(x) = hn(x);3) for t ∈ [0, 1] the following diagram commutes:

......

U2ht+2−→ U2

f ↓ ↓ f

U1ht+1−→ U1

f ↓ ↓ f

Vht−→ V.

Set then Ω =⋃n�1 V\Un = V\Kf , and Ω = V\Kf . Then there is a qc map H : Ω → Ω such that

H(x) = hn(x) for n � 1 and x ∈ V\Un and that H ◦ f |Ω∩U = f ◦H |Ω∩U, i.e. H realizes a qc-conjugacyfrom f to f off Kf . The qc constant of H is equal to C0, the qc constant of h1 on V\U.

3.2 qc-conjugacy off Kf implies qc-conjugacy

Assume now that f and f are qc-conjugate off Kf , i.e. there is a C0-qc map H : V\Kf → V\Kf

conjugating f to f . We will show now that H admits a qc extension across Kf which is again aconjugacy. This will be done in three steps.

Uniform regularity of H on ∂Kn

Proposition 3.1. Let {Kn, n � 1} be the KSS nest for f given by Theorem 2.1. There is a constantL′, such that for any n � 1, H |∂Kn admits an L′-qc extension inside Kn.

The basic step of the proof of Proposition 3.1 is the following lemma on covering maps of the unit disk.

Lemma 3.2. (See Lemma 3.2 in [1]): For every integer d � 2 and every 0 < ρ < r < 1 there existsL0 = L0(ρ, r, d) with the following property. Let g, g : (D, 0) → (D, 0) be holomorphic proper maps ofdegree d, with critical values contained in Dρ. Let η, η′ : T → T be two homeomorphisms satisfyingg ◦ η′ = η ◦ g. Assume that η admits an L-qc extension ξ : D → D which is the identity on Dr. Then η′

admits an L′-qc extension ξ′ : D→ D which is the identity on Dr, where L′ = max{L,L0}.

Proof of Proposition 3.1. We will just adapt the proof of Theorem 3.1 in [1] to our KSS nest.For all n � 1, let Kn be the puzzle piece bounded by H(∂Kn).Notice that H |∂K1 has a qc extension on a neighborhood of ∂K1. It extends thus to an L1-qc map

K1 → K1, c→ c for some L1 � 1 (see e.g. [13, Lemma C.1]).For the sequence K−

n given by Theorem 2.1, define K−n accordingly. As the operators Γ,A,B in the

definition of the KSS nest can be read off from the dynamical degree on the boundary of the puzzlepieces, and H preserves this degree information, the sequence Kn is precisely the KSS nest for f startingfrom K0. Therefore Theorem 2.1 is valid for this sequence as well, with the same constant CK = δ3, andprobably a different Δ as a lower bound for mod(Kn\K−

n ).Recall that for each i � 1, pi denotes the integer such that fpi(Ki) = Ki−1. We have fpi(Ki) = Ki−1,

and fpi : Ki → Ki−1 and fpi : Ki → Ki−1 are proper holomorphic maps of degree δ3.


Fix now n � 1. We will show that H |∂Kn has an L′-qc extension inside Kn with L′ independent of n.This will not be an induction argument on n. In other words, this extension will not be a pullback of apreviously defined extension of H inside Kn−1.

Set vn = c, and then, for i = n− 1, n− 2, . . . , 1, set consecutively vi = fpi+1+···+pn(c).Since Ki\K−

i is vacuous, the critical values of fpi+1 |Ki+1 , as well as vi, are contained in K−i , 1 � i �

n− 1.Let ψi : (Ki, vi) → (D, 0) be a bi-holomorphic uniformization, i = 1, . . . , n. For i = 2, . . . , n, let

gi = ψi−1 ◦ fpi ◦ ψ−1i . These maps fix the point 0, are proper holomorphic maps of degree δ3, with the

critical values contained in ψi−1(K−i−1).

Let ψi(K−i ) = Ωi. Since mod(D\Ωi) = mod(Ki\K−

i ) � Δ > 0 and Ωi � ψi(vi) = 0, 1 � i � n, thesedomains are contained in some disk Ds with s = s(Δ) < 1. So the critical values of gi are contained inΩi−1 ⊂ Ds, 2 � i � n.

The corresponding objects for f will be marked with a tilde. The same assertions hold for gi. Thenall the maps gi and gi satisfy the assumptions of Lemma 3.2, with d = δ3, and ρ = max{s, s}.

(D, 0)ψ1←− (K1, v1) (K1, v1)

ψ1−→ (D, 0)

g2 ↑ ↑ fp2 fp2 ↑ ↑ g2(D, 0)

ψ2←− (K2, v2) (K2, v2)ψ2−→ (D, 0)

g3 ↑ ↑ fp3 fp3 ↑ ↑ g3...

......

...

(D, 0)ψn−1←− (Kn−1, vn−1) (Kn−1, vn−1)

ψn−1−→ (D, 0)

gn ↑ ↑ fpn fpn ↑ ↑ gn(D, 0)

ψn←− (Kn, vn), (Kn, vn)ψn−→ (D, 0).

Note that each of ψi, ψi extends to a homeomorphism from the closure of the puzzle piece to D.Let us consider homeomorphisms ηi : T → T given by ηi = ψi ◦ H |∂Ki ◦ ψ−1

i . They are equivariantwith respect to the g-actions, i.e., ηi−1 ◦ gi = gi ◦ ηi.

Due to the qc extension of H |∂K1 , we know that η1 extends to an L1-qc map (D, ψ1(c))→ (D, ψ1(c)).Fix some r with ρ < r < 1. Since c ∈ K−

1 , c ∈ K−1 , we have ψ1(c), ψ1(c) ∈ Dρ. We conclude that η1

extends to an L-qc map ξ1 : D→ D which is the identity on Dr, where L depends on L1, ρ and r.Let L0 = L0(ρ, r, δ3) be as in Lemma 3.2, and let L′ = max{L,L0}. For i = 2, 3, . . . , n − 1, apply

consecutively Lemma 3.2 to the following left diagram (from top to bottom):

Tη1−→ T

g2 ↑ ↑ g2

Tη2−→ T

g3 ↑ ↑ g3...

...

Tηn−1−→ T

gn ↑ ↑ gn

Tηn−→ T,

we get

(D, 0)ξ1−→ (D, 0)

g2 ↑ ↑ g2

(D, 0)ξ2−→ (D, 0)

g3 ↑ ↑ g3...

...

(D, 0)ξn−1−→ (D, 0)

gn ↑ ↑ gn

(D, 0)ξn−→ (D, 0),

so that for i = 2, . . . , n, the map ηi admits an L′-qc extension ξi : D → D which is the identity on Dr.The desired extension of H |∂Kn inside Kn is now obtained by taking ψ−1

n ◦ ξn ◦ ψn.


Spreading principleFix now n � 1. Recall that Kn is the unique critical puzzle piece of depth Mn. Set W0 = UMn\Kn andX0 = Kn. For j � 0, define a sequence H(j) of qc maps as follows:

H(0) =

⎧⎪⎪⎨

⎪⎪⎩

H, on V\(X0 ∪W0) = V\UMn ,

the L′-qc map given by Proposition 3.1, on X0 = Kn,

a C′n-qc extension for some C′

n � 1, on W0 = UMn\Kn.

The constant C′n may depend on n.

For j � 1, setXj = {z ∈ UMn | f l(z) ∈ Kn for some 0 � l � j}

andWj = {z ∈ U | z, f(z), f2(z), . . . , f j(z) ∈W0} = UMn+j\Xj.

Notice that for any z ∈ Xj, Lz(Kn) is a connected component of Xj , and there is j � l(z) � 0 so thatf l(z) maps Lz(Kn) univalently onto Kn. On the other hand, any component S of Wj is a puzzle piece ofdepth Mn + j and f j maps S univalently onto a component of W0. Set then

H(j) =

⎧⎪⎪⎨

⎪⎪⎩

H, on V\(Xj ∪Wj),

the L′-qc map given by univalent pullback of H(0)|X0 , on Xj ,

the C′n-qc map given by univalent pullback of H(0)|W0 , on Wj .

Set Cn = max{C0, L′, C′

n}. The {H(j)}j�0 is a sequence of Cn-qc maps. Hence it is precompact in theuniform topology.

Notice that H(j) = H(j−1) outside Wj−1. Thus, the sequence {H(j)} converges pointwise outside⋂

j

Wj = {x ∈ Kf | fk(x) /∈ Kn, k � 0}.

This set is a hyperbolic subset, on which f is uniformly expanding, and hence has zero Lebesgue measure,in particular no interior. So any two limit maps of the sequence {H(j)}j�0 coincide on a dense open setof V, therefore coincides on V to a unique limit map. Denote this map by Hn. It is Cn-qc.

By construction, Hn coincides with H on V\((⋃j Xj)∪ (

⋂jWj)) is therefore C0-qc there; and is L′-qc

on⋃j Xj . It follows that the dilatation of Hn is bounded by max{C0, L

′} except possibly on the set⋂jWj . But this set has zero Lebesgue measure. It follows that the dilatation of Hn is max{C0, L

′},which is independent of n.

ConclusionThe sequence Hn : V→ V has a subsequence converging uniformly to a limit qc map H ′ : V→ V, withH ′|V\Kf

= H . Therefore H ′ is a qc extension of H . On the other hand, H ◦ f = f ◦H on U\Kf , andKf has empty interior by assumption. So H ′ ◦ f = f ◦H ′ holds on U by continuity. Therefore H ′ is aqc-conjugacy from f to f . This ends the proof of Theorem 1.2.

Acknowledgements This work was partially supported by Postdoctoral Science Foundation of China (Grant

No. 20080440270), Doctoral Education Program Foundation of China (Grant No. 20060001003), National Natural

Science Foundation of China (Grant No. 10831004). and EU Research Training Network CODY, Conformal

Structures and Dynamics. The authors would like to thank Jeremy Kahn and Cui Guizhen for inspiring discussions

and also thank Yin Yongcheng for carefully reading the paper and providing valuable suggestions. Both of the

authors also want to thank Cergy-Pontoise University where mist of this work has keen done.

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Appendix A: Proof of Theorem 1.1

For any x ∈ Kf , we say that the forward orbit of x combinatorially accumulates to c, written as x → c, if a critical

vertex appears in each row of T (x)\{0-th column}.For the unique critical point c, one of the following cases will occur.

Case a. c �→ c.

Case b. c → c and T (c) is not persistently recurrent (this is commonly called reluctantly recurrent).

Case c. c → c and T (c) is persistently recurrent.

Lemma A.1. In Case a or Case b, there are a puzzle piece Pn0(x) of depth n0 containing x ∈ V and infinitely

many integers in such that

f in (Pn0+in(c)) = Pn0(x), f in(Pn0+in+1(c)) = Pn0+1(x)

and deg(Pn0+in (c)�Pn0 (x)) = δ.

Proof. In Case a, there is an integer n0 � 0 such that each vertex of at least n0-th row in T (c)\ {0-thcolumn} is non-critical. So for each n � 1,

deg(Pn0+n(c)�Pn0 (fn(c))) = δ.

Since there are finitely many puzzle pieces in every depth, we can take a subsequence in such that

f in (Pn0+in(c)) = Pn0(x), f in (Pn0+in+1(c)) = Pn0+1(x)

for some fixed puzzle piece Pn0(x).

In Case b, since T (c) is reluctantly recurrent, there exist an integer n0 � 0 and infinitely many integers mk � 1

such that {Pn0+mk(c)}k�1 are children of Pn0(c) and then deg(Pn0+mk(c)�

Pn0 (c)) = δ. Now we take a subsequence

in of mk such that

f in (Pn0+in(c)) = Pn0(c) = Pn0(x), f in(Pn0+in+1(c)) = Pn0+1(x)


for some fixed puzzle piece Pn0+1(x).

Now we prove Kf is a Cantor set. It is equivalent to show that for each z ∈ Kf , the connected component of

Kf containing z, denoted by Kf (z), is a single point. The argument is similar as the proof of Lemma 7 in [18].

For z ∈ Kf , there are four possibilities as follows:

Case I. z �→ c.

Case II. z → c and c �→ c.

Case III. z → c, and T (c) is reluctantly recurrent.

Case IV. z → c and T (c) is persistently recurrent.

In Case I, there is an integer n1 � 0 such that each vertex of at least n1-th row in T (z)\{0-th column} is

non-critical. So for each n � 1,

deg(Pn1+n(z)�Pn1 (fn(z))) � δ.

Take a subsequence jn of n1 + n with Pn1(f jn (z)) = Pn1 for some fixed puzzle piece Pn1 . Let

νn1 = min{mod(Pn1\W ) | W a piece of depth n1 + 1 contained in Pn1} .

Then

mod(Pjn(z)\Pjn+1(z)) � νn1

δ.

Thus∑∞

n=1 mod(Pjn(z)\Pjn+1(z)) = +∞. Applying then Grotzsch’s inequality one could conclude immediately

that mod(P0(z)\Kf (z)) = +∞ and Kf (z) is a point.

In Case II or Case III, by Lemma A.1, there are a puzzle piece Pn0(x) and infinitely many integers in such

that f in(Pn0+in(c)) = Pn0(x). As in Subsection 2.1, we construct a puzzle piece Lz(Pn0+in(c)) for n � 1 and by

Lemma 2.4,

deg(Lz(Pn0+in (c))�Pn0+in (c)) � δ

and then

deg(Lz(Pn0+in (c))�Pn0 (x)) = deg(Lz(Pn0+in (c))�

Pn0+in (c)) · deg(Pn0+in (c)�Pn0 (x)) � δ2.

Let ln = |Lz(Pn0+in (c))�Pn0+in (c)|. Set

νn0 = min{mod(Pn0(x)\W ) | W a piece of depth n0 + 1 contained in Pn0(x)}.

Then

mod(Pn0+in+ln(z)\Pn0+in+ln+1(z)) � νn0

δ2.

Thus∑∞

n=1 mod(Pn0+in+ln(z)\Pn0+in+ln+1(z)) = +∞ and Kf (z) is a point.

In Case IV, we use the estimate (2.11) to obtain that Kf (z) is a point. Let Tn(z) = Lz(Kn) and T ′n(z) be the

puzzle piece such thatT ′

n(z)

|Tn(z)

andK′

n|

Kn

form a parallelogram. SinceK′

n|

Kn

is vacuous, applying tableau rule3b (see

also Figure 1.2), we know that deg(T ′n(z)�

K′n ) = deg(Tn(z)�

Kn) � δ. Then

mod(T ′n(z)\Tn(z)) � mod(K′

n\Kn)

δ� C

δ> 0.

Until now, we come to the conclusion that Kf is a Cantor set.

We may assume that {Pn0+in(c)}n�1 in Lemma A.1 is a nested sequence of critical puzzle pieces. Set Qn =

Pn0+in(c) and Q−n = Pn0+in+1(c). In the following, we will combine Lemma 3.2 and Lemma A.1 to prove

Proposition A.2. For every n � 1, H |∂Qn admits a K′-qc extension inside Qn, where K′ is a constant

independent of n.

Proof. Fix n � 1.

Let Qn, Q−n be the puzzle pieces bounded by H(∂Qn) and H(∂Q−

n ) respectively. Since H preserves the degree

information, Lemma A.1 is valid for Qn with the same δ, more precisely, for the map f , there exists a piece Pn0(x)

of depth n0 containing x ∈ V such that

f in (Pn0+in(c)) = Pn0(x), f in (Pn0+in+1(c)) = Pn0+1(x)


and

deg(Pn0+in (c)�Pn0 (x)) = δ.

Let mod(Pn0(x)\Pn0+1(x)) = Δ and mod(Pn0(x)\Pn0+1(x)) = Δ.

Let φ0 : (Pn0(x), f in(c)) → (D, 0) and φn : (Qn, c) → (D, 0) be bi-holomorphic uniformizations. Let hn =

φ0 ◦ f in ◦ φ−1n . These maps fix the point 0, are proper holomorphic maps of degree δ. The critical values of them

are contained in φ0(Pn0+1(x)) because c is the unique critical point in Qn of the map f in restricted in Qn.

Since mod(D \ φ0(Pn0+1(x))) = mod(Pn0(x)\Pn0+1(x)) = Δ > 0 and φ0(Pn0+1(x)) � φ0(fin(c)) = 0, we have

φ0(Pn0+1(x)) ⊂ Dt with t = t(Δ) < 1. So the critical values of hn are contained in φ0(Pn0+1(x)) ⊂ Dt.

The corresponding objects for f will be marked with a tilde. The same assertions hold for hn. Then all the

maps hn and hn satisfy the assumptions of Lemma 3.2, with d = δ, and ρ = max{t, t}.Note that each of φn, φn extends to a homeomorphism from the closure of the puzzle piece to D.

Let us consider homeomorphisms σ0 : T → T and σn : T → T given by σ0 = φ0 ◦ H |∂Pn0 (x) ◦ φ−10 and

σn = φn ◦ H |∂Qn ◦ φ−1n respectively. Then σ0 ◦ hn = hn ◦ σn.

Notice that H |∂Pn0 (x) has a K1-qc extension on a neighborhood of ∂Pn0(x) for some K1 � 1. Fix some r with

ρ < r < 1. We conclude that σ0 extends to a K-qc map ζ0 : D → D which is the identity on Dr, where K depends

on K1, ρ and r.

Let K0 = K0(ρ, r, δ) be as in Lemma 3.2, and let K′ = max{K, K0}. Apply Lemma 3.2 to the following left

diagram :

Tσn−→ T

hn ↓ ↓ hn

Tσ0−→ T,

we get

(D, 0)ζn−→ (D, 0)

hn ↓ ↓ hn

(D, 0)ζ0−→ (D, 0),

so that the map σn admits a K′-qc extension ζn : D → D which is the identity on Dr. The desired extension of

H |∂Qn inside Qn is now obtained by taking φ−1n ◦ ζn ◦ φn. This completes the proof of this proposition.

Now we can use the spreading principle to the sequence {Qn}n�1 as in Subsection 3.2 and conclude that

under the assumptions that c �→ c or T (c) is reluctantly recurrent, the statements B and C in Theorem 1.1 are

equivalent. Under these assumptions, it is easy to verify that the proof in Subsection 3.1 is also valid for the

equivalence of statements A and B in Theorem 1.1. Thus we complete the proof of Theorem 1.1.

Documents

Combinatorial rigidity of unicritical mapsCitation: Peng W J, Tan L. Combinatorial rigidity of unicritical maps. Sci China Math, 2010, 53(3): 831–848, doi: 10.1007/s11425-010-0022-x