SCIENCE CHINA
Mathematics
. ARTICLES .
March 2010 Vol. 53 No. 3: 831–848
doi: 10.1007/s11425-010-0022-x
Combinatorial rigidity of unicritical maps
Dedicated to Professor Yang Lo on the Occasion of his 70th Birthday
PENG WenJuan1,∗ & TAN Lei2
1School
of Mathematical Sciences, Peking University, Beijing 100871, China;
de Mathématiques, Université d’Angers, Angers 49045, France
Email: [email protected], [email protected]
2Département
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
MSC(2000):
combinatorial rigidity, unicritical maps, KSS nest
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 general
rigidity problem can be posed as
Rigidity problem [11].
Any two combinatorially equivalent rational maps are quasi-conformally
equivalent. Except for the Lattès examples, the quasi-conformal deformations come from the dynamics
of the Fatou set.
In the quadratic case, the rigidity problem is equivalent to the famous hyperbolic conjecture. The
MLC conjecture asserting that the Mandelbrot set is locally connected is stronger than the hyperbolic
conjecture (cf. [4]). In 1990, Yoccoz [7] proved MLC for all parameter values which are at most finitely
renormalizable. Lyubich [11] proved MLC for infinitely renormalizable quadratic polynomials of bounded
type. In [9], Kozlovski, Shen and van Strien gave the proof of the rigidity for real polynomials with all
critical points real. In [1], Avila, et al. proved that any unicritical polynomial fc : z → z d + c which is at
most finitely renormalizable and has only repelling periodic points is combinatorially rigid which implies
that the connectedness locus (the Multibrot set) is locally connected at the corresponding parameter
values. 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 the
quasi-conformal rigidity for rational maps with Cantor Julia sets. Kozlovski and van Strien proved that
topologically 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
c Science China Press and Springer-Verlag Berlin Heidelberg 2010
math.scichina.com
www.springerlink.com
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-critical
maps.
In the proof, we will exploit the powerful combinatorial tool called “puzzle” and a sophisticated choice
of puzzle pieces called the KSS nest constructed in [9]. We use the estimate obtained in [15] as a starting
point of the proof. To get the quasi-conformal conjugation, we adapt the analytic method in [1] (see
Lemma 3.2) to the KSS nest.
The paper is organized as follows. In Subsection 1.1, we present the main result of this paper, Theorem 1.1. In Subsection 1.2, we introduce the definitions of the puzzle and the tableau. In Subsection
1.3, we propose Theorem 1.2. We summarize the construction of the KSS nest and prove Theorem 2.1
in Section 2. The proof of Theorem 1.2 is given in Section 3. We deduce Theorem 1.2 to Theorem 1.1 in
the appendix.
1.1
Set up
V = i∈I Vi 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, f n (z) ∈ U, ∀n}.
Denote by Pm := {f j (c), j = 1, . . . , m} and by P := m1 Pm the closure of the orbit of c (the
postcritical set).
For another such map f˜ : Ũ → Ṽ, we mark the same objects with a tilde.
Two such maps (f : U → V), (f˜ : Ũ → Ṽ) are said to be c-equivalent (combinatorially equivalent), if
there is a pair of orientation preserving homeomorphisms h0 , h1 : V → Ṽ such that
⎧
⎪
h1 (U) = Ũ and h1 (P) = P̃,
⎪
⎪
⎪
⎨ h is isotopic to h rel ∂V ∪ P,
1
0
(1.1)
−1
˜
⎪
h
◦
f
◦
h
|
=
f
,
⎪ 0
1 Ũ
⎪
⎪
⎩
h1 |V\U is C0 -qc (qc is an abbreviation of quasi-conformal) for some C0 1,
h
1
Ũ ⊂ Ṽ
V ⊃ U −→
⏐
⏐
f
f˜
commutes.
in particular
V −→ Ṽ
h0
This definition is to be compared with the notion of combinatorial equivalence introduced by McMullen
in [13]. Notice that this definition is slightly different from the definitions of combinatorial eguivalence
in [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 → Ṽ\Kf˜ so that
H ◦ f = f˜ ◦ H on U\Kf ,
H
U\Kf −→ Ũ\Kf˜
⏐
⏐
f˜
i.e. f
commutes.
V\Kf −→ Ṽ\Kf˜
H
We say that f and f˜ are qc-conjugate if there is a qc map H : V → Ṽ so that H ◦ f = f˜ ◦ H on U,
H
V ⊃ U −→ Ũ ⊂ Ṽ
⏐
⏐
f˜
f
commutes.
i.e.
Ṽ
V −→
H
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
833
No. 3
Theorem 1.1. Assume that f is a map in the set up with the component of Kf containing c nonperiodic, then Kf is totally disconnected. Assume that f˜ is another map in the set up. 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.
1.2
Puzzle and tableau
The connected components of f −m (V) are called puzzle pieces of depth m for every non-negative integer
m.
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 standard
R2 with reversed orientation), with vertices indexed by −N × N, where N = {0, 1, . . .}, with the vertex
at (−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 isometric
to 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.
E
A vertical segment, say bounded by two vertices E and F , is assigned a length, denoted by | , which
F
is simply the depth difference between F and E. It is also assigned a modulus, equaling to mod(E\F ).
1
Recall that modulus of an annulus A is a conformal invariant, and is defined to be 2π
log R if A is mapped
conformally 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 a
vertical 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 lowest
vertex.
834
PENG WenJuan et al.
Figure 2
Sci China Math
March 2010
Vol. 53
No. 3
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 vertical
segments and two diagonal segments.
A vertical parallelogram D is said critically full if its two vertical edges are entirely critical, and if
every vertical critical segment that touches the top diagonal edge of D (if any) remains critical at least
until 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.
I
A vertical segment
|
is called vacuous if (I\J) ∩ (P ∪ {c}) = ∅. In other words, in any column of any
J
critical tableau T (c), whenever I appears the full vertical segment from I to J appears (otherwise the
orbit 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 f k (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), f 2 (Q), . . . ,
f k (Q) = I. We use also [Q I[ if we exclude I. Define ]Q I] and ]Q I[ accordingly. This diagonal
segment (whether it is closed, half open or open) is assigned a length, denoted by |Q I |, which is simply
the depth difference between Q and I. It is also assigned a degree, equal to deg(f k : 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.
PENG WenJuan et al.
Sci China Math
Figure 3
1.3
March 2010
Vol. 53
835
No. 3
Child
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 two
constants CK (in fact CK = δ 3 ) and Δ > 0, depending on δ and μ
(see below), and a nested sequence of
critical 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 Kn− such that mod(Kn \Kn− ) Δ and (Kn \Kn− ) ∩ P
= ∅.
Here
(2.1)
μ
= min{mod(P0 (c)\W ) | W a component of U contained in P0 (c)}.
2.1
First hits have bounded degree
x
×
|
Lx (I)
x
first hit
−−
1
I (critical)
,
×
|
first hit
−−
0
I (critical)
.
L
x (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). Then
that spot represents I and Lx (I) is the pulled-back piece by f k of I containing x. It is the lower vertex
of an upper triangle whose left edge is on the 0-th column of T (x), whose right vertex is I and whose top
edge does not contain critical spots (except at the ends).
1)
We will frequently use the word ‘spot’ for a vertex in a tableau.
836
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
Similarly we define the pullback of the first 0 hit of x to I to be a puzzle piece L
x (I) containing x
which 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 any
critical 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, and
both have a vertical left edge entirely critical. Now moving the smaller triangle to the left and applying
Rule 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. Then
2 {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 greatest
depth.
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 this
possibility. 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 );
R(J) <
(iii)
|Γ(J) J | r(Γ(J)), in particular r(Γ(J)) 2r(J).
2 · r(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 =)J k , J k−1 , . . . , J 1 , J 0 (= J) be the consecutive critical spots on the closed
l
diagonal [J J ] . Applying the above argument on each [J l+1 J [ we get
(i)
|J J | r(J k ) + r(J k−1 ) + · · · + r(J 1 ) k · r(J k ) = 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 its
length by l.
J
If W is critical, we use Rule 2 to compare T with the upper triangle in T (c) with left edge
|
Lc (J)
and
(automatically) right vertex J, one concludes that W = Lc (J) and W is the first child of J. But there
are 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 critical
spots (counting the ends). Therefore 2 · r(J) |TΓ | = |Γ(J) J |.
PENG WenJuan et al.
2.3
Sci China Math
March 2010
Vol. 53
No. 3
837
The operators A and B
Given any critical puzzle piece I, define B(I) and A(I) as follows:
Definition.
T (c)
f t (c)
I
I
B(I) = Γ(I)
first hit
−−
1
I
(2.2)
W
A(I)
Lemma 2.5.
For any critical puzzle piece I, the vertical segment
⎧
⎨ deg B(I) I = δ =: CB ;
⎩ {critical spots} ∩
I[
= 1;
[B(I) B(I)
|
A(I)
is vacuous; and
⎧
⎨ deg A(I) I = δ 2 =: CA ;
⎩ {critical spots} ∩
I[
= 2.
[A(I) I
| . Let W = L t
Proof. Let t be the length of Γ(I)
f (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 f s (c) ∈ B(I)\A(I) for some s > 0. This is seen in T (c) as below,
with E non critical:
f s (c)
first hit
I − − −− I
1
W
B(I)
|
|
row(E) = row(A(I)), row(W ) = row(W ) = row(Lc (I)) .
E
..
.
Q
Consider the two parallelograms, the first with vertical edges
edges
B(I)
|
E
and
I
|
W
B(I)
|
A(I)
and
I
|
W
, the second with vertical
. 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 from
there march diagonally left-downwards until the first critical vertex Q. Now Q cannot be on column(W )
(for otherwise Q has to be above W 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.
Therefore
2.4
B(I)
|
A(I)
is vacuous. The rest is trivial.
The KSS nest
We now construct inductively the KSS nest (Kn , Kn )n from a critical piece K0 by Kn = 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).
838
PENG WenJuan et al.
Sci China Math
Kn−1
In
B(In ) =
In
March 2010
No. 3
Kn−1
In = B(Kn−1 )
Kn
Vol. 53
degree = δ
no critical spots except at ends
degree= δ 2
one critical spot besides ends
vacuous
A(In ) = Kn
Kn
(at least three times as long as the previous step)
B 2 (Kn ) = Kn+1
vacuous
AB(Kn ) = Kn+1
The operators A, B, Γ and the KSS nest
Figure 4
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 );
1
2
4
8
r(In ) un r(Kn ), pn = qn−1 + un r(In ) + un un r(Kn );
3
3
3
3
pn+1 = qn + un+1 r(Kn ) + 3 r(In+1 ) r(Kn ) + 3 · 2 r(Kn ) = 8 r(Kn ) 3 pn ;
1
3
1
|Kn K0 | = pn + · · · + p1 < 1 + + 2 + · · · pn = pn .
3 3
2
Proposition 2.6.
Kn )n1 satisfies
(2.3)
Starting from any critical piece K0 , the sequence of pairs of critical pieces (Kn ,
⎧
⎪
⎪
⎨ K0 ⊃⊃ K1 ⊃⊃ K1 ⊃⊃ K2 ⊃⊃ K2 ⊃⊃ · · · ;
∀ n 1, both Kn and Kn are pullbacks of Kn−1 ;
⎪
⎪
⎩ ∀ n 1, (K \K ) ∩ P = ∅.
n
(2.4)
n
And, for any m 1, for the maps g, f constructed below, and for CK = δ 3 , d = δ 4 , β = δ 5 such that:
Km−1
(a) deg(Km Km−1 ) CK , deg(Km
) CK ,
ξ
(b)
deg(B Km ) = 1,
(c)
deg(A B ) d,
(d)
deg(Km+2
U ) = deg(Km+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 , Km+2
and Km+2 .
K
K
m |, M = |Km 0 |, x̂ = f ξ (c) and ŷ = f M (x̂). Set the triple U, V, g as:
Set ξ = |Km+2
V := K0 − V
| g:=f M
U := Km
PENG WenJuan et al.
Figure 5
Sci China Math
March 2010
Vol. 53
No. 3
839
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 ŷ, row(Km),
marching horizontally to the right l steps one will meet again Km . If l = 0, i.e. ŷ ∈ Km , set B = Km
and B = Km
. If l > 0, let B ⊃⊃ B be the pair of puzzle pieces on column(ŷ) forming a parallelogram
with Km , Km (i.e. ŷ ∈ B and f l (B) = Km ). In both cases let A ⊃⊃ A be the pair of puzzle pieces on
column(x̂) forming a parallelogram with B , B.
Proof.
Points (a) and (d) follow from
deg Kn Kn−1 = CA · CB = δ 3 = CK ;
deg
Kn
Kn−1
= CB2 < CK ;
(Kn \K n ) ∩ P = ∅;
(2.5)
(2.6)
Kn−1 ) = deg(Kn+1
Kn ) · deg(Kn Kn−1 ) = δ 2 · δ 3 = β .
deg(Kn+1
(2.7)
Point (b) is obvious.
Starting from T (x̂), 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 the
first 0 hit of a critical spot, and set κj+1 to be the corresponding column number. Let now L be the
minimal integer so that κL > M := |Km K0 |.
Therefore κL−1 M . This implies that (L − 1)um M < 32 pm 32 · 43 um = 2 um. So L 2.
Now deg A B (deg Km Im )L = (deg A(Im ) Im )L (CA )2 = δ 4 = d.
\K m contains no postcritical points), this gives Point (c).
But deg(A B ) = deg(A B ) (as Km
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 edge
Km
|
f ξ (Km+2 )
(resp.
Km
|
A
) on column(x̂).
At first, recall that B = L
ŷ (Km ), so κL is actually greater or equal to the column number (in T (x̂))
of the right vertex of T . Also, by Rule 1, the number of critical spots on row(Km) from the κj -th
840
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
column (excluded) to the κj+1 -th column (included) is at most equal to the number of critical spots on
Im ]
, which is equal to 2 by Lemma 2.5. Therefore, denoting by top(T ) the top edge of in T
[Km =A(Im ) including the end vertices,
{critical spots ∈ top(T )} − 1
|Γ(Km ) Km
def. of L
L · 2 4,
| · ({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 ) =
(2.3)
r(Im+2 )
r(Im+1 ) 3 · |Γ(Km ) Km |,
r(Im+1 )
the one marked by ∗ is due to the definition of R(Km ) as the maximal possible length between two
consecutive critical spots on row(Km); the one marked by ∗∗ is due to the
that the
fact
left edge of T
Km+2
Km+2
B(Im+2 ) forms a parallelogram with
(see Figure 2.4), therefore |T | = | = | , which in turn is
|
A(I
)
K
K
m+2
m+2
m+2
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
Kn
|
−
Kn
For any n 1, let Kn− be the critical puzzle piece so that
Figure 6.
Corollary 2.7.
For every n 1, the segment
Kn+1
|
−
Kn+1
>
Kn
|
−
Kn
and
Kn
|
−
Kn
is vacuous,
;
and
Kn+1
|
Kn+1
>
Kn
|
Kn
B(Kn−1 )
|
A(Kn−1 )
form a parallelogram. See
,
(2.8)
\Kn−1 )
mod(Kn−1
.
3
CA CB
−
mod(Kn+1 \Kn+1
)>
(2.9)
The vacuousness follows from the property of A and B (Lemma 2.5). Now
⎧ ⎪
⎪ B(Kn−1 ) Kn ⎪
⎪
⎪
|
⎪
= | ⎪
− ⎨
Kn
A(Kn−1 )
Cor.2.4.(iii)
r(In ) = r(Γ(Kn−1 ))
>
R(Kn−1 ) .
K ⎪
⎪
n−1 ⎪
⎪
⎪
R(In−1 ) | ⎪
⎪
⎩
Kn−1 Proof.
Hence
Kn+1
|
−
Kn+1
=
B(Kn )
|
A(Kn )
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
(2.10)
r(Kn ) r(In ) >
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
Kn
|
−
Kn
Kn−1
|
Kn−1
,
Kn+1
|
Kn+1
(2.10) r(In+1 ) > (2.10)
Kn
|
Kn
.
PENG WenJuan et al.
Sci China Math
Figure 6
K
n−1
|
Kn−1
<
K
n+1
|
−
Kn+1
,
|
Vol. 53
No. 3
841
Control of moduli
Kn−1
J
Let J be the puzzle piece so that
March 2010
and
form a parallelogram. See Figure 6. As
|
−
Kn+1
Kn−1
J
|
K−
n+1
=
we know that J ⊂ Kn+1 . Therefore
−
−
) mod(J\Kn+1
)
mod(Kn+1 \Kn+1
vacuous
=
mod(Kn−1
mod(Kn−1
\Kn−1 )
\Kn−1 )
.
3C
deg(K − Kn−1 )
CA
B
n+1
2.6
Proof of Theorem 2.1
In the proof, we will use the following lemma:
Lemma 2.8 (Kahn-Lyubich Covering Lemma [18]).
that: given any g : U → V ,
Fix η > 0 and D ∈ N. There is ε(η, D) > 0 such
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| ) d,
⎪
⎪
⎩ mod(B \B) η · mod(U \A),
A
either
=⇒ mod(U \A)
> ε(η, D) > 0;
or
η
>
mod(V \B).
2d2
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 \Kn− is vacuous is proved in Corollary 2.7.
We will prove that
∃ C = C(δ, μ
) > 0, ∀ n, mod(Kn \K n ) C.
(2.11)
It follows then from Corollary 2.7 that mod(Kn \Kn− ) 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 = 2d3 β 2 + 1 = 2δ 22 + 1. Set μn = mod(Kn \K n ), n 1.
842
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
∀ m <kn
Case 1. Assume there is an increasing sequence kn → ∞ such that μm
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 ).
is chosen so that f M
= g
(A ) and
(U ) = V
; B
Set then V
= Km−Z ; g
= f M : U → V
where M
=
B
g(A).
Z
We then apply the estimates in Proposition 2.6. Note that, by (a) of Proposition 2.6, deg(
g) CK
=: D
with D independent of m. Note also
(c)
deg(A B ), deg(B
B ) deg(A B ) d,
(2.12)
(b)
deg(B
Km ) = deg(B
B ) deg(B Km ) d .
(2.13)
We have (see Figure 2.4)
μm+2 mod(U \f ξ (Km+2 )) (e) mod(U \A)
;
deg(Km+2
U)
β
(d)
no P
\B)
=
mod(B
μm
deg B
Km
choice of kn
(2.13)
(2.14)
μm+2 (2.14) mod(U \A)
=: η · mod(U \A),
d
dβ
1
with η := dβ
independent of m.
Therefore for any n n0 and m = kn − 2, we may then apply Kahn-Lyubich Covering Lemma to
, B),
to conclude that,
g
: (U, A , A) → (V
, B
either
(2.14)
β · μm+2 mod(U \A)
> ε(η, D) > 0;
or
>
\B)
η·mod(V
2
B
A 2 deg(
)
(∗)
η
2d2 (μm
choice of kn
+ · · · + μm−Z+1 )
ηZ
2d2 μm+2 ,
where the inequality (∗) is proved as follows: For j = m, m−1, . . . , m−Z+1, and for lj 0 minimal so that
f lj (
g (x̂)) ∈ Kj , denote by Bj , resp. Bj , the connected component of f −lj (Kj ), resp. f −lj (Kj ), containing
μj
with mod(B \Bj ) =
= μj
g
(x̂). Then {B \Bj }j are pairwise disjoint essential annuli in V
\B
j
j
(due to (Kj \K j ) ∩ P = ∅ and the fact that deg(Bj → Kj ) = 1).
deg(Bj →Kj )
2
By our choice of Z we have Z > 2d3 β 2 = 2dη β . Hence the second line above is impossible. So
μkn = μm+2 > β1 · ε(η, D) > 0. Therefore ∀ l ∈ N, μl limn→∞ μkn β1 · 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 β1 · ε(η, D).
Case 2.2. k0 − 2 < Z. Then k0 Z + 1 = 2δ 22 + 2. Notice that Kk 0 is mapped by some iterate of
Z+1
f onto K0 = P0 (c) with degree bounded by CK
, due to (a) of Proposition 2.6. The same iterate of f
1
1
mod(P0 (c)\W ) C Z+1
μ
.
maps Kk0 into a component W of U contained in P0 (c). So μk0 C Z+1
We have now proved (2.11) for C = min{ β1 · ε(η, D),
3
1
Z+1
CK
μ
}.
K
K
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-equivalent
and qc-conjugate off Kf .
PENG WenJuan et al.
3.1
Sci China Math
March 2010
Vol. 53
No. 3
843
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 → Ṽ be an isotopy path linking h0 to h1 .
Then there is a unique continuous extension (t, z) → h(t, z), [0, ∞[×V → Ṽ such that
0) 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:
..
.
U2
..
.
ht+2
−→
↓ f˜
f↓
U1
ht+1
−→
Ũ1
↓ f˜
f↓
V
Ũ2
h
t
−→
Ṽ.
Set then Ω = n1 V\Un = V\Kf , and Ω̃ = Ṽ\Kf˜. Then there is a qc map H : Ω → Ω̃ such that
H(x) = hn (x) for n 1 and x ∈ V\Un and that H ◦ f |Ω∩U = f˜ ◦ H|Ω̃∩Ũ , i.e. H realizes a qc-conjugacy
from 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 → Ṽ\Kf˜
conjugating f to f˜. We will show now that H admits a qc extension across Kf which is again a
conjugacy. 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 constant
L , 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 exists
L0 = L0 (ρ, r, d) with the following property. Let g, g̃ : (D, 0) → (D, 0) be holomorphic proper maps of
degree d, with critical values contained in Dρ . Let η, η : T → T be two homeomorphisms satisfying
g̃ ◦ 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 K̃n 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 → K̃1 , c → c̃ for some L1 1 (see e.g. [13, Lemma C.1]).
For the sequence Kn− given by Theorem 2.1, define K̃n− 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 puzzle
pieces, and H preserves this degree information, the sequence K̃n is precisely the KSS nest for f˜ starting
from K̃0 . Therefore Theorem 2.1 is valid for this sequence as well, with the same constant CK = δ 3 , and
probably a different Δ̃ as a lower bound for mod(K̃n \K̃n− ).
Recall that for each i 1, pi denotes the integer such that f pi (Ki ) = Ki−1 . We have f˜pi (K̃i ) = K̃i−1 ,
and f pi : Ki → Ki−1 and f˜pi : K̃i → K̃i−1 are proper holomorphic maps of degree δ 3 .
844
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
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 a
previously defined extension of H inside Kn−1 .
Set vn = c, and then, for i = n − 1, n − 2, . . . , 1, set consecutively vi = f pi+1 +···+pn (c).
Since Ki \Ki− is vacuous, the critical values of f pi+1 |Ki+1 , as well as vi , are contained in Ki− , 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 ◦ f pi ◦ ψi−1 . These maps fix the point 0, are proper holomorphic maps of degree δ 3 , with the
−
critical values contained in ψi−1 (Ki−1
).
Let ψi (Ki− ) = Ωi . Since mod(D\Ωi ) = mod(Ki \Ki− ) Δ > 0 and Ωi ψi (vi ) = 0, 1 i n, these
domains are contained in some disk Ds with s = s(Δ) < 1. So the critical values of gi are contained in
Ωi−1 ⊂ Ds , 2 i n.
The corresponding objects for f˜ will be marked with a tilde. The same assertions hold for g̃i . Then
all the maps gi and g̃i satisfy the assumptions of Lemma 3.2, with d = δ 3 , and ρ = max{s, s̃}.
(D, 0)
ψ1
←−
(K1 , v1 )
(K̃1 , ṽ1 )
↑ f p2
f˜p2 ↑
g2 ↑
(D, 0)
ψ2
←−
ψn−1
←−
(D, 0)
−→
(K̃2 , ṽ2 )
↑ f p3
..
.
f˜p3 ↑
..
.
(Kn−1 , vn−1 )
(K̃n−1 , ṽn−1 )
↑ f pn
f˜pn ↑
gn ↑
ψn
←−
(Kn , vn ),
(D, 0)
↑ g̃2
ψ̃2
(K2 , v2 )
g3 ↑
..
.
(D, 0)
ψ̃1
−→
(D, 0)
↑ g̃3
..
.
ψ̃n−1
−→
(D, 0)
↑ g̃n
ψ̃n
−→
(K̃n , ṽn )
(D, 0).
Note that each of ψi , ψ̃i extends to a homeomorphism from the closure of the puzzle piece to D.
Let us consider homeomorphisms ηi : T → T given by ηi = ψ̃i ◦ H|∂Ki ◦ ψi−1 . They are equivariant
with respect to the g-actions, i.e., ηi−1 ◦ gi = g̃i ◦ ηi .
Due to the qc extension of H|∂K1 , we know that η1 extends to an L1 -qc map (D, ψ1 (c)) → (D, ψ̃1 (c̃)).
Fix some r with ρ < r < 1. Since c ∈ K1− , c̃ ∈ K̃1− , we have ψ1 (c), ψ̃1 (c̃) ∈ 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, δ 3 ) 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):
T
η1
−→
g2 ↑
T
η2
−→
ηn−1
−→
T
↑ g̃n
ηn
−→
T,
ξ1
−→
g2 ↑
T
↑ g̃3
..
.
gn ↑
T
(D, 0)
↑ g̃2
g3 ↑
..
.
T
T
(D, 0)
we get
↑ g̃2
ξ2
−→
g3 ↑
..
.
(D, 0)
(D, 0)
↑ g̃3
..
.
ξn−1
−→
gn ↑
(D, 0)
(D, 0)
(D, 0)
↑ g̃n
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 ψ̃n−1 ◦ ξn ◦ ψn .
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
845
No. 3
Spreading principle
Fix now n 1. Recall that Kn is the unique critical puzzle piece of depth Mn . Set W0 = UMn \Kn and
X0 = Kn . For j 0, define a sequence H (j) of qc maps as follows:
⎧
⎪
on V\(X0 ∪ W0 ) = V\UMn ,
⎪
⎨ H,
(0)
H =
the L -qc map given by Proposition 3.1, on X0 = Kn ,
⎪
⎪
⎩ a C -qc extension for some C 1,
on W = U \K .
n
n
0
Mn
n
Cn
may depend on n.
The constant
For j 1, set
Xj = {z ∈ UMn | f l (z) ∈ Kn for some 0 l j}
and
Wj = {z ∈ U | z, f (z), f 2(z), . . . , f j (z) ∈ W0 } = UMn +j \Xj .
Notice that for any z ∈ Xj , L
z (Kn ) is a connected component of Xj , and there is j l(z) 0 so that
f l(z) maps L
z (Kn ) univalently onto Kn . On the other hand, any component S of Wj is a puzzle piece of
depth Mn + j and f j maps S univalently onto a component of W0 . Set then
⎧
⎪
on V\(Xj ∪ Wj ),
⎪
⎨ H,
(j)
(0)
H =
the L -qc map given by univalent pullback of H |X0 , on Xj ,
⎪
⎪
⎩ the C -qc map given by univalent pullback of H (0) | , on W .
W0
n
j
Set Cn = max{C0 , L , Cn }. The {H (j) }j0 is a sequence of Cn -qc maps. Hence it is precompact in the
uniform topology.
Notice that H (j) = H (j−1) outside Wj−1 . Thus, the sequence {H (j) } converges pointwise outside
Wj = {x ∈ Kf | f k (x) ∈
/ Kn , k 0}.
j
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) }j0 coincide on a dense open set
of 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 ) ∪ ( j Wj )) is therefore C0 -qc there; and is L -qc
on
X . It follows that the dilatation of Hn is bounded by max{C0 , L } except possibly on the set
j j
j Wj . But this set has zero Lebesgue measure. It follows that the dilatation of Hn is max{C0 , L },
which is independent of n.
Conclusion
The sequence Hn : V → Ṽ has a subsequence converging uniformly to a limit qc map H : V → Ṽ, with
H |V\Kf = H. Therefore H is a qc extension of H. On the other hand, H ◦ f = f˜ ◦ H on U\Kf , and
Kf has empty interior by assumption. So H ◦ f = f˜ ◦ H holds on U by continuity. Therefore H is a
qc-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.
References
1 Avila A, Kahn J, Lyubich M, Shen W X. Combinatorial rigidity for unicritical polynomials. Ann of Math, 2009, 170:
783–797
846
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
2 Branner B, Hubbard J H. The iteration of cubic polynomials, Part II: Patterns and parapatterns. Acta Math, 1992,
169: 229–325
3 Cui G Z, Tan L. A characterization of hyperbolic rational maps. ArXiv:math/0703380v2 [math.DS] 30 Oct, 2007
4 Douady A, Hubbard J H. Étude dynamique des polynômes complexes I, II. Publ Math d’Orsay, 1984–1985
5 Graczyk J, Smirnov S. Non-uniform hyperbolicity in complex dynamics. Invent Math, 2009, 175: 335–415
6 Haissinsky P. Rigidity and expansion for rational maps. J London Math Soc, 2001, 63: 128–140
7 Hubbard J H. Local connectivity of Julia sets and Bifurcation loci: three theorems of J.-C. Yoccoz. In: topological
Methods in Mordern Mathematics, edited by L. R. Goldberg and A. V. Phillps, 467–511. Houston: Publish or Perish,
Inc., 1993
8 Khan J, Lyubich M. The quasi-additivity law in conformal geometry. Ann of Math, 2009, 169: 561–593
9 Kozlovski O, Shen W X, van Strien S. Rigidity for real polynomials. Ann of Math, 2007, 165: 749–841
10 Kozlovski O, van Strien S. Local connectivity and quasi-conformal rigidity of non-renormalizable polynomials. Proc
London Math Soc, 2009, 99: 275–296
11 Lyubich M. Dynamics of quadratic polynomials, I-II. Acta Math, 1997, 178: 185–247, 247–297
12 McMullen C T. Complex Dynamics and Renormalizations. In: Ann of Math Stud., No. 135. Princeton: Princeton
University Press, 1994
13 McMullen C T. Self-similarity of Siegel disks and Hausdorff dimension of Julia sets. Acta Math, 1998, 180: 247–292
14 Milnor J. Dynamics in One Variable. In: Annals of Mathematics Studies, Vol. 160. Princeton: Princeton University
Press, 2006
15 Peng W J, Qiu W Y, Roesch P, et al. A tableau approach of the KSS nest. Conformal Geometry and Dynamics, to
appear
16 Przytycki F, Rohde S. Rigidity of holomorphic Collet-Eckmann repellers. Ark Math, 1999, 37: 357–371
17 Tan L, Yin Y C. Unicritical Branner-Hubbard conjecture. In Complex Dynamics: Family and Friends. Schleicher D.
ed. Welleskey: A K Deters Ltd, 2009
18 Yin Y C, Zhai Y. No invariant line fields on Cantor Julia sets. Forum Mathematicum, to appear
19 Zhai Y. Rigidity for rational maps with Cantor Julia sets. Sci China Ser A, 2008, 51: 79–92
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-th
column} is non-critical. So for each n 1,
deg(Pn0 +n (c) Pn0 (f
n
(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)}k1 are children of Pn0 (c) and then deg(Pn0 +m (c) Pn0 (c) ) = δ. Now we take a subsequence
k
in of mk such that
f in (Pn0 +in (c)) = Pn0 (c) = Pn0 (x),
f in (Pn0 +in +1 (c)) = Pn0 +1 (x)
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
847
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,
n
deg(Pn1 +n (z) Pn1 (f (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 Grötzsch’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) ) δ 2 .
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 Tn (z) be the
puzzle piece such that
(z)
Tn
|
Tn (z)
and
Kn
|
Kn
form a parallelogram. Since
also Figure 1.2), we know that deg(Tn (z) Kn
Kn
|
Kn
is vacuous, applying tableau rule3b (see
) = deg(Tn (z) Kn ) δ. Then
mod(Tn (z)\Tn (z)) C
mod(Kn \Kn )
> 0.
δ
δ
Until now, we come to the conclusion that Kf is a Cantor set.
We may assume that {Pn0 +in (c)}n1 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.
independent of n.
For every n 1, H|∂Qn admits a K -qc extension inside Qn , where K is a constant
Proof. Fix n 1.
−
Let Q̃n , Q̃−
n be the puzzle pieces bounded by H(∂Qn ) and H(∂Qn ) respectively. Since H preserves the degree
information, Lemma A.1 is valid for Q̃n with the same δ, more precisely, for the map f˜, there exists a piece P̃n0 (x̃)
of depth n0 containing x̃ ∈ Ṽ such that
f˜in (P̃n0 +in (c̃)) = P̃n0 (x̃),
f˜in (P̃n0 +in +1 (c̃)) = P̃n0 +1 (x̃)
848
PENG WenJuan et al.
Sci China Math
March 2010
Vol. 53
No. 3
and
deg(P̃n
0 +in
(c̃) P̃n0 (x̃)
) = δ.
Let mod(Pn0 (x)\Pn0 +1 (x)) = Δ and mod(P̃n0 (x̃)\P̃n0 +1 (x̃)) = Δ̃.
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−1
n . 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 (f in (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 h̃n . Then all the
maps hn and h̃n satisfy the assumptions of Lemma 3.2, with d = δ, and ρ = max{t, t̃}.
Note that each of φn , φ̃n extends to a homeomorphism from the closure of the puzzle piece to D.
and
Let us consider homeomorphisms σ0 : T → T and σn : T → T given by σ0 = φ̃0 ◦ H|∂Pn0 (x) ◦ φ−1
0
−1
σn = φ̃n ◦ H|∂Qn ◦ φn respectively. Then σ0 ◦ hn = h̃n ◦ 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 :
T
σ
n
−→
hn ↓
T
T
↓ h̃n
σ0
−→
T,
(D, 0)
we get
ζn
−→
hn ↓
(D, 0)
(D, 0)
↓ h̃n
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 φ̃−1
n ◦ ζn ◦ φn . This completes the proof of this proposition.
Now we can use the spreading principle to the sequence {Qn }n1 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.