Topics in One-Dimensional and Holomorphic DynamicsCUNY Graduate Center, Department of Mathematics

MATH 82000 [20038], 4.5 credits

Yunping Jiang

January 28-May 15, 2013

(This note is used only for this class, not for distribution)

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Lecture 1. Introduction

• Topology: f : X → Y , where the domain X and the range Yare two spaces and f is a map.• Dynamics: f : X → X, where the domain and the range are

the same space X and f is a self-map.

Therefore, the study of dynamics can be thought as a part of topology.However, since we can iterate f to get

f n = f · · · f︸ ︷︷ ︸n

for each integer n in dynamics, we have a group Gf = fn∞n=−∞ iff is one-to-one and onto or a semi-group SGf = fn∞n=0 if f is non-invertible. We simple use fn to denote f n and f 0 always means theidentity map from X onto itself. The group Gf or the semi-group SGfis called a (discrete) dynamical system. The space X is called a phasespace.

One can also consider a continuous dynamical system f tt∈R wheref t is a family of one-to-one and onto map from X into itself parameter-ized by real number t such that f t+s = f tf s. A continuous dynamicalsystem is a solution of an ordinary differential equation. For example,suppose X = Rn and suppose f t is differentiable at t = 0, then we havea vector field

V (x) = limt→0

f t(x)− xt

, x ∈ X.

The continuous dynamical system f t is a solution of the ordinarydifferential equation

x = V (x),

that is,

f t(x) =df t(x)

dt= lim

∆t→0

f t+∆t(x)− f t(x)

∆t

= lim∆t→0

f∆t(f t(x))− f t(x)

∆t= V (f t(x)), t ∈ R, x ∈ X.

Conversely, a solution of ordinary differential equation above is a con-tinuous dynamical system.

Given a continuous dynamical system f tt∈R, we can consider thetime-one map f = f 1. Then Gf = fn∞n=−∞ is a part of the continuousdynamical system f tt∈R. However, it is not true, in general, that agroup Gf = fn∞n=−∞ can be always thought as a group generated bythe time-one map of a continuous dynamical system. It is actually aninteresting problem:

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Problem 1. Under what conditions, can a one-to-one and onto mapf : X → X be embedded into a continuous dynamical system as thetime-one map?

It is a hard problem if we insist to work on the same phase space X.However, if we consider another phase space with one-dimension higherthan X, we can always embed f into a continuous dynamical system.This is called suspension technique. Consider X × [0, 1] and glue (x, 1)with (f(x), 0). We get a new phase space Mf = X × [0, 1]/ ∼ whichhas one dimension higher than X. Consider the vertical vector fieldV (x, t) = ∂/∂t on Xf . The suspension flow f t(x, t) is just the solutionof the ordinary differential equation

˙(x, t) = V (x, t) = (0, 1)

on Xf . So f 1(x, 0) = (x, 1) = (f(x), 0). Then f tt∈R is a continuousdynamical system and f is its time-one map. Basically, we can usea discrete dynamical system to study a continuous dynamical system.One can refer to the following book for the embedding problem.

• P. Hartman, Ordinary Differential Equations (second edition),Birkhauser, Boston (1982)

In this class, I will concentrate on a discrete dynamical system.Moreover, we will mostly concentrate on a semi-group SG = fn∞n=0

where f is a one-dimensional map from an interval I of the real lineinto itself (in this case, we call it a one-dimensional dynamical system)or a holomorphic map from a region of the complex plane into itself(in this case, we call it a complex dynamical system or a holomorphicdynamical system).

A basic problem in dynamics is to understand the future of the orbitO(x0) = xn = fn(x0)∞n=0 for a given initial data x0.

Definition 1. The ω-limit set of x0 is the set of all limiting points ofconvergent subsequences of O(x0).

If X is a good topological space (for example, a smooth Riemannianmanifold), we can prove that ω(x0) is a closed subset of X. The setω(x0) is the future of the orbit O(x0).

Definition 2. A point x0 ∈ X is called a fixed point if f(x0) = x0.

It is clear that if x0 is a fixed point, then ω(x0) = x0.

Definition 3. A point x0 ∈ X is called a periodic point of periodn > 0 if fn(x0) = x0 and f i(x0) 6= x0 for all 1 ≤ i < n.

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It is clear that if x0 is a periodic point of period n, then ω(x0) =x0, f(x0), · · · , fn−1(x0). In general, ω(x0) could be a very complicatesubset of X.

To better understand the future, we need to obtain knowledge of thehistory of x0, that is, the set of ∪∞n=0f

−n(x0). In particular, when f isnon-invertible, f−n(x0) may contain more than one points. Then wecan define a life tree Treef (x0) as follows. The starting point is x0.The first level vertices are all points y in f−1(x0) and then use an edgeto connect x0 to y. Suppose the nth-level of tree is defined. Then the(n + 1)th-level of tree is defined by all points in f−1(y) for all y in thenth-level and use an edge to connect y and z if f(z) = y.

The life trees for all x0 ∈ X define an inverse limiting space

X∞,f = . . . xnxn−1 . . . x1x0 | xn ∈ X, f(xn) = xn−1, n ≥ 0,

which gives the full history of the life of x0. Therefore, we can use theknowledge of the full history of the life to study the future ω(x0) forevery x0 ∈ X.

Example 1. Let S1 = z ∈ C | |z| = 1 be the unit circle where Cis the complex plane. Suppose f(z) = z2 : S1 → S1. Find the inverselimiting space S1

∞,f .

Proof. Since |z| = 1, we write z = e2πiθ where 0 ≤ θ < 1. Then

f−1(z) = z0 = eπiθ, z1 = eπi(θ+1).

Thus

S12,f = z0z, z1z

becomes a 2-fold of circle. Similarly,

f−1(z0) = z00 = eπiθ2 , z10 = e

πi(θ+2)2

and

f−1(z1) = z01 = eπi(θ+1)

2 , z11 = eπi(θ+3)

2 .

Thus

S13,f = z00z0z, z10z0z, z01z1z, z11z1z

becomes a 4-fold of circle. Similarly, S1n,f is a 2n−1-fold of circle for

any n ≥ 1. They are still connected. The inverse limiting space S1∞,f

is actually very complicated (for example, it is disconnected). Forevery point z ∈ S1, the life tree Treef (z) is a uncountable and totally

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disconnected (we will prove it later), thus it is a Cantor set which canbe identified with the symbolic space

Σ− =−1∏

n=−∞

0, 1 = w− = . . . jnjn+1 . . . j−2j−1 | jn = 0 or 1.

Note that 0, 1 is a set with discrete topology and Σ− has the producttopology (we will construct this product topology later).

For any 0 ≤ θ ≤ 1, we have a 2-adic expansion

θ =∞∑m=0

im2m+1

, im = 0 or 1.

Then we consider the symbolic space (with the product topology)

Σ+ =∞∏m=0

0, 1 = w+ = i0i1 . . . im−1im . . .

and a map

π+(w+) = e2πiθ, θ =∞∑m=0

im2m+1

.

Then π+ is one-to-one except for 000 . . . and 111 . . ., 1000 . . . and0111 . . ., and w+

m = i0 . . . im−1im100 . . . and v+m = i0i1 . . . im−1im011 . . .

on which π+ is two-to-one (for example, 000 . . . and 111 . . . are mappedto 1, 1000 . . . and 0111 . . . are mapped −1, and w+

0 = i0100 . . . andv+

0 = i0011 . . . are mapped i if i0 = 0 and −i if i0 = 1). Thus wecan think S1 = Σ+/000 . . . ∼ 111 . . . , 1000 . . . ∼ 0111 . . . , w+

m ∼ v+m.

This gives us a picture of the inverse limiting space S1∞ which can be

written as

S1∞,f = Σ/000 . . . ∼ 111 . . . , 1000 . . . ∼ 0111 . . . , w+

m ∼ v+m

where

Σ = Σ−⊕

Σ+ = w = w−.w+ = . . . jnjn+1 . . . j−2j−1.i0i1 . . . im−2im−1 . . .,

where n < 0 and m ≥ 0. We call S1∞ a solenoid.

Define σ : Σ→ Σ as

σ(w) = . . . j′nj′n+1 . . . j

′−2j′−1.i

′0i′1 . . . i

′m−2i

′m−1 . . .

= . . . jnjn+1 . . . j−2j−1i0.i1 . . . im−2im−1 . . . .

Then σ is invertible. As you can see, from a non-invertible dynamicalsystem f(z) = z2, we construct an invertible dynamical system whichcontains the full history and the future for every z ∈ S1.

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Exercise 1. Let S1 = z ∈ C | |z| = 1 be the unit circle where C isthe complex plane. Suppose qd(z) = zd : S1 → S1, d ≥ 3. Find theinverse limiting space S1

∞,qd.

Definition 4. Given a (discrete) dynamical system f : X → X. Asubset Λ ⊂ X is called an invariant subset of f if f(Λ) ⊆ Λ.

If Λ is an invariant subset of f , then f : Λ → Λ is a new (discrete)dynamical system.

Definition 5. An invariant subset Λ is called completely invariant iff(Λ) = f−1(Λ) = Λ.

Definition 6. An invariant set Λ is called minimal if there is no propersubset of Λ which is also invariant.

Example 2. Suppose f : X → X. Suppose x0 is a periodic point ofperiod n ≥ 1. Then the set Λ = x0, . . . , f

n−1(x0) is an invariant setand minimal. But it may not be a complete invariant set. However,if we consider f : Λ → Λ as a new dynamical system, then Λ is acompletely invariant set.

Example 3. Let C∗ = C \ 0. Suppose f(z) = zd : C∗ → C∗. d > 1.Then S1 is a completely invariant set of f but it is not minimal.

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Lecture 2. Markov partition and counting problemand zeta function

Given a dynamical system f : X → X. Let Pn denote the set of allperiodic points of period n ≥ 1. Let Fn be the set of all fixed points offn. Then it is clear that

Pn = Fn \ (∪1≤p<n,p|nFp).

An important problem is to count numbers of Fn, therefore, numbersof Pn. Actually, we are more interested in the asymptotical growth rateof numbers of Fn which can be defined as

h(f) = limn→∞

1

nlog #(Fn).

(Later, this number is also called the topological entropy.) What wecan do is to put all numbers #(Fn) into a power series and define the(unweighted) dynamical zeta function

ζ(z) = exp( ∞∑n=1

#(Fn)

nzn).

(It definitely has some relation with the classical Riemann zeta func-tion. Refer to

• Dieter H. Mayer, The thermodynamic formalism approach toSelberg’s zeta function for PSL(2,Z), Bull. Amer. Math. Soc.(N.S.), Volume 25, Number 1 (1991), 55-60.

)Then function ζ(z) is holomorphic in a disk centered at 0 of radius

r ≥ 0. the logarithm of the reciprocal of the convergence radius rmaxof ζ(z) is the growth rate h(f). One of most important and interestingquestion is that

Problem 2. Can this zeta function be extended to an meromorphic orentire function on the whole complex plane?

To study this problem, we should first study an important methodin dynamics called Markov partitions.

Suppose X is a compact metric space and d(·, ·) is the metric.

Definition 7. A partition X = X0 ∪ . . . ∪ Xd−1 is called a Markovpartition for f if

(1) Each Xi is a non-empty compact subset of X;(2) Xi ∩Xj = ∅ for any 0 ≤ i 6= j ≤ d− 1;(3) f |Xi is one-to-one for every 0 ≤ i ≤ d− 1;

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(4) f(Xi) = ∪mik=1Xik for every 0 ≤ i ≤ d− 1.

More generally, we replace (2) and (3) in the above definition as

Definition 8. A partition X = X0 ∪ . . . Xd−1 is called a Markovpartition for f if

(1) Each Xi is a non-empty compact subset of X;(2’) mathringXi ∩Xj = ∅ for any 0 ≤ i 6= j ≤ d− 1;

(3’) f |Xi is one-to-one for every 0 ≤ i ≤ d− 1;(4) f(Xi) = ∪mik=1Xik for every 0 ≤ i ≤ d− 1,

where Xi means the interior of Xi.

In this case, we have to take care of more on the boundary of Xi.

Example 4. Let I = [0, 1] and I0 = [0, 1/3] and I1 = [2/3, 1]. Letf(x) = 3x for x ∈ I0 and f(x) = 3(x − 2/3) for x ∈ I1. Let Λ =x ∈ I | fn(x) ∈ I, ∀n > 0. The set Λ is called the non-escaping setfor f . Then f : Λ → Λ is a dynamical system. Let Λ0 = Λ ∩ I0 andΛ1 = Λ ∩ I1. The Λ0,Λ1 gives a Markov partition for f .

Example 5. Let S1 = z ∈ C | |z| = 1 be the unit circle of thecomplex plane. Let qd(z) = zd : S1 → S1. Then it is a dynamicalsystem. Let Ik = [e2πik/d, e2πi(k+1)/d] for 0 ≤ k < d. Then I0, . . . , Id−1gives us a Markov partition for qd in the meaning of Definition 8.

If f : X → X has a Markov partition, we use

η0 = X0, . . . , Xd−1to denote the initial partition. Associate with this initial partition, wehave a d× d matrix A = (aij) where

aij =

1, if f(Xi) ⊃ Xj;0, otherwise.

We also have a symbolic space

ΣA = w = i0i1 . . . ik−1ik . . . | ik ∈ 0, . . . , d− 1, aik−1ik = 1, ∀k ≥ 1and a shift map

σA : w = i0i1 . . . ik−1ik . . .→ σA(w) = i1 . . . ik−1ik . . . .

Here w = i0i1 . . . ik−1ik . . . is called an admissible sequence (with re-spect to A) of 0, . . . , d− 1.

We can also get a sequence of nested pull-back partitions from theinitial partition, ηk = f−k(η0) which consists of all non-empty compactsubsets Y of X such that fk : Y → Xi is one-to-one and onto (or fk :

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Y → Xi is one-to-one and onto in Definition 8) for some 0 ≤ i ≤ d− 1.Therefore, we can label subsets in ηk as

ηk = Xi0i1...il...ik−1ik

so that f l(Xi0i1...il...ik−1ik) ⊂ Xil for all 0 ≤ l ≤ k. Note that each ηk isalso a Markov partition. Thus we have a sequence of Markov partitionswith the property that

f(Xi0i1...il...ik−1ik) = Xi1...il...ik−1ik

and

Xi0i1...il...ik−1ik ⊂ Xi0i1...il...ik−1.

Since

Xi0i1...il...ik−1ik ⊂ Xi0i1...il...ik−1⊂ Xi0i1...il...ik−2

⊂ · · ·Xi0i1 ⊂ Xi0 ⊂ X

and since each subset is non-empty and compact, for any admissiblesequence w = i0i1 . . . ik−1ik . . ., we have a non-empty subset

Xw = ∩∞k=0Xi0i1...il...ik−1ik .

Definition 9. We say f : X → X is topologically expansive if everyXw = xw contains only one point.

For every finite admissible sequence w0k = i00 . . . i

0k−1i

0k, where ii ∈

0, . . . , d− 1, 0 ≤ i ≤ k, we call

[w0k] = w = w0

kik+1 . . . il . . . | il ∈ 0, 1, . . . , d− 1, l ≥ k + 1a cylinder. Since 0, . . . , d − 1 has the discrete topology, ΣA has theproduct topology, that is, the topology generated by all cylinders [w0

k].We also can put a metric on ΣA as

d(w,w′) =∞∑k=0

|ik − i′k|dk+1

where w = i0i1 . . . ik−1ik . . . and w′ = i′0i′1 . . . i

′k−1i

′k . . .. The topology

induced by this metric on ΣA is the same topology generated by theset of cylinders.

For a topologically expansive f : X → X with the initial Markovpartition η0 and the associative matrix A, we then have a homeomor-phism

πf (w) = xw : ΣA → X

if we are in Definition 7. (However, if we are in Definition 8, πf isone-to-one and onto except for all boundary points of Xi.) Moreover,σA : ΣA → ΣA and f : X → X are topologically conjugate by πf in

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Definition 7 (or they are topologically semi-conjugate in Definition 8),that is,

f πf = πf σA.

Definition 10. Two dynamical systems f : X → X and g : Y → Yare said to be topologically conjugate if there is a homeomorphism h :Y → X such that

h g = f h.Here h is called the conjugacy between f and g. If h is just continuousand onto, we call f and g are topologically semi-conjugate. In this case,h is called the semi-conjugacy.

Proposition 1. Suppose f : X → X and g : Y → Y are topologicallyconjugate and h is the conjugacy. Then p ∈ Y is a periodic point ofperiod n ≥ 1 of g if and only if q = h(p) is a periodic point of periodn ≥ 1 of f . Thus #(Pn(f)) = #(Pn(g)) and #(Fn(f)) = #(Fn(g)).

Proof. The point p ∈ Y is a periodic point of period n ≥ 1 of g ifgj(p) 6= p for 1 ≤ j < n but gn(p) = p. Since f = h g h−1,we have that fn(q) = h(gn(h−1(h(p)))) = h(gn(p)) = h(p) = q andf j(q) = h(gj(h−1(h(p)))) = h(gj(p)) 6= h(p) = q for 1 ≤ j < k. Thus qis a periodic point of period k ≥ 1 of f . By using h−1, we get the “ifand only if”. We have bijective maps

h : Pn,g → Pn,f and h : Fn,g → Fn,f .

This implies #(Pn,g) = #(Pn,f ) and #(Fn,g) = #(Fn,f ).

For an expansive f : X → X with the initial Markov partition η0 (inDefinition 7) and the transitive matrix A, we have that

#(Fn,f ) = #(Fn,σA) = trace(An).

Use this relation, we can have a full calculation of the dynamical zetafunction of f as follows:

ζ(z) = exp( ∞∑n=1

#(Fn)

nzn)

= exp( ∞∑n=1

trace(An)

nzn).

By using the relation det(B) = exp(trace(logB)) for a square matrix,we have that

ζ(z) = exp(−trace(log(I − zA))) =1

det(I − zA).

This implies that ζ(z) can be extended analytically to a rational map.In the calculation of ζ(z) we have a non-negative matrix. So let us

to learn some facts in theory of matrices.

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SupposeV = v = (v1, . . . , vn)T | vi ∈ R

is the n-dimensional vector space. We will use the norm

(1) ‖v‖ =n∑i=1

|vi|, v = (v1, · · · , vn)T ∈ V .

For any linear map L : V → V , let A be the corresponding n×n-matrixfor L, that is, L(v) = Av. Define

‖A‖ = sup‖v‖=1

‖Av‖.

The spectral radius ρ(A) of A can be calculated as

ρ(A) = limn→∞

n√||An|| ≥ 0.

A square matrix A = (aij)n×n is said to be positive if all aij > 0for 1 ≤ i, j ≤ n. An eigenvalue is called simple if the correspondingeigenspace is one-dimensional.

Theorem 1 (Perron-Frobenius Theorem). If a n× n matrix A is pos-itive, then A has a unique, simple, positive, maximal eigenvalue λ witha positive eigenvector v0 = (v1, · · · , vn)T , i.e, vi > 0 for all 1 ≤ i ≤ n.Here the term “unique” means that all other eigenvalues µ of A satisfythat

|µ| < λ.

Proof. Let

S = v = (v1, . . . , vn)T ∈ V | ‖v‖ = 1, vi ≥ 0, 1 ≤ i ≤ nbe a convex compact subset in V . Define

f(v) =Av

‖Av‖: S → S.

The Brouwer fixed point theorem: every continuous function whichmaps a convex compact subset into itself must have a fixed point,implies that f has a fixed point v0 ∈ S, that is, f(v0) = v0. Then wehave that λ = ‖Av0‖ is an eigenvalue with an corresponding eigenvectorv0. Suppose v0 = (v1, . . . , vn)T . Then

λvi =n∑j=1

aijvj.

This implies that λ > 0 and vi > 0 for all 0 ≤ i ≤ n. Clearly,λ = ρ(A) (note that ‖Av‖ ≤ ‖A|v|‖ for every v = (v1, . . . , vn) where|v| = (|v1|, . . . , |vn|)T .

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Now we show the eigenspace Eλ is one-dimensional. Suppose u =(u1, . . . , un)T 6= 0 ∈ Eλ. Let w = tv0 − u ≥ 0 and at least onecomponent of w be zero, that is,

t = minuivi

∣∣∣ 1 ≤ i ≤ n.

Then from Aw = λw, we get w = 0. That is u = tv0. So Eλ =spanv0. Similarly, we also showed that no other positive vector notin Eλ is an eigenvector.

There is no negative eigenvalue r such that |r| = λ. Otherwise,Au = −ru and A2u = r2u for some u 6= 0 and u 6∈ Eλ. Since A2 isalso a positive matrix and λ2 is a positive eigenvalue with the eigenspaceEλ, this is a contradiction. This implies that all real eigenvalues r 6= λof A must have |r| < λ.

There is one more property about f as follows: For any v ∈ S,fk(v) → v0 as k → ∞. This property can be proved by using theHilbert projective metric on the positive cone

V+ = v = (v1, . . . , vn)T | vi ≥ 0, 1 ≤ i ≤ n.

Refer to, for example,

• G. Birkhoff, Extensions of Jentzchs theorem. Trans. Amer.Math.Soc.,85 (1957), 219-227.• Y. Jiang, Nanjing Lecture Notes In Dynamical Systems. Part

One: Transfer Operators in Thermodynamical Formalism. FIMPreprint Series, ETH-Zurich, June 2000.• C. MaCarthy, The Hilbert projective metric, multi-type branch-

ing processes and mathematical biology: a model of the evolu-tion of resistance. Ph.D thesis, CUNY Graduate Center, 2010.

Now we use this property to show that A has no complex eigenvaluez such that |z| = λ. We prove it by contradiction. First, the propertyimplies that Akx tends to Eλ as k goes to ∞ for any x > 0. Supposewe have a complex number z with |z| = λ such that Au = zu for acomplex vector u = x + iy. Let z = λe2πiθ where 0 < θ < 1. Let t1 bea real number such that t1v0−x ≥ 0 with at least one zero componentand t2 be another real number such that t2v0−y ≥ 0 with at least onezero component. Note that x = t1v0 − x and w = t2v0 − y can not bein Eλ (otherwise, from Ax = λx, we get x = 0 and a similar argumentholds for y, this implies that Au = λu = zu but z 6= λ). Consider Akxand Aky, we have the equation,

Akx + iAky = λk(t1v0 − e2πkiθx) + λk(t2v0 − e2πkiθy)i.

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The real part and the imaginary part of the left side both tend to theline Eλ. Now let us examine the real part and the imaginary part ofthe right side. If θ = p/q is a rational number, then we have a sequenceof integers kl = lq such that e2πikl = 1. If θ is an irrational number,then the set kθ (mod 1) | k ∈ Z is dense in the interval [0, 1]. Weconcluded that for any θ, we have a sequence of integers kl tending to∞ such that e2πiklθ → 1 as l → ∞. This implies that the real part ofthe right side tends to the line spanned by x = t1v0 − x as l → ∞,which is not the line Eλ. Similar, the imaginary part of the right sidetends to the line spanned by y = t2v0 − y as l →∞, which is not theline Eλ. This is a contradiction. The contradiction implies that for anycomplex eigenvalue z of A, we have that |z| < λ.

One can check references (the third reference will be also used later)for other proofs of the Perron-Frobenius theorem and its generalizationto infinite-dimensional transfer operators.

1. F. R. Gantmacher, Theory of matrices. Chelsea, 1959.2. G. Strang, Linear Algebra and Its Applications, third edition,

1988. (A simple proof for the finite-dimensional case.)3. Y. Jiang, A proof of existence and simplicity of a maximal eigen-

value for Ruelle-Perron-Frobenius operators. Letters in Math.Phys. 48 (1999), no. 3, 211-219. (A simple proof for the infinite-dimensional case.)

A square matrix A = (aij)n×n is said to be non-negative if all aij ≥ 0for 1 ≤ i, j ≤ n. A non-negative n × n matrix A is called irreducibleif no permutation of the indices places the matrix in a block lower-triangular form. More precisely, there is no permutation matrix P ,which is a matrix consisting of 0 and 1 such that each row or eachcolumn contains one and only one 1, such that

PAP−1 =

(A11 A12

0 A22

),

where A11 and A22 are square matrices.If A is a non-negative but reducible, then we can find a permutation

matrix P such that

PAP−1 =

A11 ∗ ∗ · · · ∗0 A22 ∗ · · · ∗...

......

... ∗0 0 · · · 0 Akk

,

and such that all Aii are irreducible matrices for 1 ≤ i ≤ k. The set ofeigenvalues of A is the union of sets of eigenvalues of Aii, 1 ≤ i ≤ k.

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For example,

I =

(1 00 1

),

is reducible with one positive maximal eigenvalue λ = 1 of multiple 2.

J =

(0 11 0

),

is irreducible with one positive simple maximal eigenvalue 1 and onenegative simple eigenvalue −1. Note that J2 = I.

K =

(1 11 1

),

is irreducible with one positive simple maximal eigenvalue 2 and onesimple eigenvalue 0.

An equivalent definition of irreducibility is that for any 1 ≤ i, j ≤ n,there exists a 0 ≤ q = q(i, j) ≤ n such that the ij-th entry of Aq ispositive.

Theorem 2 (Perron-Frobenius Theorem). If A is irreducible, ρ(A)is a simple, positive, maximal eigenvalue with a positive eigenvectorv = (v1, · · · , vn), i.e, vi > 0 for all 1 ≤ i ≤ n.

Remark 1. If A is a non-negative matrix, then

PAP−1 =

A11 ∗ ∗ · · · ∗0 A22 ∗ · · · ∗...

......

... ∗0 0 · · · 0 Akk

,

Then we can apply the Perron-Frobenius theorem on Aii to get the sim-ple, positive, maximal eigenvalue ρ(Aii). Then the maximal eigenvalueof A is ρ(A) = max1≤i≤k ρ(Aii).

However, even if A is irreducible, there may exist another eigenvalueµ 6= ρ(A) but |µ| = ρ(A). For example, consider

J =

(0 11 0

).

It is an irreducible matrix. The spectral radius is 1 which is a simple,positive, maximal eigenvalue with an eigenvector v1 = (1, 1). However,−1 is also an eigenvalue with an eigenvector v = (1,−1).

Now let us return to zeta function ζ(z). If f is topologically expansivewith a Markov partition η0, then ζ(z) can be extended analytically toa rational function. The smallest pole of ζ(z) is the reciprocal of the

15

maximal eigenvalue ρ(A) of A. Notice that according to the Perron-Frobenius theorem A has a positive maximal eigenvalue ρ(A) since allentries of A are non-negative. Then we have that the growth rate ofnumbers of Fn,f is h(f) = log ρ(A). In particular, when A is a positivematrix, then 1/ρ(A) is the simple smallest pole of ζ(z). For example,if

A = I =

(1 00 1

),

ζ(z) = 1/(1− z)2. If

A = J =

(0 11 0

),

ζ(z) = 1/(1− z2) = 1/(1− z)(1 + z). And if

A = K =

(1 11 1

),

ζ(z) = 1/(1− 2z).

Example 6. Suppose I = [0, 1] and I0 = [0, a] and I1 = [b, 1] for0 < a < b < 1. Suppose f : I0 ∪ I1 → I is a map satisfying thatf0 = f |I0 : I0 → I is a homeomorphism f1 = f |I1 : I1 → I is ahomeomorphism and that f(0) = f(b) = 0 and f(a) = f(1) = 1. Thenon-escaping set for f is, by definition,

Λ = x ∈ [0, 1] | fn(x) ∈ [0, 1], ∀n ≥ 0.

It is clearly,

Λ = ∩∞n=0f−n(I).

Then f : Λ → Λ is a Markovian dynamical system with the initialpartition η0 = Λ0,Λ1 where Λ0 = Λ ∩ I0 and Λ1 = Λ ∩ I1 and thetransitive matrix

A =

(1 11 1

).

We assume that f is expansive. So #(Fn,f ) = trac(An) = 2n. Thenmaximal positive eigenvalue of A is 2. The growth rate of numbers ofFn is log 2. Furthermore, we have that

1

det(I − zA)=

1

1− 2z.

Example 7. Suppose S1 = z = e2πθ; 0 ≤ θ ≤ 1 and f(z) = z2. I0 =z = e2πθ; 0 ≤ θ ≤ 1/2 and I1 = z = e2πθ; 1/2 ≤ θ ≤ 1. Then f is

16

a Markovian dynamical system with the initial partition η0 = I0, I1in Definition 8. The transitive matrix is still

A =

(1 11 1

)The map πf : ΣA → S1 is one-to-one except for all endpoints Iwk forwk = i0i1 . . . il . . . ik−1ik for k ≥ 0.

So #(Fn,f ) = #(σnA) − 1 = trac(An) − 1 = 2n − 1 for n ≥ 1. Thenmaximal positive eigenvalue of A is still 2. The growth rate of numbersof Fn is log 2. Furthermore, we have that

ζ(z) = exp( ∞∑n=1

#(Fn)

nzn)

= exp( ∞∑n=1

trac(An)− 1

nzn)

= exp( ∞∑n=1

trac(An)

nzn)· exp

(−∞∑n=1

1

nzn)

=1

det(I − zA)exp

(log(1− z)

)=

1− z1− 2z

.

Definition 11. A dynamical system f : X → X is said to be metri-cally expanding if there are constants 0 < a < 1, C > 0, and λ > 1such that

d(fn(x), fn(y)) ≥ Cλnd(x, y)

for all x, y ∈ X with d(f i(x), f i(y)) ≤ a, 0 ≤ i ≤ n− 1 and n > 0.

Theorem 3. Suppose (X, d) is a compact metric space. Any metricallyexpanding dynamical system f : X → X has a Markov partition.

Proof. Suppose 0 < ε ≤ a/2 is a real number such that f is injectiveon any ball B(x, ε) of radius ε and centered at x. Let δ < 2C(λ −1)ε/λ. Consider a cover of X by balls B1, · · · , Bn of radius δ, thatis, X = ∪nk=1Bk. For f(B1) let B1jk be its minimal covers andS1 = ∪jkBjk . Let B11 be the preimage of S1. Then the diameterof B11 is ≤ δ + δ/(Cλ). Now consider new B2 which is the old B2

minus B11 ∪ f(B11). Consider the minimal cover B2jk of B2. LetS2 = ∪kB2jk . Let B22 be the preimage of S2. Do this for all Bi we getBii with diameter ≤ δ+ δ/(Cλ). Inductive, we can construct Bin withdiameter ≤

∑n−1i=0 δ/(Cλ

i). By taking limit, we have a partition of X

by Xi = Bi∞ such that Xi ∩ Xj = ∅ for 1 ≤ i 6= j ≤ n and f(Xi) isjust a union of some Xj. Since the diameter of xi ≤

∑∞i=0 δ/(Cλ

i) =(δ/C) · λ/(λ − 1) ≤ 2ε ≤ a, we have that f is injective on Xi. Andf(Xi) = ∪mik=1Xik .

17

Remark 2. Thus if a map f : X → X is metrically expanding, thenit is topologically expanding.

18

Lecture 3. Non-linear dynamical systems and dis-tortions

Refer to Chapter 1 of the following book for this lecture.

• Y. Jiang, Renormalization and Geometry in One-Dimensionaland Complex Dynamics. Advanced Series in Nonlinear Dynam-ics, Vol. 10 (1996) World Scientific Publishing Co. Pte. Ltd.,River Edge, NJ, xvi+309 pp. ISBN 981-02-2326-9.

Suppose f : I0 ∪ I1 → I is an expanding map. Let G = (a, b) be thecomplement of I0 ∪ I1 in I. A number x in I is said to be escaping toG if f k(x) is in G for some integer k ≥ 0 (where f 0 is the identity).The set Ω ⊆ I of escaping points is an open subset of the real line. Thecomplement Λ of Ω in I is called the non-escaping set under f . It is acompact (closed and bounded) subset of the real line R.

Example 8 (13-Cantor set). Suppose a = 1/3 and b = 2/3. Define

f(x) =

3x, 0 ≤ x ≤ a;3x− 2, b ≤ x ≤ 1.

Then f is an expanding map for which the non-escaping set Λ under fis the famous 1

3-Cantor set.

One can see that f : Λ → Λ is a Markov partition with the initialpartition

Λ0 = [0, 1/3] ∩ Λ and Λ1 = [2/3, 1] ∩ Λ.

In general we have Λwk for all wk = i0 . . . ik−1ik. The length |Λwk | =1/3k+1. Clearly Λ has zero Lebesgue measure since |Λ| ≤

∑wk|Λwk | ≤

2k+1/3k+1 for all k ≥ 0. Let s > 0 be real number such that∑wk

|Λwk |s = 1.

Then we can see that 2k+13−s(k+1) = 1. This implies that

s =log 2

log 3.

This number is called the Hausdorff dimension of Λ (we will define itlater). The reason we can do all these calculations is that this is alinear dynamical system.

Now we are going to see how to study a typical non-linear dynamicalsystem. Let X be an interval I = [a, b] or the circle S1 = z = e2πiθ.Let f(x) be a self-map of X. The dynamical system generated by fis the semigroup f n∞n=0 of iterations of f . For semplice, we call fis a dynamical system. Linear dynamical systems are among simplest

19

ones to study. For a dynamical, an important problem one would liketo study is to estimate how far is it derived from linear dynamicalsystems? An important analytic characterization of a linear dynamicalsystem f is

log∣∣∣(f n)′(x)

(f n)′(y)

∣∣∣ = 0, x, y ∈ I, n ≥ 0.

Therefore, for a dynamical system f , we call the set of ratioslog∣∣∣(f n)′(x)

(f n)′(y)

∣∣∣the distortion. The distortion of a linear dynamical system containsonly zero. Indeed, estimating the distortion of a dynamical systemis useful and important. In this series of lectures, we will study dy-namics of circle homeomorphisms and dynamics of interval mappings.Through this study we will also learn several important techniques inone-dimensional dynamical system which are widely used to estimatedistortions of one-dimensional dynamical systems.

One technique, going back to Denjoy, estimates the distortion ofa one-dimensional dynamical system without critical points. We cansummarize this technique as follows.

Let f be a function defined on a set U of the real line R. It is saidto be C1 (or C1+α for 0 < α ≤ 1 or C1+bv) if it can be extended to adifferentiable function defined on an open set containing U and if thederivative of the extension is continuous (or is α-Holder continuous oris of bounded variation). Suppose f is a C1 function on a set U ⊂ Rand X = xini=1 and Y = yini=1 are two sequences of points in U .The number

log∣∣∣ n∏i=1

f ′(xi)

f ′(yi)

∣∣∣is called the distortion of f along X and Y .

Lemma 1 (Navie Distortion Lemma). If

κ = infx∈U|f ′(x)| > 0,

then the distortion of f along X and Y can be estimated as∣∣∣log∣∣∣ n∏i=1

f ′(xi)

f ′(yi)

∣∣∣∣∣∣ ≤ 1

κ

n∑i=1

|f ′(xi)− f ′(yi)|.

Moreover if f is of C1+α for some 0 < α ≤ 1, let

ι = supx 6=y∈U

( |f ′(x)− f ′(y)||x− y|α

)<∞,

20

then the distortion of f along X and Y is bounded by (ι/κ)∑n

i=1 |xi −yi|α, that is, ∣∣∣log

∣∣∣ n∏i=1

f ′(xi)

f ′(yi)

∣∣∣∣∣∣ ≤ ι

κ

n∑i=1

|xi − yi|α.

Lemma 2. If f is of C1+bv, then there is a constant C > 0 so thatthe distortion of f along X and Y is bounded by C whenever the openintervals Ii, bounded by xi and yi, for i = 1, . . ., n, are pairwise disjoint,that is, ∣∣∣log

∣∣∣ n∏i=1

f ′(xi)

f ′(yi)

∣∣∣∣∣∣ ≤ C.

These distortion results are not difficult-they are just obtained fromsome direct calculations from Calculus. However, using them wisely ina research problem is a big challenge in the study of one-dimensionaldynamical systems.

Suppose I = [0, 1] is the unit interval and I0 = [0, a] and I1 = [b, 1]are two subintervals where 0 < a < b < 1. A C1 map f defined onI0 ∪ I1 is said to be (degree two, for any degree it is similar) expandingif f |Ii from Ii to I is bijective for i = 0, 1 and if there are constantsC > 0 and λ > 1 so that

|(f n)′(x)| ≥ Cλn

whenever f i(x) are in I0 ∪ I1 for all i = 0, 1, . . ., n− 1.A C1 map f defined on an interval I is said to have Holder continuous

derivative if there are constant 0 < α ≤ 1 and C > 0 such that

|f ′(x)− f ′(y)| ≤ C|x− y|α, ∀x, y ∈ I.More precisely, we call f C1+α. A C1 map f of I is expanding if thereare constants C > 0 and λ > 1 such that

(fn)′(x)| ≥ Cλn, ∀x ∈ I, n > 0.

Remark 3. If a C1 map f : I0 ∪ I1 → I is expanding, then f : Λ→ Λis metrically expanding, thus topologically expanding.

Topologically, a Cantor subset of the real line R is compact, uncount-able, totally disconnected, and perfect. We will see later that Cantorsets or Cantor-like sets will appear again and again in the study ofdynamical systems.

Theorem 4 (Cantor set). Suppose f : I0 ∪ I1 → I is a topologicallyexpansive map. Then the non-escaping set Λ = Λf under f is a Cantorset.

21

Proof. Let fi be the restriction of the function f to Ii, and gi = f−1i :

I → Ii be the inverse of fi for i = 0 or 1. We can consider compositions

gwn = gi0 gi1 · · · ginfor all strings wn = i0i1 . . . in of 0′s and 1′s.

Suppose wn = i0i1 . . . in is a string of 0′s and 1′s. Let Iwn = gwn(I)be the image of I under gwn , and let Gwn = gwn(G) be the set of allthe points escaping to G under g−1

wn . The union ∪wnIwn is the set of allpoints not escaping to G under the iterates f k for k = 0, 1, . . ., n,where wn runs over all the strings of 0′s and 1′s of length n + 1. Theset Iwn is a collection of pairwise disjoint closed intervals and one toone correspondence with the set wn of all the strings of 0′s and 1′sof length n + 1. Hence Λ = ∩∞n=0 ∪wn Iwn , where wn runs over all thestrings of 0′s and 1′s of length n+ 1.

Let us first prove that Λ is uncountable. For a string wn = i0i1 . . . inof 0′s and 1′s and a digit in+1 = 0 or 1, Iwnin+1 ⊆ Iwn since Iin+1 ⊆ I.This implies that

· · · ⊆ Ii0i1...in ⊆ · · · ⊆ Ii0i1 ⊆ Ii0

and thatIw = ∩∞n=0Ii0i1...in

is a non-empty closed subset for any infinite string w = i0i1 . . . of 0′sand 1′s. Hence the set Iw is a collection of pairwise disjoint non-empty closed subsets and is in one to one correspondence with theuncountable set w = i0i1 . . . of all infinite strings of 0′s and 1′s.Hence the set Iw is uncountable. So is the set Λ because Λ = ∪wIwwhere w = i0i1 . . . runs over all infinite strings of 0′s and 1′s.

Since f is topologically expansive, Iw = xw contains only onepoint. The map π(w) = xw from w to Λ is bijective. We use this toprove that Λ is totally disconnected, that is, every (connected) com-ponent Π of Λ contains only one number. Suppose there is a compo-nent Π of Λ which contains two different numbers xw and xw′ wherew = i0i1 . . . inin+1 . . . and w′ = i0i1 . . . ini

′n+1 . . . where in+1 6= i′n+1.

Both xw and xw′ are in Iwn where wn = i0i1 . . . in. The set Iwn isthe union of an open interval Gwn and two closed intervals Iwnin+1 andIwni′n+1

which are on different sides of Gwn . The numbers xw and xw′

are in Iwnin+1 and Iwni′n+1, respectively. Take a point z in Gwn . Then

Π =(

Π ∩ (−∞, z))∪(

Π ∩ (z,∞)).

This contradicts the statement that Π is a component of Λ and provesthat Λ is totally disconnected.

22

Since Λ is closed, the set Λ′ of limit points of Λ is contained inΛ. To prove that Λ is a perfect set, we only need to show that Λ iscontained in Λ′. Let xw be a number in Λ and w = i0i1 . . . inin+1 . . ..Let r(i) = i + 1 (mod2); r(i) is 1 for i = 0 and 0 for i = 1. Takew(n) = i0 . . . in−1r(in)in+1 . . .; w

(n) differs from w at (n+ 1)th position.Then xw(n) 6= xw and both of them are in Ii0...in−1 . Since the length ofIi0...in−1 tends to zero as n goes to infinity, xw(n) tends to xw as n goesto infinity. This says that xw is a limit point of Λ. So Λ is containedin Λ′. Hence Λ is a Cantor set.

Exercise 2. Construct a topologically expansive map f : I0 ∪ I1 → Isuch that Λ has positive Lebesgue measure.

Theorem 5 (C1+α-hyperbolic Cantor set, Lebegue measure). Supposef : I0 ∪ I1 → I is a C1+α expanding map for some 0 < α ≤ 1. Thenthe non-escaping set Λ = Λf under f has zero Lebesgue measure.

Proof. Letm(·) mean the Lebesgue measure and let |J |mean the lengthof an interval. An inequality which can be easily obtained is

m(Λ) ≤∑wn

|Iwn| < C2n+1µn,

where wn runs over all the strings of 0′s and 1′s of length n + 1. Thisinequality is true because Iwn is a cover of Λ and the total number ofthe strings of 0′s and 1′s of length n+ 1 is 2n+1. If µ < 1/2, it is mucheasier to see m(Λ) = 0. However, to prove that the Lebesgue measureof Λ is zero for any 0 < µ < 1, we need help from Lemma 1

Suppose wn = i0 . . . in is a string of 0′s and 1′s of length n + 1.The map fn+1 from Iwn to I is a monotone function and its inverse isgwn . For any two numbers x and y in Iwn , let xi = f (n−i+1)(x) andyi = f (n−i+1)(y) for i = 0, 1, . . ., n + 1. By the mean value theoremand the chain rule, |xi − yi| < Cµi and

∑n+1i=0 |xi − yi|α < C/(1− µα).

According to Lemma 1, the distortion of f along X = xi and Y =yi is bounded by the constant C ′ = (ι/κ)

(C/(1− µα)

), that is,

∣∣∣log∣∣∣(f (n+1)

)′(x)(

f (n+1))′

(y)

∣∣∣∣∣∣ ≤ C ′,

where ι is the Holder constant of f ′ on I0∪I1 and κ = infx∈I0∪I1 |f ′(x)|.This implies that

|Gwn||Iwn|

≥ c = e−C′|G|,

23

since G = f (n+1)(Gwn) and I = f (n+1)(Iwn). Now we have that

|Iwn0|+ |Iwn1| ≤ (1− c)|Iwn|because Iwn = Iwn0 ∪Gwn ∪ Iwn1; moreover,

m(Λ) ≤∑wn+1

|Iwn+1| =∑wn

(|Iwn0|+|Iwn1|) ≤ (1−c)∑wn

|Iwn | ≤ · · · ≤ (1−c)n+1

for all positive integers n. Hence the Lebesgue measure of Λ is zero.

Problem 3. Can you construct a C1 expanding map f : I0 ∪ I1 → Isuch that Λ has positive Lebesgue measure?

24

Lecture 4. The Mandelbrot set and hyperbolic holo-morphic dynamical systems

4.1. Quadratic polynomials.

Let C be the complex plane and C = C ∪ ∞ be the extendedcomplex plan, that is, the Riemann sphere. Suppose f(z) = az2 +bz + d is a quadratic polynomial, a 6= 0. By a linear conjugacy h(z) =(1/a)(z− b/(2a)), we can assume qc(z) = h−1 f h(z) = z2 + c, wherec = d − b2/(4a). The map qc(z) has a unique critical point 0 in thecomplex plane. We consider the critical orbit O(0) = qnc (0)∞n=0. TheMandelbrot set

M = c | O(0) is bounded..

Theorem 6. The set M is a closed bounded simply connected subsetcontained in the disk of radius 2 centered 0 in the complex plane. AndM∩ R = [−2, 1/4].

Proof. If |c| > 2, then |q2c (0)| = |qc(c)| = |c2 + c| ≥ |c|(|c| − 1) > |c|.

Inductively, |qnc (0)| ≥ |c|(|c| − 1)2n−1. This implies that

M⊆ D2 = |c| ≤ 2.

Thus it is bounded. Now consider |c| ≤ 2 and there is an integerm > 0 such that |qmc (0)| = 2 + t, t > 0. Then we have that |qm+1

c (0)| ≥(2 + t)2 − 2 ≥ 2 + 4t. Inductively, we have that |qm+n

c (0)| ≥ 2 + 4nt.This implies qnc (0) → ∞ as n → ∞. So c 6∈ M. This implies that Mis closed. This also implies that C\M has no bounded component (bythe maximum principle) and thus is connected. This implies that Mis simply connected.

Consider qc(z) = z2 + c, we have z2 − z + c = 0. This implies

z =1±√

1− 4c

2

So qc always has two fixed points

p0 =1 +√

1− 4c

2and p1 =

1−√

1− 4c

2.

When c = 1/4, this becomes a one fixed point of multiplier 2. Whenc and x are real and c > 1/4, qc has no real fixed point. qc(x) > x forall x > 0. So qnc (0) → ∞ as n → ∞. When c is real and −2 ≤ c ≤1/4, the largest fixed point p0 ≥ |c| = |qc(0)|. If |qmc (0)| ≤ p0, then|qm+1c (0)| ≤ p2

0 + c = p0. Thus c ∈M. We get M∩R = [−2, 1/4].

25

4.2. Local dynamics of holomorphic maps, easy cases.

Let U be a domain in the complex plane C. Let f : U → C be aholomorphic map. Suppose z0 ∈ U is a fixed point of f , i.e., f(z0) = z0.Then λ = f ′(z0) is called the multiplier of f at z0. The number λ is aconformal invariant, that is, if g : V → C is another holomorphic mapand w0 is a fixed point of g and if h : U → V is a conformal map suchthat w0 = h(z0) and such that g = h f h−1, then g′(w0) = f ′(z0).

We can classify z0 by λ as follows:

(1) Attracting if 0 < |λ| < 1;(2) Repelling if |λ| > 1;(3) Super-attracting if λ = 0;

(4) Rational neutral if λ = e2π pq where p and q are integers;

(5) Irrational neutral if λ = e2πθ where 0 < θ < 1 is an irrationalnumber.

Conjugating by a simple linear transformation w → w− z0, that is, byconsidering g(w) = f(w+ z0)− z0, we will assume that z0 = 0. We use

∆δ = z ∈ C | |z| < δto denote the disk centered at 0 with radius δ > 0. In particular,∆ = ∆1 is the unit disk.

4.2.1. Attracting case.Suppose 0 ∈ U is an attracting fixed point of f . Let λ = f ′(0) with

0 < |λ| < 1. Then f(z) = λz or its Taylor expansion of f at 0 is

f(z) = λz + apzp + · · ·

where ap 6= 0 and p ≥ 2. In the first case f(z) = λz is a linear mapwhose dynamics is easy to understand. In the second case, we willsee that under an appropriate holomorphic change of coordinate, it isagain a linear map.

Let |λ| < ρ < 1. There is a domain 0 ∈ U ′ ⊂ U such that

|f(z)| ≤ ρ|z|, z ∈ U ′.This implies

|fn(z)| ≤ ρn|z|, n > 0, z ∈ U ′.

Theorem 7 (Konigs’ Theorem). There is a conformal map φ definedon ∆δ for some δ > 0 such that φ(0) = 0 and

φ f φ−1(z) = λz, z ∈ φ(∆δ).

The conjugacy φ is unique up to multiplication by a non-zero factor.

26

Proof. Let us first prove the “uniqueness”. Suppose φ1 defined on ∆δ

is another conjugacy such that φ1(0) = 0 and

φ1 f φ−11 (z) = λz, z ∈ φ1(∆δ).

Then

ψ = φ φ−11 : φ1(∆δ)→ φ(∆δ)

is also a conformal map and commutes with λz, i.e.,

ψ(λz) = λψ(z), z ∈ φ1(∆δ).

So

ψ′(z) = ψ′(λnz) = ψ′(0), z ∈ φ1(∆δ).

This implies that ψ = c is a constant function. Thus φ = cφ1.Now let us prove the “existence”. Define

φn(z) = λ−nfn(z), z ∈ U ′.We have

φn f(z) = λ−nfn+1(z) = λφn+1(z).

If φn → φ as n→∞ in some ∆δ, then φ f = λφ.So we need to show that φn∞n=0 is a Cauchy sequence on ∆δ ⊂ U ′

for some δ > 0. First there are constants δ′ > 0 and C ′ > 0 such that

|f(z)− λz| ≤ C ′|z|2, for all |z| < δ′.

This implies

|f(z)| ≤ |λ||z|+ C ′|z|2 ≤ (|λ|+ C ′δ′)|z|.By taking δ′ small, we assume that ρ = |λ|+ C ′δ′ < 1. So

|f(z)| < ρ|z|.And,

|fn(z)| < ρn|z|, n ≥ 0, |z| < δ′.

Again by taking δ′ small, we assume µ = ρ2/|λ| < 1. Let C =C ′(δ′)2/λ. Then

|φn+1(z)− φn(z)| =∣∣∣fn+1(z)

λn+1− fn(z)

λn

∣∣∣ =|fn+1(z)− λfn(z)|

|λn+1|

≤ C ′|fn(z)|2

|λn+1|≤ ρ2nC ′|z|2

|λ|n+1= Cµn, |w| < δ.

So φn converges uniformly to an analytic map φ on ∆δ′ . Since φf(z) =λφ(z) on ∆δ′ , φ is not a constant. Since φ′n(0) = 1, so φ′(0) = 1. Thereis a 0 < δ ≤ δ′ such that φ is conformal on ∆δ.

27

4.2.3. Repelling case.If 0 is a repelling fixed point of f : U → C. Then there is another

domain V ⊂ U such that f |V is injective. Its inverse f−1 on f(V ) is alsoholomorphic and 0 is an attracting fixed point of f−1 : f(V )→ C. Letλ = f ′(0) be the multiplier of f at 0. Then λ−1 is the multiplier of f−1

at 0. Following the theorem in the attracting case, there is a conformalmap φ defined on ∆δ ⊂ f(V ) such that φ f−1 φ−1(z) = λ−1z. So

φ f φ−1(z) = λz, z ∈ φ(f−1(∆δ)).

Since φ(f−1(∆δ)) is an open domain containing 0, we have

Theorem 8. There is a conformal map φ defined on ∆δ0 for someδ0 > 0 such that φ(0) = 0 and

φ f φ−1(z) = λz, z ∈ φ(∆δ0).

The conjugacy φ is unique up to multiplication by a non-zero factor.

4.2.4. Super-Attracting case.Suppose 0 is an isolated super-attracting fixed point of f : U → C.

Then f(z) = apzp or the Taylor expansion of f at 0 is

f(z) = apzp + aqz

q + · · · , ap, aq 6= 0, q > p ≥ 2.

By changing the coordinate ξ = φ0(z) = bz with bp−1 = ap, we have

φ0 f φ−10 (ξ) = ξp or φ0 f φ−1

0 (ξ) = ξp + aqb−q+1ξq + · · · .

So we assume

f(z) = zp + aqzq + · · · , aq 6= 0, q > p > 1.

Theorem 9 (Bottcher’s Theorem). There is a conformal map φ de-fined on ∆δ ⊂ U such that

φ f φ−1(z) = zp, w ∈ φ(∆δ).

The conjugacy φ is unique up to multiplication by a (p − 1)th-root ofthe unity.

Proof. Let us first prove the “existence”. Consider

f(z) = zp(1 +O(z))

Then

fn(z) = f(fn−1(z)) = (fn−1(z))p(1 +O(fn−1(z))) = . . .

= zpn

(1 +O(z))pn−1 · · · (1 +O(fn−2(z)))p(1 +O(fn−1(z)))

There are a 0 < ρ < 1 and ∆δ for some δ > 0 such that

|f(z)| ≤ ρ|z|, z ∈ ∆δ.

28

This implies|fn(z)| ≤ ρn|z|, n ≥ 1, |z| < δ.

So fn(z) tends to 0 as n goes to infinity. The image of ∆δ underevery (1 + O(fk(z))) avoids 0. Thus we has a well-defined branch

(1 +O(fk(z)))p−k−1

on ∆δ for all k > 0. Let

gn(z) = (1 +O(z))p−1 · · · (1 +O(fn−2(z)))p

−n+1

(1 +O(fn−1(z)))p−n

be a fixed branch andφn(z) = zgn(z).

Then φn is holomorphic and

φn(f(z)) = (φn+1(z))p, z ∈ ∆δ′ .

If φn tends φ uniformly as n goes to infinity, then φ f = φp. To showφn → φ uniformly, we note that

φn+1(z) =(φ1(fn(z))

)p−n=(fn(z)(1 +O(|fn(z)|))

)p−n.

So there are constants C ′, C ′′ > 0 such that∣∣∣φn+1(z)

φn(z)

∣∣∣ ≤ (1+C ′|fn(z)|)p−n ≤ (1+C ′ρn|z|

)p−n ≤ 1+C ′′ρnp−n|z|, |z| ≤ δ′.

Thus we have another constant C > 0 such that∣∣∣φn+1(z)

φn(z)

∣∣∣ ≤ 1 + Cp−n, |z| ≤ δ′.

Therefore,

φn(z) = φ1(z)∞∏n=1

φn+1(z)

φn(z)→ φ(z) n→∞

uniformly on ∆δ′ . Since φn(0) = 0 and |φ′n(0)| = 1, we have φ(0) = 0and |φ′(0)| = 1. So φ is conformal on ∆δ for some 0 < δ ≤ δ′.

Now let us see the “uniqueness”. If φ1 and φ2 are both conjugacies,i.e.,

φ1 f = φp1 and φ2 f = φp2,

thenφ1φ

−12 (zp) =

(φ1φ

−12 (z)

)p.

Consider the Taylor expansion

φ1φ−12 (z) = az + a2z

2 + · · · .We have

azp + a2z2p + · · · = (az)p + · · · .

This implies that a1 = ap1 and ai = 0 for all i ≥ 2. So φ1 = a1φ2 withap−1

1 = 1.

29

4.3. Basins of ∞ and Julia sets.Given a quadratic polynomial qc(z) = z2 + c. Then ∞ is a super-

attractive fixed point. According to Theorem 9, there is a conformalmap φc from the neighborhood U0 of ∞ to a neighborhood V0 of ∞such that

φc qc φ−1c (z) = z2 = q0(z), z ∈ V0.

Let V1 = q−10 (V0) and, inductively, let Vn = q−1

0 (Vn−1) for n > 0. Thenit is clear

C \∆ = ∪∞n=0Vn.

Now let U1 = q−1c (U0) and, inductively, let Un = q−1

c (Un−1) for n > 0.Define

Ac(∞) = ∪∞n=0Un and Kc = C \ Ac(∞).

Definition 12. The domain Ac(∞) is called the basin of ∞ for qc.Then set Kc is called the filled Julia set of qc. The boundary

Jc = ∂Kc = ∂Ac(∞)

is called the Julia set of qc.

If 0 6∈ Un, then Un is simply connected and we can use

φc(z) = q−10 (φc(qc(z))),

to extend φc to Un. That is, using the diagram

Unφc−→ Vn

qc ↓ q0 ↓Un−1

φc−→ Vn−1

to lift φc on Un−1 to Un since in this case qc : Un → Un−1 and q0 : Vn →Vn−1 are both covering maps. Then we need to discuss the followingtwo cases:

(1) c ∈M and(2) c 6∈ M.

In the first case, 0 6∈ Un for all n ≥ 0. Then Ac(∞) = ∪∞k=0Un is asimply connected domain and we have a conformal map

φc(z) : Ac(∞)→ C \∆

such that φc(∞) =∞ and φ′c(∞) = 1 and such that

φc qc φ−1c (z) = z2.

Recall the Riemann mapping theorem in complex analysis. A domainΩ is an open and (path) connected set in C. A domain Ω is simplyconnected if every closed curve γ in Ω can be deformed in Ω to a point.

30

Theorem 10 (Riemann Mapping Theorem). There is a conformalmapping φ from Ω onto the unit disk ∆ for every simply connecteddomain Ω having at least two boundary points.

The conformal mapping φc we constructed above is a Riemann map-ping.

Definition 13. The analytic curve

Sr = φ−1c (c = re2πiθ ∈ C | 0 ≤ θ < 1), r > 1,

is called the equipotential curve for qc and the analytic ray

Eθ = φ−1c (c = te2πiθ ∈ C | t > 1), 0 ≤ θ < 1,

is called the external ray for qc.

We have the following property

Proposition 2.

qc(Sr) = Sr2 , and qc(Eθ) = E2θ (mod 1).

Now the problem is that

Problem 4. Can the inverse φ−1c : ∆ → Ac(∞) of the Riemann

mapping φc be extended to the boundary Jc continuously?

This problem is related with the topological property of the set Jc.In the case c ∈M, Kc is connected and thus Jc is also connected.

Definition 14. A connected set S ∈ C is called locally connected iffor every point p ∈ S and any neighborhood V about p, there is aneighborhood p ∈ U ⊂ V such that S ∩ U is connected.

Example 9. The set

S1 = x sin(1/x) | 0 < x ≤ 1/2πis locally connected. The set

S2 = sin(1/x) | 0 < x ≤ 1/2πis not locally connected.

Theorem 11 (Caratheodory’s Theorem). Suppose Ω is a simply con-nected domain having at least two boundary points. The inverse φ−1 :∆ → Ω of the Riemann mapping φ : Ω → ∆ can be extended to acontinuous map φ : ∆→ Ω if and only if S = ∂Ω is locally connected.

Thus Problem 4 is equivalent to the following problem.

31

Problem 5. For which c ∈M, the Julia set Jc is locally connected.

We say an external ray Eθ(t) lands at Jc if limt→1+ Eθ(t) ∈ Jc exists.Thus if Jc is locally connected, then every external ray lands at Jc.

Definition 15. A quadratic polynomial qc(z) = z2 + c is called hyper-bolic if there is a neighborhood U of Jc and two constants C > 0 andλ > 1 such that

|(qnc )′(z)| ≥ Cλn

for any z ∈ U such that qkc (z) ∈ U , 1 ≤ k < n − 1. This is equivalentto say that there is a constant 0 < a ≤ 1 such that for any z, w ∈ Uwith |qkc (z)− qkc (w)| ≤ a for all 0 ≤ k ≤ n− 1, then

|qnc (z)− qnc (w)| ≥ Cλn|z − w|.

Example 10. If |c| is small, then qc is hyperbolic and Jc is a Jordancircle.

More general, we define the main cardioid

M0 = c ∈M | qc has an attractive fixed point.For any c ∈M0, qc is hyperbolic and Jc is a Jordan circle.

Theorem 12. Suppose qc is hyperbolic. Then φ−1c : ∆ → Ac(∞)

can be extended to a continuous map φc : ∆ → Ac(∞) ∪ Jc such thatφc(S

1) = Jc. Thus Jc is locally connected.

Proof. Without lost of generality, we assume C = 1. Otherwise, weconsider an integer k > 1 such that k

√Cλ > 1 and consider qkc instead

of qc. Fix a real number R > 1. Consider the equipotential curveparametrized as γn(θ) = φ−1

c (R2−ne2πiθ). Then qc(γn(θ)) = γn−1(2θ).This implies that

|γn(θ)− γn+1(θ)| ≤ λ−1|γn−1(2θ)− γn(2θ)|.Thus γθ converges uniformly to a continuous function γ(θ) from S1 toJc. The function γ extends φ−1

c continuously to the boundary of theopen unit disk.

Later, after we learn the global theory, we will give a characterizationof hyperbolic quadratic polynomial as that

Theorem 13. A quadratic polynomial is hyperbolic if it has an attrac-tive periodic point.

The above proof can be applied to a slightly more general quadraticpolynomial called a sub-hyperbolic quadratic polynomial.

32

Definition 16. A quadratic polynomial qc(z) = z2 + c is called sub-hyperbolic if its critical orbit qnc (0)∞n=0 is finite but is not a periodicorbit.

In the sub-hyperbolic case, 0 is not periodic but some pk = qkc (0),k > 0, is periodic, that is, qk+m

c (0) = qkc (0). Let pi = qic(0). Then thecritical orbit is

0→ p1 → · · · → pk → pk+1 → · · · → pk+m−1 → pk.

In this case all pi ∈ Jc. Let U be a neighborhood of Jc. Let ρ(z) bea function smooth at all point in U but pi’s. At a small neighborhoodabout pi, let ρ(z) = 1/

√|z − pi|. Then ρ(z)|dz| is a new metric on U

equivalent to |dz|. The expanding condition as defined in Definition 15holds for this metric. Replace the Lebesgue metric | · | by ρ(z)|dz| inthe proof of Theorem 12, we have

Theorem 14. Suppose qc is sub-hyperbolic. Then φ−1c : ∆ → Ac(∞)

can be extended to a continuous map φc : ∆ → Ac(∞) ∪ Jc such thatφc(S

1) = Jc. Thus Jc is locally connected.

Example 11. Let q−2(z) = z2−2. Then p1 = −2 and p2 = q−2(p1) = 2which is fixed by q−2. Thus q−2 is sub-hyperbolic. The Julia set J−2 =[−2, 2] which is clearly locally connected. Then ρ(z) = 1/

√|z2 − 4|.

One can check that the absolute value of the derivative of q−2 withrespect to the new metric ρ(z)|dz| is 2, that is

|Dρ(q−2)(z)| =ρ(q−2(z))q′−2(z)

ρ(z)= 2.

The study of Problem 5 (equivalently, Problem 4) can be dividedinto the following cases according to c ∈M:

(1) hyperbolic point;(2) sub-hyperbolic point;(3) rational neutral point;(4) irrational neutral Cremer point(5) irrational neutral Siegel point;(6) finitely renormalizable point;(7) infinitely renormalizable point.

A lot of research has been done. However, for 7) there are still somepoints which have not too good and not too bad combinatorics whichare not fully understood. Refer to

• J. Milnor, Dynamics in One Complex Variable: IntroductoryLectures. Third Edition. Annals of Mathematics Studies.

33

• J. H. Hubbard, Local connectivity of Julia sets and bifurcationloci: three theorems of J.-C. Yoccoz. Topological Methods inModern Mathematics, A Symposium in Honor of John Milnor’sSixtieth Birthday (1993).• Y. Jiang, Renormalization and Geometry in One-Dimensional

and Complex Dynamics, Advanced Series in Nonlinear Dynam-ics 10, World Sci. Publ., 1996.• Y. Jiang, Infinitely renormalizable quadratic polynomials. Trans.

Amer. Math. Soc., 352(2000), no. 11, 5077-5091.• M. Lyubich, Dynamics of quadratic polynomials, I-II. Volume

178, Issue 2 , pp 185-297.• G Levin and S Van Strien, Local connectivity of the Julia set

of real polynomials. Annals of mathematics, 1998, 471-541.

4.4. Structures of the complement of the Mandelbrot set.Now let us consider c 6∈ M. From the proof of Theorem 9, we can

write φc as

φc(z) = limn→∞

qc(z)2−n = z∞∏n=1

( qnc (z)

(qn−1c (z))2

)2−n

= z∞∏n=0

(1+

c

(qnc (z))2

)2−n−1

.

The functions φc(z) is holomorphic on both variables z and c and isdefined up to the domain Un such that c ∈ Un. Thus Φ(c) = φc(c) isdefined for c ∈ C \M. Since

Φ(c) = c∞∏n=0

(1 +

c

(qnc (c))2

)2−n−1

,

it has a simple pole at ∞ and Φ′(∞) = 1. Douady and Hubbard,independently, Sibony, proved that Φ : C \M → C \∆ is a conformalmapping (by using some properties of Green’s function G(c) = |Φ(c)|which can be extended to the whole C as a harmonic function suchthat G(c) = 0 for c ∈M which we will study later). Refer to

• L. Carleson and T. Gamelin, Complex Dynamics, Springer 1993.• A. Douady and J. Hubbard, Iteration des polynomes quadra-

tiques complexes, C.R. Acad. Sc. Paris, t. 294 (18 janvier1982).

This implies that

Theorem 15. The Mandelbrot set M is connected.

Another way to prove the above theorem topologically is to consider

Φn(c) = qn+1c (0).

34

It is a polynomial of c of degree 2n. Remember that we use D2 = c ∈C | |c| ≤ 2 to denote the closed disk of radius 2 centered at 0. Define

Mn = Φ−1n (D2).

Then from the proof of Theorem 6, we knew that

M = ∩∞n=1Mn.

Kahn showed that Mn is connected for all n > 0. ThusM is connected.Refer to

• J. Kahn, The Mandelbrot set is Connected: A topological proof.Preprint.

The analytic curve

SR = Φ−1(c = Re2πiθ ∈ C | 0 ≤ θ < 1), R > 1,

is called the equipotential curve of M and the analytic ray

Eθ = Φ−1(c = te2πiθ ∈ C | t > 1), 0 ≤ θ < 1,

is called the external ray ofM. We say an external ray Eθ(t) lands atM if limt→1+ Eθ(t) ∈M exists.

Conjecture 1. Every external ray Eθ(t) lands.

Moreover, we have that

Conjecture 2. The inverse

Φ−1(c) : C \∆→ C \Mof the Riemann mapping can be extended to the circle S1 continuously.

Equivalently, we have

Conjecture 3 (MLC conjecture). The Mandelbrot set ∂M is locallyconnected.

Another conjecture as a conclusion of the MLC conjecture is thefollowing famous hyperbolicity conjecture. Suppose

H = c ∈M | qc is hyperbolic = c ∈M | qc has an attractive periodic pointIt is clear that H is an open set in C.

Conjecture 4 (Hyperbolicity Conjecture). The set H is dense in M.

The study of the MLC conjecture was divided into the followingcases:

(1) rational neutral points;

35

(2) irrational neutral points;(3) finitely renormalizable points;(4) infinitely renormalizable points.

A lot of research has been done in this direction but there are still someremaining unknown points which are infinitely renormalizable points.Refer to

• J. H. Hubbard, Local connectivity of Julia sets and bifurcationloci: three theorems of J.-C. Yoccoz. Topological Methods inModern Mathematics, A Symposium in Honor of John Milnor’sSixtieth Birthday (1993).• Y. Jiang, Y. Jiang, Renormalization and Geometry in One-

Dimensional and Complex Dynamics, Advanced Series in Non-linear Dynamics 10, World Sci. Publ., 1996.• Y. Jiang, Local connectivity of the Mandelbrot set at certain in-

finitely renormalizable points. Complex Dynamics and RelatedTopics, New Studies in Advanced Mathematics, 2004, The In-ternational Press, 236-264.• J. Kahn and M. Lyubich, A priori bounds for some infinitely

renormalizable quadratics, I. Bounded primitive combinatorics,ArXiv Preprint; II. Decorations. Annales Scientifiques de lEcoleNormale Superieure; III. Molecules. Complex Dynamics, theproceedings of the conference in honor of J. H. Hubbard’s 60thbirthday.

4.6. Julia sets for c 6∈ M.For c 6∈ M, let

SR(c) = φ−1c (R(c)e2πiθ | 0 ≤ θ < 1)

be the equipotential curve passing c. Let

Eθ(c) = φ−1c (te2πiθ(c) | t ≥ R(c))

be the external ray landing at c. The preimage of SR(c) under qc is a8-shape analytic curve

S√R(c)

= φ−1c (

√R(c)e2πiθ | 0 ≤ θ < 1)

intersecting at 0. Two external rays

Eθ(c)/2 = φ−1c (teπiθ(c) | t ≥

√R(c))

and

E(θ(c)+1)/2 = φ−1c (teπi(θ(c)+1) | t ≥

√R(c))

land at the intersection point 0.

36

Now consider another equipotential curve SR(c)−ε for ε > 0. Let V bethe open domain bounded by SR(c)−ε. Let U = q−1

c (V ). We have a small

ε > 0 such that U ⊂ V . The set U has two disjoint simply connecteddomains V0 and V1 such that qc|V0 and qc|V1 are one-to-one and ontoV . Let J = Jc be the non-escaping set for qc : U = V0 ∪ V1 → V . TheJ = Jc = ∂Kc is the non-escaping set and the Julia set.

The map qc : J → J is a Markovian dynamical system with theinitial partition η0 = J0, J1, where J0 = J ∩ V0 and J1 = J ∩ V1,such that f(J0) = J and f(J1) = J . Thus we have a homeomorphismπ = πc : Σ+ =

∏∞n=00, 1 → J such that qc π = π σ.

Before a further study of this dynamical system, we study someproperties in complex analysis and in hyperbolic geometry. Let ∆ =z ∈ C | |z| < 1 be the open unit disk. Let ∆r = z ∈ C | |z| < rfor 0 < r < 1.

Lemma 3 (Schwarz Lemma). For all |z| < 1, suppose that f(z) =a1z+a2z

2+· · · converges, and that |f(z)| ≤ 1. Then for all 0 < |z| < 1,

|f(z)| ≤ |z| and |f ′(0)| = |a1| ≤ 1

with equalities unobtainable except if f(z) = eiαz for some real numberα.

Proof. Define the function

g(z) =f(z)

z= a1 + a2z + . . . .

By the maximum modulus principle, g(z) cannot achieve a maximumof |g(z)| in ∆r. Thus

|g(z)| ≤ 1

rfor z in ∆r. This holds for any z ∈ ∆ and any |z| < r < 1. Therefore,

|g(z)| ≤ 1,

that is,|f(z)| ≤ |z| and |a1| = |g(0)| ≤ 1

for |z| < 1.If the equality sign holds for some 0 < |z0| < 1 (or if |a1| = 1),

then |f(z0)| = |z0| (or |g(0)| = 1). Unless g is a constant function, gmust map a neighborhood about z0 (or 0) onto a neighborhood aboutg(z0) (or g(0)) since an analytic function maps an open set to an openset. Since |g(z0)| = 1 (or |g(0)| = 1), this would imply that there arepoints z arbitrarily close to z0 (or 0) for which |g(z)| > 1, which wouldcontradict the first part of this argument. Hence, if the equality sign

37

holds anywhere in 0 < |z| < 1 (or if |a1| = 1), then g must be a constantfunction. In this case, |g(z)| = 1 everywhere, whence f(z) = eiαz on∆ for some real number α.

An analytic function f from ∆ into C is called a schlicht function (ora univalent function or a conformal mapping) if it is one-to-one. Thisis equivalent to saying f ′(z) 6= 0 for all z in ∆.

Lemma 4. Every conformal map f from ∆ onto itself is a Mobiustransformation

f(z) = az + b

1 + bzwhere a and b are complex numbers with |a| = 1 and |b| < 1.

Proof. Suppose f(0) = 0; by applying Lemma 3 to f and to f−1, weget

|f(z)| ≤ |z| and |z| ≤ |f(z)|.Thus,

f(z) = az

with |a| = 1. In general, consider

g = γ f,where

γ(z) =z − f(0)

1− f(0)z.

Then g is a schlicht function from ∆ onto ∆ such that g(0) = 0. Thusg(z) = az with |a| = 1, whence

f(z) = γ−1(az) =az + f(0)

1 + af(0)z= a

z + b

1 + bz

where b = af(0).

Definition 17. The hyperbolic disk, which is still denoted as ∆, is theopen unit disk ∆ equipped with the metric

dHs = ρ(z)|dz| = |dz|1− |z|2

.

The corresponding distance dH is

dH(z1, z2) = infl

∫l

dHs = infl

∫l

|dz|1− |z|2

where l are over all C1 curves in ∆ connecting z1 and z2.

38

Remark 4. The metric dHs = ρ(z)|dz| is called the Poincare metric(or hyperbolic metric). The distance dH is called the (induced) hyper-bolic distance.

Theorem 16. Suppose f is an analytic function from ∆ into ∆. Thenfor all z1 and z2 in ∆,

dH(f(z1), f(z2)

)≤ dH(z1, z2)

with equality unobtainable except if

f(z) = az + b

1 + bz

for complex numbers a and b with |a| = 1 and |b| < 1.

Proof. For a fixed z in ∆, let

g(ξ) =f( ξ+z

1+zξ)− f(z)

1− f(z)f( ξ+z1+zξ

).

Then g(0) = 0 and |g(ξ)| ≤ 1. Thus we have |g′(0)| ≤ 1, whence

|f ′(z)|1− |f(z)|2

≤ 1

1− |z|2.

The equality sign holds if and only if g(ξ) = cξ with |c| = 1, which isequivalent to saying f takes a Mobius transformation. Take two pointsz1 and z2 in ∆. For any C1 curve l in ∆ connecting z1 and z2,

dH(f(z1), f(z2)) ≤∫f(l)

|dz|1− |z|2

=

∫l

|f ′(z)|1− |f(z)|2

|dz| ≤∫l

|dz|1− |z|2

.

Therefore,

dH(f(z1), f(z2)

)≤ dH(z1, z2).

The equality sign holds for some z1 and z2 if and only if f is a Mobiustransformation.

Suppose Ω is a simply connected domain. Let g : Ω → ∆ be theRiemann map which maps a basepoint z0 ∈ Ω to 0. The hyperbolicmetric dH,Ωs on Ω is

dH,Ωs = ρ(g(z))|g′(z)||dz|.All results (with appropriate constants) in this section applies to an an-alytic function defined on a simply connected domain Ω whose bound-ary ∂Ω contains at least two points. In particular, if

Ω = H = z = x+ yi ∈ C | y > 0

39

is the upper-half plane, then the hyperbolic metric is

dH,Hs =|dz||z − z|

=|dz|2y

.

Any analytic function f : H→ H contracts this hyperbolic metric.For the disk ∆r, the hyperbolic metric

dH,∆rs =r|dz|r2 − z2

and the hyperbolic distance is dH,∆r(·, ·).Now for any c 6∈ M, let SR(c) be the equipotential curve passing c.

Let D(c) be the domain bounded by SR(c). Let V be a domain boundedby an equipotential curve SR(c)−ε for some small ε > 0. Consider g0

and g1 are two inverse branches of q−1c on D(c). Then we have

g0 : D(c)→ g0(D(c)) ⊂ D(c)

andg1 : D(c)→ g1(D(c)) ⊂ D(c).

It is clear that both g0 and g1 are not Mobius transformations. Thusg0 and g1 contract the hyperbolic distance dH,D(c)(·, ·), that is,

dH,D(c)(gi(z), gi(w)) < dH,D(c)(z, w), z, w ∈ D(c)

for i = 0, 1. Since V is a compact subset of D(c), we have a constant0 < τ < 1 such that

dH,D(c)(gi(z), gi(w)) ≤ τdH,D(c)(z, w), z, w ∈ V.for i = 0, 1. Thus we have that

dH,D(c)(gwn(z), gwn(w)) ≤ τn+1dH,D(c)(z, w), z, w ∈ V.where wn = i0 . . . in−1in and gwn = gi0 · . . . ·gin−1 ·gin . On V , dH,D(c)(·, ·)is equivalent to the Lebesgue metric, that is, there is a constant C > 0such that

C−1|z − w| ≤ dH,D(c)(z, w) ≤ C|z − w|, ∀z, w ∈ V.This implies that

|g′wn(z)| ≤ Cτn+1, ∀z ∈ V, ∀wn, ∀n ≥ 0,

for some positive constant which we still denote as C. Equivalently, wesay that qc : J → J is a hyperbolic dynamical system, that is, thereare constants C > 0 and λ > 1 such that

|(qn+1c )′(z)| ≥ Cλn+1, ∀z ∈ J, n ≥ 0.

Similar to the proof of Theorem 4

40

Theorem 17. For any c 6∈ M, qc is hyperbolic and its Julia set Jc isa Cantor set.

Next we will study the Lebesgue measure for the hyperbolic dynam-ical system qc : J → J for c 6∈ M. This study works for any hyperbolicrational map which we will introduce later. For this study, we needsome results in complex analysis which has been obtained in 60’s, referto Chapter V of

• G.M.Goluzin, Geometric Theory of Functions of a ComplexVariable, Translation of Mathematical Monographs, Vol 26,AMS, 1969.

A domain D in the Riemann sphere C is called double connected if

its complement C \D consists of two disjoint connected sets.

Theorem 18. Suppose D is a double connected domain. Then thereis a conformal map f from D to one of the following domains:

(1) C∗ = C \ 0;(2) ∆∗ = ∆ \ 0;(3) AR = z ∈ C | 1 < |z| < R.

And no one of the above three domains is conformally equivalent toanother one.

Proof. If C \D contains only two points a, b, then we have a Mobiustransformation f which maps a to 0 and b to ∞. It is the first case.

If one component of C \ D contains one point a and the other Econtains at least two points. By the Rimeann mapping theorem, there

is a conformal map f1 : C \ E → ∆. Then we have a Mobius trans-formation f2 : ∆ → ∆ such that f2(f1(a)) = 0. Then f = f2 f1 is a

conformal map from C \D to ∆∗. This is the second case.

Now let us prove the third case: both components E1 and E2 of C\Dcontain more than two points. By the Rimeann mapping theorem,

there is a conformal map f1 : C \E2 → ∆. Again, we have a conformal

map f2 : C \ f1(E1)→ C \∆. Then f2(S1) is an analytic closed curve

γ(t) in C \∆. The unit circle and the closed curve bounded a doublyconnected domain D′.

The analytic map f3(z) = ez maps a region D′ bounded by theimaginary axis and an analytic curve γ(t) in the right-plane into D′,where γ(t) → −i∞ as t → −∞ and γ(t) → i∞ as t → ∞. It isa universal covering. Using the Riemann mapping theorem again, wehave a conformal mapping f4 from D′ to the strip Sd = z = x+iy | 0 <y < d such that it can be extended to the boundary ∂D′ with f4(0) =

41

0, f4(−i∞) = −i∞ and f4(i∞) = i∞. By picking an appropriate d,we can have f4(2πi) = 2πi. The conformal map f4(z+ 2πi)− 2πi fromD′ to Sd also fixes 0, −i∞,and i∞. By the uniqueness in the Riemannmapping theorem, we have f(z + 2πi) = f(z) + 2πi.

Now f3 maps the strip Sd to the round annulus AR for R = log d.Now f = f3f4f−1

3 f2f1 is a conformal map from D onto AR, wheref−1

3 = log z is the inverse of f3 : D′ ∩ z = x+ iy | 0 ≤ y < 2π → D′.It is clear that three of domains are not conformally equivalent each

other.

Remark 5. By using the Schwarz reflection theorem, AR is unique fora given D in the third case.

Definition 18. In the third case, m(D) = (1/2π) logR is called themodulus of D. In the first case and the second case, m(D) =∞.

Remark 6. The modulus is a conformal invariant, that is, there is aconformal map f : D1 → D2 between two doubly connected domainsD1 and D2 if and only if m(D1) = m(D2).

Remark 7. The proof of Theorem 18 is only for the doubly connectedplanar domain where only the Riemann mapping theorem is used inthe proof. It actually holds for any double connected Riemann sur-face. A Riemann surface is double connected if its fundamental groupis Z. However, the known proof of Theorem 18 for any Riemann sur-face using the uniformization theorem which says that any simply con-nected Riemann surface is conformally equivalent to either the Riemannsphere C, or the complex plane C, or the open unit disk ∆. Refer to

• W. Abikoff, The Uniformization Theorem. The American Math-ematical Monthly, Vol. 88, No. 8 (Oct., 1981), pp. 574-592.

Given a round annulus AR. Suppose f is an analytic map on AR.Then we have the Laurent expansion

f(z) =∞∑

n=−∞

cnzn.

Let Γr = z ∈ C | 1 < |z| = r < R be the circle in AR. Then f(Γr)is a closed curve in C. Let Ωr be the region bounded by f(Γr) andζ = f(z). Then from Green’s theorem,

Area(Ωr) =1

2i

∮∂Ωr

ζdζ =1

2i

∮|z|=r

f(z)f ′(z)dz

42

=1

2i

∮|z|=r

(∞∑

n=−∞

cnzn)(

∞∑n=−∞

ncnzn−1)dz

=1

2

∫ 2π

0

∞∑n,k=−∞

(ncnckrn+ke2π(n−k)iθ)dθ = π

∞∑n=−∞

n|cn|2r2n.

Now suppose Ω1 and Ω2 are two bounded simply connected domains.Suppose Ω1 ⊂ Ω2. Then D = Ω2 \ Ω1 is a doubly connected domain.Let m(D) = (1/(2π)) logR be the modulus of D.

Lemma 5.

R2 ≤ Area(Ω2)

Area(Ω1).

The equality holds if and only if

Ω1 = z ∈ C | |z| < R1+ c0 and Ω2 = z ∈ C | |z| < R2+ c0

for R1 < R2 and R = R2/R1.

Proof. Let f : AR → D be an conformal map from AR onto D. Let1 < r < s < R be two real numbers. Let

Γr = z ∈ C | |z| = r and Γs = z ∈ C | |z| = s

be circles in AR. Then f(Γr) and f(Γs) are closed curves in D. Let Ωr

be the region bounded by f(Γr) and let Ωs be the region bounded byf(Γs) Then

Area(Ωs)

Area(Ωr)− s2

r2=

∑∞n=−∞ n|cn|2s2n∑∞n=−∞ n|cn|2r2n

− s2

r2

=r2∑∞

n=−∞ n|cn|2s2n − s2∑∞

n=−∞ n|cn|2r2n

r2∑∞

n=−∞ n|cn|2r2n

=

∑∞n=−∞ n|cn|2r2s2(s2(n−1) − r2(n−1))

r2∑∞

n=−∞ n|cn|2r2n> 0.

Let r → 1 and s → R. We have that Area(Ωs) → Area(Ω2) andArea(Ωr)→ Area(Ω1). Then

Area(Ω2)

Area(Ω1)−R2 ≥ 0.

This proves the inequality.The equality holds for all n 6= 0, 1, cn = 0. That is f(z) = c1z +

c0.

43

We study a more general case than the quadratic case, that is, d = 2in the following notation.

Suppose Ω is a bounded simply connected region. Suppose Ω0, . . .,Ωd−1 are bounded simply connected regions such that

Ωi ∩ Ωj = ∅, ∀ 0 ≤ i 6= j ≤ d− 1

and

∪d−1i=0 Ωi ⊂ Ω.

Let f : ∪d−1i=0 Ωi → Ω be a holomorphic map, that means that f is

holomorphic on a neighborhood of ∪d−1i=0 Ωi. Suppose f |Ωi : Ωi → Ω is

conformal for every 0 ≤ i ≤ d− 1. Let

J = ∩nn=0f−n(Ω)

be the non-escaping set of f (we also call it the Julia set later). Thenfrom the hyperbolic geometry we know that f : J → J is an expandingMarkovian map with the initial partition

η0 = J0, . . . , Jd−1

where Ji = J ∩ Ωi. Then we have a homeomorphism

π = πf : Σd =∞∏n=0

0, . . . , d− 1 → J.

Thus J is a Cantor set (by using the same proof as the proof of Theo-rem 4).

Theorem 19. The Lebesgue measure Area(J) = 0.

Proof. Let B ⊂ Ω be a simply connected region such that

∪d−1i=0 Ωi ⊂ B and B ⊂ Ω.

Then G = Ω \B is a doubly connected domain. Let

m(G) =1

2πlogR

be the modulus where R > 1. Let

gi = (f |Ωi)−1 : Ω→ Ωi, 0 ≤ i ≤ d− 1.

For any wn = i0 . . . in−1 of a sequence of 0, . . . , d− 1,let

gwn = gi0 . . . gin−1 : Ω→ Ωwn = gwn(Ω).

Then we have

Bwn = gwn(B) ⊂ Ωwn and ∪d−1i=0 Ωwni ⊂ Bwn = gwn(B).

44

The region Gwn = Ωwn \ Bwn is a double connected domain with thesame modulus of G, that is,

m(Gwn) = m(G) =1

2πlogR.

From Lemma 5,

Area(J) ≤d−1∑i=0

Area(Ωwni) ≤ Area(Bwn) ≤ 1

R2Area(Ωwn) =

1

R2Area(Ωwn).

Inductively, we have that

Area(J) ≤ 1

R2(n+1)Area(Ω).

Thus Area(J) = 0.

More general, suppose U = ∪nk=1Uk is a union of pairwise disjointconnected regions. Suppose f on U is a holomporphic map such thatV = f(U) ⊃ U . Then we can define the non-escaping set K =z | fn(z) ∈ U, ∀n > 0. We say that f : K → K is expandingif there are two constants C > 0 and λ > 1 such that

(fn)′(z) ≥ Cλn, ∀n > 0

Following a similar proof of Theorem 19, we have that

Corollary 1. Suppose f : K → K is expanding. Then it is anMarkovian dynamical system and J = ∂K = K and Area(J) = 0.

For example, for c ∈ M0, the main cardioid of the Mandelbrot set.The Julia set J = Jc of qc is a Jordan circle. Then we have a doublyconnected domain U ⊃ J such that V = qc(U) ⊃ U . We have ahomeomorphism h : S1 → Jc such that qc h = h q0. Let

J0 = h(e2πiθ | 0 ≤ θ ≤ 1/2) and J1 = h(e2πiθ | 1/2 ≤ θ ≤ 1).Then

η0 = J0, J1is a Markov partition for qc : J → J . Let

η1 = f−1(η0) = J00, J01, J10, J11be the second Markov partition. Let Ui1i0 be a simply connected do-main containing Ji1i0 . Then Vi0 = qc(Ui1i0) ⊃ U i1i0 and qc : Ui1i0 → Vi0is conformal. Let Bi0 ⊃ Ji1i0 be a simply connected domain such thatBi0 ⊂ Vi0 . Then Di0 = Vi0 \ Bi0 is a doubly connected domain. Thenusing the exact the same argument as that of Theorem 19, we havethat Area(J) = 0.

45

In general, we can consider a polynomial of degree d ≥ 2,

P (z) = adzd + ad−1z

d−1 + · · ·+ a1z + a0, ad 6= 0.

Since

P (z) = adzd(

1 +ad−1

adz+ · · ·+ a1

a− dzd−1+

a0

adzd

),

we have a disk U = ∆r for r > large enough such that V = f(U) ⊃ U .The filled Julia set

K = ∩∞n=1P−n(U)

is a compact subset. The boundary J = ∂K is the Julia set of P .Let C be the set of critical points of P , that is,

P ′(c) = adcd + ad−1c

d−1 + · · ·+ a1c+ a0 = 0, c ∈ C.

Then P has d − 1 critical points in the complex plane (counted bymultiplicity) and∞ is another critical point with the multiplicity d−1So P has 2d− 2 critical points (counted by multiplicity). We write

C = c1, . . . , cd−1

be the set of all critical point in the complex plane. Consider criticalorbits Oi = ci,n = P n(ci)∞n=0. If all ci,n →∞ as n→∞, then we cansee that, from the hyperbolic geometry, J = K is a Cantor set and Pon J is an expanding Markovian hyperbolic dynamical system, that is,there are two constants C > 0 and λ > 1 such that

(P n)′(z) ≥ Cλn, ∀ z ∈ J, n > 0

and we have a partition J = J1 ∪ · · · ∪ Jn such that f |Ji is one-to-onefor every 1 ≤ i ≤ n and f(Ji) = ∪si=1Jki . Indeed, Let D be a largeopen disk containing J such that D ⊂ P (D). Choose an integer m > 0such that Pm(c) ∈ C \ D for every critical point c of P . Then Dcontains no critical values P n(C) for n ≥ m. All inverse branches R−n

are defined in D and map D conformally into D. Let Dni , 0 ≤ i < dn,

be the components of R−n(D). Then each Dni is a topological disk

such that Dni ⊂ D and Rn : Dn

i → D is conformal. Suppose z ∈ J ,then P n(z) ∈ J . We define fn as the inverse branch of P n which mapsP n(z) to z. Then fnn≥m are uniformly bounded in a neighborhoodof D. So it is a normal family. Since d(fn(w), J) → 0 as n → ∞ forany w ∈ D ∩B∞, any limiting function f of a convergent subsequenceof fnn≥m maps D ∩ B∞ into J . But J has no interior point, so f isa constant function. This implies that |fn(D)| → 0 as n → ∞. Sincefn(∂D) ∩ J = ∅, z must be a connected component of J . So J is

46

totally disconnected. But we know J is perfect. So J is a Cantor set.Following a similar proof of Theorem 19, we have that

Theorem 20. If all critical orbits of P tend to ∞, then P |J is anexpanding Markovian dynamical system and the Julia set J is a Cantorset with Area(J) = 0.

Suppose there is no finite critical point of P in B∞. Then we canextended the Bottcher coordinate to the whole B∞ as we have discussedfor a quadratic polynomial. Let φ : B∞ → C \ ∆ be the Bottchercoordinate, that is, it is the 1-1 analytic (or conformal) conjugacy withφ(∞) =∞ and φ′(∞) = 1 such that

φ(P (z)) = φ(z)d, z ∈ B∞Then G(z) = log |φ(z)| is the Green function of (B∞,∞) satisfying

G(P (z)) = dG(z), z ∈ B∞.The Green function G can be also defined as

G(z) = max0, limn→∞

1

dnlog |P n(z)|.

Define

Sr = z ∈ C | G(z) = log r = φ−1(z ∈ C) | |z| = r, 1 < r <∞and

Eθ = z ∈ C | arg φ(z) = 2πθ = φ−1(z = re2πiθ | 1 < r <∞), 0 ≤ θ < 1.

We call Sr the equipotential circle of radius r and Eθ the external rayof angle θ. It is clear that

P (Sr) = Srd and P (Eθ) = Edθ (mod 1).

If B∞ contains no finite critical point of P . Then B∞ is simply con-nected. Thus we have that

Theorem 21. The filled-in Julia set K (or the Julia set J) is connectedif and only if there is no finite critical point of P in B∞.

47

Lecture 5. Koebe distortions

Let C be the complex plane and let

∆ = z ∈ C | |z| < 1be the open unit disk of the complex plane. A function defined on ∆is analytic if it can be written as a convergent power series, that is,

f(z) =∞∑n=0

anzn

for z in ∆. We first prove a theorem obtained by Koebe in 1907.

Theorem 22 (Koebe’s 14-Theorem). Suppose f : ∆→ C is a schlicht

function. Then f(∆) contains an open disk centered at f(0) with radius|f ′(0)|/4.

First we prove a lemma.

Lemma 6. Suppose

f(z) = z + a2z2 +

∞∑n=3

anzn : ∆→ C

is a schlicht function. Then

|a2| ≤ 2.

Proof. Suppose ∆r = z ∈ C | |z| < r for 0 < r < 1. Then Gr =f(∆r) is a bounded subset of the complex plane C. Let G′r be thecomplement of Gr. For w = ReΘi in G′r,∫ ∫

G′r

wtwtR dR dΘ > 0

provided the integral converges, whence

limr→1

∫ ∫G′r

wtwtR dR dΘ ≥ 0.

The integral certainly converges if t < −1. Let us set z = reθi andw = f(z). On the boundary of G′r, which is the image of |z| = r underf , we set R = R(θ) and Θ = Θ(θ). Then∫ ∫

G′r

wtwtR dR dΘ =

∫ ∫G′r

R2t+1 dR dΘ = −∫ 2π

0

R2t+2

2t+ 2

dΘ

dθdθ > 0.

We have

Θ = argw =logw − logw

2i.

48

ThereforedΘ

dθ=w′wz + w′wz

2ww,

and ∫ 2π

0

R2t+2

2t+ 2

dΘ

dθdθ =

∫ 2π

0

R2t(w′wz + w′wz)

4(t+ 1)dθ < 0.

This is easily transformed into∫ 2π

0

(wt+1z

dwt+1

dz+ wt+1z

dwt+1

dz

)dθ < 0.

But

wt+1 = zt+1(

1 + (t+ 1)a2z +(

(t+ 1)a3 +t(t+ 1)

2a2

2

)z2 + · · ·

).

We substitute this series into the last integral and integrate term byterm, observing that for any non-zero integer k,

∫ 2π

0ekiθdθ = 0. Thus

we obtain

1 + (t+ 1)(t+ 2)|a2|2r2 + (t+ 1)(t+ 3)∣∣∣a3 +

t

2a2

2

∣∣∣2r4 + · · · > 0.

For r → 1, this yields

1 + (t+ 1)(t+ 2)|a2|2 + (t+ 1)(t+ 3)∣∣∣a3 +

t

2a2

2

∣∣∣2+ · · ·+ (t+ 1)(t+ k)

∣∣∣ak + gk(a2, . . . , ak−1)∣∣∣2 + · · · ≥ 0.

In particular, we take t = −3/2; then

1− |a2|2

4≥ 0,

whence|a2| ≤ 2.

Remark 8. In general, |an| ≤ n for all n ≥ 2. This was the famousBieberbach conjecture formulated in 1916 and proved by De Branges in1984.

Proof of Theorem 22. Let f(z) 6= c for all |z| < 1. The function

f1(z) =f(z)− f(0)

f ′(0)= z + a2z

2 + · · · 6= c− f(0)

f ′(0)

for all |z| < 1. Set

f2(z) =f(0)− cf ′(0)

f(z)− f(0)

f(z)− c= z +

(a2 −

f ′(0)

f(0)− c

)z2 + · · · .

49

Both of f1 and f2 are schlicht functions from ∆ into C. Thus

|a2| ≤ 2 and∣∣∣a2 −

f ′(0)

f(0)− c

∣∣∣ ≤ 2.

Therefore,

|c− f(0)| ≥ |f′(0)|4

.

We have thus proved that no boundary point, of the image of |z| < 1under the mapping f , has distance from f(0) less than |f ′(0)|/4.

Lemma 7. Suppose f(z) = z +∑∞

n=2 anzn : ∆ → C is a schlicht

function. Then

1− |z|(1 + |z|)3

≤ |f ′(z)| ≤ 1 + |z|(1− |z|)3

.

Proof. For a point z in ∆, let

g(ξ) =f( ξ+z

1+zξ)− f(z)

f ′(z)(1− zz).

This is a schlicht function from ∆ into C and has a power series ex-pansion valid in |ξ| < 1:

g(ξ) = ξ + b2ξ2 + . . . ,

where

b2 =1

2

(f ′′(z)(1− zz)

f ′(z)− 2z

).

Therefore, ∣∣∣f ′′(z)(1− zz)

f ′(z)− 2z

∣∣∣ ≤ 4.

Hence ∣∣∣zf ′′(z)

f ′(z)− 2|z|2

1− |z|2∣∣∣ ≤ 4|z|

1− |z|2.

Thus2|z|2 − 4|z|

1− |z|2≤ <

(zf ′′(z)

f ′(z)

)≤ 2|z|2 + 4|z|

1− |z|2.

Since

<(zf ′′(z)

f ′(z)

)= |z| ∂

∂|z|

(<(

log f ′(z)))

= |z| ∂∂|z|

(log |f ′(z)|

),

we obtain

2|z| − 4

1− |z|2≤ ∂

∂|z|

(log(|f ′(z)|

))≤ 2|z|+ 4

1− |z|2.

50

Integration now yields

1− |z|(1 + |z|)3

≤ |f ′(z)| ≤ 1 + |z|(1− |z|)3

.

Theorem 23 (Koebe Distortion Theorem). Suppose f : ∆ → C is aschlicht function. Then(1− r

1 + r

)4

≤ |f′(z1)||f ′(z2)|

≤(1 + r

1− r

)4

for all |z1| ≤ r < 1 and all |z2| ≤ r < 1.

Proof. Let g(z) = (f(z)−f(0))/f ′(0). Then g′(z) = f ′(z)/f ′(0). Thus,this theorem follows from Lemma 7.

Corollary 2. Suppose f(z) = a1z + a2z2 + · · · + anz

n + · · · : ∆ → Cis a schlicht function, then

|f ′(0)| |z|(1 + |z|)2

≤ |f(z)| ≤ |f ′(0)| |z|(1− |z|)2

.

Remark 9. The above theorems, lemmas, corollary are are the sharpestpossible. One sees this with the example

f(z) =z

(1− z)2.

51

Lecture 6. Theory of Fatou and Julia

Recall that C is the Riemann sphere. Let

dSP s =2|dz|

1 + |z|2

is the spherical metric. Then the spherical distance dSP (z1, z2) is

dSP (z1, z2) = infl

∫l

dSP s

where inf takes all smooth curves connecting z1 and z2.Suppose F is a family of meromorphic functions defined on a domain

U in C. Then F is said to be normal if every sequence fn∞n=0 in Fcontains a subsequence that converges uniformly in the spherical metricon compact subsets of U . Following Arzela-Ascoli theorem, the familyF is normal if and only if it is equicontinuous on every compact subsetof U (under spherical metric). Note that we allow fn → ∞ in thedefinition of a normal family. Thus it is more convenient to considerthe function f ≡ ∞ as a meromorphic function.

Theorem 24. A family F of analytic functions on a domain U in Cwhich is bounded by some fixed constant is a normal family.

Proof. Since U can be covered by disks, it is sufficient to prove thetheorem for a disk. We may assume that U is the unit disk. SupposeM is a fixed constant such that |f(z)| ≤ M for all f ∈ F and |z| < 1.Cauchy’s theorem implies that |f ′(z)| ≤M/(1− r) on each closed diskz||z| ≤ r < 1. Therefore, F is equicontinuous on every compact setof U . So F is normal.

Let U be the set of all univalent functions f on the unit disk ∆ =z ∈ C | |z| < 1 such that f(0) = 0 and f ′(0) = 1.

Corollary 3. The family U is normal and any limit of any sequencein U is still in U .

Proof. By the Koebe distortion theorem,

1− |z|(1 + |z|)3

≤ |f ′(z)| ≤ 1 + |z|(1− |z|)3

, z ∈ ∆.

So U is uniformly bounded on any compact set of ∆. Suppose fn isa convergent sequence in U . Let f be its limiting function. FollowingHurwitz’s theorem and the normalization f ′n(0) = 1, f is either univa-lent or a constant function. But |f ′(z)| ≥ (1 − |z|)/(1 + |z|)3. So f isunivalent.

52

3

Theorem 25 (Montel’s Theorem). Let F be a family of meromorphic

functions on a domain U of C. If there are three fixed values that areomitted by every f ∈ F , then F is a normal family.

Proof. Following the same reason as that in the proof of the previoustheorem, we assume that U is a disk. Furthermore, by conjugating by aMobius transformation, we assume that the functions in F do not takethe values 0, 1, and ∞. Let V = C \ 0, 1. Then V is a hyperbolicsurface. Its universal cover is the unit disk ∆. Let φ : ∆ → V bea universal covering map. Let f : ∆ → ∆ be a lift of f ∈ F , i.e.,f φ = f . Then the family F = f ; f ∈ F is normal. So F is normaltoo.

Now suppose

f(z) =p(z)

q(z)

is a rational function, where

p(z) = anzn + · · ·+ a0 and q(z) = bmz

m + · · ·+ b0

are two complex polynomials of degrees n and m, respectively. Thedegree of f is d = maxn,m. Let fn mean the nth-iterate of f .

Definition 19. A point z ∈ C is called a Fatou point if there is aneighborhood U about z such that fn|U∞n=0 is a normal family. Theset F of all Fatou points is called the Fatou set of f . The complement

J = C \ F is called the Julia set of f .

From the definition, the Fatou set F is an open set and the Julia setJ is a compact set.

Theorem 26. If the degree d > 1, then the set J is not empty.

Proof. Since the degree d of f is greater than one, suppose J = ∅, then

fn is normal on the whole C and there is a subsequence fni con-

verges to a rational function f0 on C. If f0 is a constant function, then

fni(C) is contained in a small neighborhood of c, but it is impossible

since Rni : C→ C is onto. If f0 is not a constant function, then fni hasthe same number of zeros as f0 for j large enough (apply the argumentprinciple), which is also impossible since fni has degree dni .

In the following, we always assume that the degree d of f is greaterthan 1.

53

Theorem 27. The Julia set J and the Fatou set F are both completelyinvariant, i.e., f(J) = f−1(J) = J and f(F ) = f−1(F ) = F . More-over, Jfn = Jf and Ffn = Ff .

Proof. The proof is from the definition directly.

Example 12. Let f(z) = zn where n > 1 is an integer. Then the Juliaset J is the unit circle and the Fatou set F = ∆ ∪∆

c, where

∆ = z ∈ C | |z| < 1 and ∆c

= z ∈ C | |z| > 1are basins of 0 and ∞, respectively.

Example 13. Let f(z) = z2 − 2. Take z = h(w) = w + 1/w. It maps

∆ and ∆c

onto C \ [−2, 2] and the unit circle S1 onto [−2, 2]. Sinceh−1 f h(w) = w2, the Julia set and the Fatou set of f are J = [−2, 2]

and F = C \ [−2, 2], which is the basin of ∞.

The following example is little complicate but fun to know.

Example 14. Let f(z) = λz + z2 with |λ| = 1. Suppose there is adomain V0 containing 0 such that there is a conformal map φ : V0 → ∆making φ f φ−1(z) = λz on ∆. Let B∞ be the basin of ∞ forf . Then B∞, V0 ⊂ F and V0 ∩ B∞ = ∅ since fn(V0) 6→ ∞ as ngoes to infinity. Let U0 and U∞ be the components of the Fatou set Fcontaining V0 and B∞. Then we have that U∞ = B∞ and fn|U0∞n=1

is uniformly bounded. The domain U0 is simply connected. The reasonis the following. For any curve γ in U0, let Dγ be the bounded domainbounded by γ. If U0 is not simply connected, then there is a such γsuch that Dγ contains some point from Julia set. Then ∪∞n=0f

n(Dγ) ⊇C \ one point. But Dγ ∩ U∞ = ∅. This is a contradiction. Thesimply connected domain U0 is the maximal domain on which φ can beextended such that φ f φ−1(z) = λw. Here U0 is called the Siegeldisk of f . Let U0 = ∪∞n=0f

−n(U0), It is clear now that F = U0 ∪ U∞and J = ∂U0 = ∂B∞.

Let z ∈ J . Suppose U is a neighborhood of z. By Montel’s theorem,

Ez = C \ ∪∞n=0fn(U) contains at most two points.

Theorem 28. The set Ez is independent of z. So we can denote it asE. If E = a contains only one point, we can use a Mobius trans-formation to move a to infinity, then f is a polynomial. If E = a, bcontains two points, we can use a Mobius transformation to move a toinfinity and b to zero, then f(z) = Czd or Cz−d. In any cases, E is asubset of F .

54

Proof. By the definition, f−1(Ez) ⊆ Ez. If Ez = a contains only onepoint, then R(a) = a. Assume a = ∞. Since f−1(∞) = ∞, f has noother poles. So f is a polynomial. Clearly, Ez is independent of z. IfE = a, b, then either (1) f(a) = a and f(b) = b or (2) f(a) = b andf(b) = a. Assume a = 0 and b =∞. In the first case, f is a polynomialand f(0) = 0 and f−1(0) = 0. So f(z) = Czd. In the second case,similarly we can get, f(z) = Cz−d.

The set E is called the exceptional set of f . It follows immediatelyif z 6∈ E, then

∪∞n=0f−n(z) ⊃ J.

This implies the following two theorems.

Theorem 29. The backward iterates ∪∞n=0f−n(z) of any point z ∈ J

is dense in J .

Theorem 30. Any non-empty completely invariant set of J is densein J .

Since the union of the boundaries of all components of the Fatou setF is closed and completely invariant, it is the Julia set.

Theorem 31. The Julia set J is a perfect set, that is, J contains noisolated points.

Proof. Take z ∈ J and U a neighborhood about z. Suppose z is not aperiodic point of f . Then there is a point z1 such that f(z1) = z andfn(z) 6= z1 for all n. Since z1 ∈ J ,

∪∞n=0f−n(z1) = J.

Therefore, there is a point ξ ∈ U such that fn(ξ) = z1. Thus ξ ∈ J ∩Ubut ξ 6= z.

If z is periodic, let m > 0 be the minimal integer such that fm(z) =z. If z is the only point in f−m(z), then z is super-attracting. Thiscontradicts to that z ∈ J . So there is a point z1 6= z = fm(z) inf−m(z). Furthermore, f j(z) 6= z1, 1 ≤ j < m. (Otherwise, f j(z) =f j+m(z) = fm(z1) = z, contradicts to the minimal property of m.)Similarly, there is a point ξ ∈ U such that fn(ξ) = z1 since

∪∞n=0f−n(z1) = J .

This ξ ∈ J ∩ U but ξ 6= z.

Theorem 32. If J has an interior point, then J = C.

55

Proof. Suppose U is a domain contained in J . Then fn(U) ⊂ J . Since

∪∞n=0fn(U) = C \ E and J is closed set, J = ∪∞n=0f

n(U) = C.

Theorem 33. Let Γ be the set of all repelling periodic points of f .Then Γ = J .

Proof. Suppose U is a domain such that U ∩ J 6= ∅ and such thatU ∩Γ = ∅. Assume U contains no poles and no critical points of f . Letf1 and f2 be two inverse branches of f |U . Since there is no solution offm(z) = z in U ,

gn(z) =fn(z)− f1(z)

fn(z)− f2(z)

z − f2(z)

z − f1(z)

omits the values 0, 1,∞. By Montel’s theorem gn is a normalfamily and hence so is fn|U. This is a contradiction. Therefore theset of all periodic points of f is dense in J . Since there are only finitenumber of attracting, super-attracting, and neutral periodic points (seethe next section). So Γ = J .

Concluding our discussion above, we have that the Julia set is non-empty, compact, perfect and is the closure of ∪∞n=0f

−n(z) for any point

z in the Julia set. Moreover, either it is the whole Riemann sphere Cor has no interior point and the set of all repelling periodic points isdense in the Julia set.

Recall that a point z is called a periodic point of period n of f iffn(z) = z and f i(z) 6= z for i = 1, · · · , n− 1. In the previous lectures,we classified a periodic point into attractive, repelling, super-attractive,rational neutral, and irrational neutral. Directly from the definition,we know that an attractive or super-attracting periodic point z is aFatou point. An repelling periodic point z is in the Julia set.

Next we will see when a neutral periodic point is in the Fatou set orin the Julia set. By the definition, a neutral periodic point of period

n is a point p ∈ C such that fn(p) = p and |(fn)′(p)| = 1. Then wedivide it into rationally neutral if (fn)′(p) = e2πip/q with integers p ≥ 0and q ≥ 1, (p, q) = 1, and irrationally neutral if (fn)′(p) = e2πiθ withan irrational number 0 < θ < 1.

In the irrational case, we divide it into two cases, Siegel and Cremer,as follows. We call p is a Siegel periodic point if there is a domain

U ⊂ C about p and a conformal map φ : U → ∆ such that φ(p) = 0 andφ fnφ−1(z) = e2πiθz. The maximal such domain U is called the Siegeldisk at p. Otherwise, p is called a Cremer periodic point. Withoutloss of generality, we assume p = 0 and f(0) = 0. Let λ = f ′(0) with

56

λ = e2πiθ with an irrational number 0 < θ < 1. Consider the Taylorexpansion

f(z) = λz + a2z2 + a3z

3 + · · ·in a domain U about 0. To decide 0 is a Siegel fixed point or Cremerfixed point we need to see if the Schroder equation

f(h(z)) = h(λz), h(0) = 0, h′(0) = 1,

has a solution or not in some ∆r, r > 0. The reason is that

Lemma 8. Any analytic solution h of the above Schroder equation isconformal, i.e., h is one-to-one.

Proof. Suppose h(z1) = h(z2) for z1, z2 ∈ ∆r. Then h(λnz1) = h(λnz2)for all n. Since θ is irrational, nθ (mod 1) is dense on [0, 1]. So λnis dense on the unit circle S1. By continuity, h(z1e

2πiτ ) = h(z2e2πiτ ) for

all τ . Since h(z1·) and h(z2·) are analytic, we have h(z1z) = h(z2z) forall z ∈ ∆r. From h′(0) = 1, we have z1 = z2.

The characterization of the Schroder equation has a solution is that

Theorem 34. The Schroder equation has a solution if and only iffn|K is uniformly bounded for any compact set K in some neighbor-hood domain U of 0

Proof. If the Schroder equation has a solution h : ∆→ U , then fn(z) =h(λnh−1(z)). It is clearly that fn|K is uniformly bounded for anycompact set K in U .

On the other hand if there is a constantM > 0 such that |fn(z)| ≤Mfor all n ≥ 0 and all z in some open domain U ′ of 0. Define

φn(z) =1

n

n−1∑k=0

λ−kfk(z).

Then φn is a uniformly bounded sequence of analytic functions onU ′. So it contains a convergent subsequence. Since

φn f = λφn +O(1

n),

any such a limit φ satisfies

φ f = λφ.

Because λ = f ′(0), we have φ′n(0) = 1 and φ′(0) = 1. So we can havea ∆r for some 0 < r < 1 such that h = φ−1 : ∆r → U = h(∆r) is asolution of the Schroder equation.

57

Since the linear map z 7→ λz has no periodic point in any smallneighborhood about 0, we have that

Theorem 35. If f has a periodic point in any small neighborhood about0, then f can not be linearizable

Applying this theorem, one can construct an example of f such that0 is a Cremer fixed point of f .

Theorem 36. There exists a λ = e2πiθ, θ ∈ [0, 1] \ Q such that theSchroder equation has no solution for any polynomial f of degree d.

Proof. Let f(z) = λz + · · · + adzd, ad 6= 0. Suppose there is a disk

∆r such that f is conjugating to λz by a conformal map h : ∆r → U .Consider the dn fixed points of fn, that is, all roots of

fn(z)− z = zdn

+ · · ·+ (λn − 1)z = 0.

One of these root is 0. Let z1, · · · , zdn−1 be all the rest roots. Sincefn(z) = h(λnh−1(z)) has only fixed point 0 in ∆r, zi 6∈ ∆r for all0 ≤ i ≤ dn − 1. Thus

rdn ≤

dn−1∏i=1

|zi| = |1− λn|.

We now consider a λ such that the last inequality is impossible for alln.

Take a sequence of positive integers

q1 < q2 < · · · .Let θ =

∑∞k=1 2−qk and λ = e2πiθ. Then

|1− λ2qk | ∼ 2qk−qk+1 .

The last inequality implies that there is a universal constant C > 0such that

rd2qk ≤ C2qk−qk+1 .

So

d2qk log2 r ≤ log2C + qk − qk+1.

Equivalently,

qk+1 ≤ log2C + qk − d2qk log2 r = d2qk( log2C + qk

d2qk− log2 r

)≤ C ′d2qk .

If we take qk growing fast like kd2qk , then the Schroder equation hasno solution.

58

Remark 10. An example in the above theorem is given in 1917 byG. A. Pfeifer. The work was continuous by H. Cremer who proved in1938 that |λ| = 1 and lim infn→∞ |1−λn| = 0, then there is an analyticfunction f(w) = λz + a2z

2 + · · · such that Schroder equation has nosolution. Refer to

• J. Milnor, Dynamics in One Complex Variable, IntroductoryLectures. Vieweg, 2nd Edition, 2000.

Concluding from the above discussion, we have that a Siegel periodicpoint is in the Fatou set and a Cremer periodic point is in the Juliaset. We have seen an example of rational neutral fixed point which is aCremer fixed point. An example of a Siegel fixed point can be obtainedfrom the following theorem, which we will prove in a later lecture.

A real number 0 < θ < 1 is called Diophantine if there are constantsC > 0 and 0 < µ <∞ such that

|θ − p

q| ≥ C

qµ

for all p and q 6= 0. This condition is equivalent to

|λn − 1| ≥ Cn1−µ

for all n ≥ 1. For a fixed µ > 2, |θ− pq| ≥ C/qµ holds for almost all real

irrational number θ ∈ [0, 1]. Indeed if E is the set of irrational numberθ ∈ [0, 1] such that |θ − p/q| < q−µ infinitely many often, then

m(E) ≤∞∑q=n

2q−µ+1 = O(n2−µ)→ 0.

In particular, almost all irrational numbers in [0, 1] are Diophantine.

Theorem 37 (Siegel, 1942). If 0 < θ < 1 is Diophantine and iff(z) = λz + a2z

2 + . . . is a convergent power series in a neighborhoodU of 0 such that λ = e2πiθ, then there is a solution to the Schroderequation, that is, f can be conjugated near 0 to the linear map z 7→ λz.

Now let us come to the rational neutral case. Suppose 0 ∈ U is arational neutral fixed point of a holomorphic map f : U → C. Let

λ = f ′(0) with λ = e2πi pq , p ≥ 0 and q ≥ 1 are integers. Suppose p and

q are relatively prime if p 6= 0. Then f(z) = λz or its Taylor expansionof f at 0 is

f(z) = λz + an+1zn+1 + · · ·

where an+1 6= 0 and n ≥ 1. In the first case f is a linear map. Themost interesting case is the second case. Let us discuss it in three cases:

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(1) λ = 1 and n = 1;(2) λ = 1 and n ≥ 2;(3) λ 6= 1.

In the case (1), conjugating by z 7→ a2z, we may assume that a2 = 1.Then f(z) = z + z2 + · · · . First we move 0 to ∞ by the conjugacyw = α(z) = −1/z, i.e., considering

g(w) = α0 f α−10 (w) = w + 1 +

b

w+O(

1

w2), w ∈ V = α(U).

There is a large constant C0 > 0 such that

<gn(w) > <w +n

2, <w > C0, n ≥ 1.

(In general n/2 < |gn(w)| ≤ |w|+ 2n for <w > C0.) Take

φn(w) = gn(w)− n− b log n, <w > C0, n ≥ 1.

Since

gn+1(w) = gn(w) + 1 +b

gn(w)+O(

1

n2), <w > C0, n ≥ 1,

φn+1(w)−φn(w) = b(

log n−log(n+1))+

1

gn(w)+O(

1

n2) = O(

1

n), <w > C0, n ≥ 1.

So

|φn(w)− w| ≤ |φ1(w)− w|+n−1∑k=1

|φk+1(w)− φk(w)|

= O(log n), <w > C0, n ≥ 1.

Therefore,

φn+1(w)− φn(w) = b(

log n− log(n+ 1))

+ gn+1(w)− gn(w)− 1

= − bn

+b

gn(w)+O(

1

n2)

= b( 1

n+ b log n+ φn(w)− 1

n

)+O(

1

n2)

=b

n2O(|b log n+ φn(w)|) +O(

1

n2)

= O(log n

n2), <w > C0, n ≥ 1.

Hence∞∑n=1

|φn+1(w)− φn(w)| <∞, <w > C0.

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This implies that φn(w) converges to φ(w) as n goes to infinity. Thisconvergence is uniformly on any compact set in <w > C0. So φis conformal in <w > C0. Since φn is conformal and φn(g(w)) =φn+1(w) + 1 + b log(1 + 1

n), we have

φ(g(w)) = φ(w) + 1.

We can extend φ analytically to any domain B+ such that g(B+) ⊆ B+

and gn(w) → ∞ as n → ∞ for w ∈ B+. The map ψ+ = φα thenconjugates f to w 7→ w + 1, i.e.,

ψ(f(z)) = ψ(z) + 1, z ∈ D+ = α−1(B+).

Similarly we have a conformal map φ defined on any domain B− suchthat g−1(B−) ⊆ B− and g−n(w) → ∞ for w ∈ B−. The map φconjugates g−1 to w 7→ w − 1, i.e.,

φ(g−1(w)) = φ(w)− 1, w ∈ B−.The map ψ− = φα then conjugates f−1 to w 7→ w − 1, i.e.,

ψ−(f−1(z)) = ψ−(z)− 1, z ∈ D− = α−1(B−).

Now let us look at the dynamics of f around 0. Suppose U is abounded domain. Consider the domain V = α(U) ∪ ∞ about ∞.Let a be a real number such that a and −a are in V and such thatcurve r1 : x = −y2 + a and r2 : x = y2 − a are all contained in V .Let B+ ⊂ V be the domain bounded by r1 (the right domain cut byr1) and let B− ⊂ V1 be the domain bounded by r2 (the left domaincut by r2). Let T (w) = w + 1 be the shift map. Then it is clear thatT (B+) ⊂ B+ and that T−1(B−) ⊂ B−. Let B = B+∪B−. Then it is a

domain about ∞ in C. Let D+ = α−1(B+) and D− = α−1− (B−). They

are domains in U bounded by analytic curves tangent at 0. Moreover,f(D+) ⊂ D+ and f−1(D−) ⊂ D−. Here D+ is called an attractive petalof f at 0 and D− is called a repelling petal of f at 0. Maps ψ+ and ψ−define an analytic map ψ on D = D+ ∪D− such that

ψ(f(z)) = ψ(z) + 1, z ∈ D.Let K ′ be the set of all horizontal lines contained in B and let K =ψ−1(K ′). Then K is a compact set and f(K) = f−1(K) = K sinceT (K ′) = T−1(K ′) = K ′. Conclude the above discussion, we have that

Theorem 38. There are two domains D+ and D− and a compact setK satisfying that

(1) D+ is bounded by an analytic curve tangent at 0 such thatf(D+) ⊂ D+ ∪ 0;

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(2) D− is bounded by an analytic curve tangent at 0 such thatf−1(D−) ⊂ D− ∪ 0;

(3) 0 is an interior point of D = D+ ∪D−;(4) K ⊂ D and ∂K ∩ ∂D 6= ∅;(5) f(K) = f−1(K) = K.(6) K consists of two closed domains K+ and K− touching at 0,

each of them is foliated by completely f -invariant analytic curvestangent at 0, that is, K = K+ ∪K− and K+ ∩K− = 0 andK± = ∪tlt,± where lt,± are analytic curves tangent at 0 andf(lt,±) = f−1(lt,±) = lt,±.

In the case (2), conjugating by z 7→ bz where bn = 1/an, we mayassume that an = 1. Then f(z) = z + zn+1 + · · · . Let

Sj = z = re2πiθ | jn≤ θ <

j + 1

n, j = 0, 1, · · · , n− 1.

For each 0 ≤ j < n, considering the map βj(z) = zn : Sj → C. Then

f(z) = βj f β−1j (z) = (z

1n + z

n+1n + · · · )n = z + nz2 + o(|z|2)

From Case (1), f has an attracting petal D+ bounded by an analyticcurve tangent with the positive real line at 0, a repelling petal D−

bounded by an analytic curve tangent with the negative real line at0, and a totally invariant compact set K foliated by analytic curvestangent with the real line at 0. Let D+

j = β−1j (D+), D−j = β−1

j (D−),

and Kj = β−1j (K). Then D+

j , j = 0, · · ·n − 1, are attracting petals

of f at 0 and D−j , j = 0, · · ·n − 1, are repelling petals of f at 0, and

K = ∪n−1j=0Kj is a completely f -invariant set. Therefore, we have that

Theorem 39. In any neighborhood U of 0 where f is defined, f has

(1) n attracting petals D+j , j = 0, · · · , n − 1, such that f(D

+

j ) ⊂D+j ∪ 0;

(2) n repelling petals D−j , j = 0, · · · , n − 1, such that f−1(D−j ) ⊂D−i ∪ 0; and

(3) a completely f -invariant compact set K, i.e., f(K) = f−1(K) =K, consisting of n petals Kj, j = 0, · · · , n− 1.

Moreover, 0 is an interior point of D = ∪n−1j=0D

+j ∪D−j , each petal Kj

is foliated by completely f -invariant analytic curves, this means thatKj = ∪tlt,j where lt,j are analytic curves and f(lt,j) = f−1(lt,j) = lt,j,and ∂Kj ∩ ∂D 6= ∅.

62

At last, we discuss the case (3). In this case, let λ = e2πip/q, where0 < p < q and (p, q) = 1. Then f q(z) = z or f q(z) = z+an+1z

n+1 + · · ·with an+1 6= 0 and n ≥ 1.

Theorem 40. Suppose f q(z) = z+ an+1zn+1 + · · · with an+1 6= 0 and

n ≥ 1. In any neighborhood U of 0 where f is defined, f has

(1) n = kq attracting petals D+j,l, j = 0, · · · , q − 1, l = 1, · · · , k,

such that

f(D+j,l) ⊂ D+

j+p (mod q),l ∪ 0;

(2) n = kq repelling petals D−j,k, j = 0, · · · , q−1, l = 1, · · · , k, suchthat

f−1(D−j,l) ⊂ D−j+p (mod q),l ∪ 0;and

(3) a completely f -invariant compact set K, i.e., f(K) = f−1(K) =K, consisting of n = kq petals Kj, j = 0, · · · , q−1, l = 1, · · · k.

Moreover, 0 is an interior point of D = ∪kl=1∪q−1i=0 D

+j,l ∪D

−j,l, each petal

Kj,l is foliated by analytic curves, this means that Kj,l = ∪tlt,j,l wherelt,j,l are analytic curves and f(lt,j,l) = f−1(lt′,j,l) = lt,j+p (mod q),l, and∂Kj,l ∩ ∂D 6= ∅.

Corollary 4. If f has a rational neutral fixed point, then either fn = idfor some n > 0 or f can not be linearizable (even topologically).

Concluding from our discussion in this section, we have that

Theorem 41. Suppose f is a rational map with the degree d > 1.Then the Julia set J of a rational map f contains all repelling periodicpoints of f and all rational neutral periodic points and all irrationalCremer neutral periodic points. The Fatou set contains all attractiveand super-attractive periodic points and all irrational Siegel neutral pe-riodic points.

Now let us have a complete picture about the local structures aroundneutral fixed points (for both rational and irrational neutral fixed points).

• R. Perez-Marco, Fixed points and circle maps. Acta Math.,179 (1997), 243-297.

Theorem 42. Suppose f is holomorphic and univalent on a domainV . Suppose 0 ∈ V is a fixed point of f , f(z) = λz+a2z

2 + · · · , |λ| = 1.For any domain 0 ∈ U ⊂ U ⊂ V , there is a set K such that

(1) K is compact and connected;

63

(2) K is full, i.e., C \K is connected;(3) 0 ∈ K;(4) K ∩ ∂U 6= ∅;(5) f(K) = f−1(K) = K;(6) f is linearizable at 0 if and only if there is a U such that K has

0 as an interior point.

Corollary 5. There is a unique maximal set Kmax in U with the prop-erty: w ∈ K implies that fn(w) ∈ K for all n ≥ 0.

Proof of Theorem 42 (6.) If f is linearizable at a small neighborhoodU , then there is a conformal map ψ : U → ∆ = z ∈ C | |z| < 1 suchthat φ(f(z)) = λψ(z) for z ∈ U . So φ−1(∆) ⊂ U is an invariant setof f . Let K be the maximal f -invariant set in U containing φ−1(∆).Then K has 0 as an interior point.

If there is a neighborhood U about 0 such that 0 is an interior pointof K. Let W be the connected component of K containing 0. ThenK is a simply connected domain such that f(W ) = W . The map f onW is univalent. Let φ : W → ∆ be the Riemann mapping, φ(0) = 0.Then

f = φ f φ−1 : ∆→ ∆, f(0) = 0,

is an isomorphism. The Schwarz lemma (Lemma 3) implies

f(z) = e2πiθz.

Suppose U is a bounded domain in C. Let S be the set of all compactsets K in U . Let

Nε(K) = z ∈ C | there is a w ∈ K such that d(w, z) < ε.be the ε-neighborhood of K. Let

d(K1, K2) = infε | K1 ⊂ Nε(K2), K2 ⊂ Nε(K1).Then d(·, ·) defines a topology in S, which is called the Hausdorff topol-ogy in S.

Lemma 9. The Hausdorff topology is compact.

Proof. Let Kn∞n=1 be a sequence in S. For each n > 0, let An =xnm∞m=1 be a dense subset in Kn. The for each fixed m > 0, xnm∞n=1

has a convergent subsequence in U . Therefore, we can find a subse-quence ni of integers such that xnim is convergent for every m withlimit points xm in U . Let K be the closure of xm∞m=1. Then K is acompact set in S such that d(Kn, K)→ 0 as n→∞.

64

Lemma 10. Suppose fn is a sequence of analytic maps defined onU0 and f is an analytic map defined on U0. Let U ⊂ U ⊂ U0. SupposeKn is a sequence of compact sets in U and K is a compact set in U .If fn → f uniformly in U and Kn → K in the Hausdorff topology, thenfn(Kn)→ f(K) in the Hausdorff topology.

Proof the Rest of Theorem 42. Suppose

f(z) = λz + a2z + · · · , λ = e2πiθ, θ ∈ [0, 1] \QTake θn ∈ Q such that θn → θ as n→∞. Define

fn(z) = e2πi(θn−θ)f = e2πiθnz + a2z2 + · · · .

Then 0 is a rational neutral fixed point of f . Following Theorem 40,for each n, there is a compact and connected set Kn ⊂ U such that Kn

is completely fn-invariant, i.e., fn(Kn) = f−1n (Kn) = Kn, Kn∩∂U 6= ∅.

Since the Hausdorff topology is compact, there is a subsequence Knisuch that Kni → K in the Hausdorff topology as i→∞. But it is clearthat fni → f uniformly in U as i→∞. So Kni = fni(Kni)→ f(K) =K in the Huasdorff topology as i → ∞. Since Kni is compact andconnected, K is compact and connected. Since 0 ∈ Kni and Kni∩∂U 6=∅, 0 ∈ K and K ∩ ∂U 6= ∅. If K is not full, then C \ K has morethat one components. Let B = A1 ∪ · · · ∪ An be the union of allbounded components and let A∞ be the unbounded component. Thenf(K ∪ B) = K ∪ B. So K ∪ B is full and completely f -invariant. Wecomplete the proof.

Now let us see some applications of Theorem 42. Suppose f is aholomorphic function defined on a domain U0 containing 0. Suppose 0is a neutral fixed point of f , i.e., f(0) = 0 and λ = f ′(0) with |λ| = 1.Let ∆r = z ∈ C | |z| < r and ∆ = ∆1. Suppose ∆r ⊂ U0. Let Kr be

the maximal completely f -invariant set in ∆r. Let φ : C \Kr → ∆ bethe Riemann mapping with φ(∞) = 0. Let A = φ(U \Kr) ⊂ ∆ be an

annulus. Then f = φ f φ−1 : A→ ∆ is a holomorphic map. By theSchwarz reflection principle, we can define a holomorphic map

F : A ∪ αA ∪ S1 → F (A ∪ A ∪ S1)

where α(z) = 1/z. Then g = F |S1 : S1 → S1 is an analytic diffeomor-phism.

Theorem 43. Suppose λ = e2πiθ, θ ∈ [0, 1). Then the rotation numberρ(g) = θ.

65

Before to prove this theorem, we discuss the Caratheodory- topologyon domains. Let Un be a sequence of domains in C and U is anotherdomain in C. Suppose all Un have a common interior point w0. Wesay Un tends to U in kernal with respect to w0 if

(i): U = w0 or U is a domain containing w0 and for any z ∈ U ,there is a neighborhood V about z and N > 0 such that V ⊂ Unfor all n > N ,

(ii): for any z ∈ ∂U , there is a point zn ∈ ∂Un such that zn tendsto z as n goes to infinity.

Suppose fn : ∆→ Un is univalent such that fn(0) = w0 and f ′n(0) > 0for every n > 0. Suppose f : ∆→ U is univalent such that f(0) = w0

and f ′(0) > 0. If fn tends to f uniformly on compact sets as n goes toinfinity, then Un tends to U in kernal as n goes to infinity. Conversely,if Un tends to U in kernal as n goes to infinity, then fn tends to funiformly on compact sets as n goes to infinity. Note that if fn tendsto f uniformly on compact sets as n goes to infinity, then f−1

n tends tof−1 uniformly on compact sets as n goes to infinity. Suppose Kn isa sequence of compact sets in C. If Kn tends to K in the Hausdorfftopology as n goes to infinity, then C \ Kn tends to C \ K in kernalas n goes to infinity. The Caratheodory topology on domains may notdescribe as how close of domains in distance. This can be seen by thefollowing example. Let Un = C\

((−∞,−1/n]∪[1/n,∞)

). Let U be the

upper-half plane. Let w0 ∈ U . Then one can see that Un tends to U inkernal with respect to w0 as n goes to infinity. Suppose fn : ∆→ Un isthe Riemann map such that fn(0) = w0 and f ′n(0) > 0. Let f : ∆→ Ube the Riemann map such that f(0) = w0 and f ′(0) > 0. Then fn tendsto f uniformly on compact sets. Note that Jn = f−1

n (lower-half plane)tends to a point in S1 as n goes to infinity. We use the Caratheodorytopology in the proof of Theorem 43.

Proof of Theorem 43. Suppose f(z) = λz + anzn+1 + · · · , n ≥ 1. If all

an = 0, then it is clear. Suppose an 6= 0 for the smallest n ≥ 1. Ifθ ∈ Q, then f has a completely f -invariant set Kr in ∆r made by mpetals ending at 0. Let γ be a curve in ∆r \ Kr connecting 0 and aboundary point of U . If θ = 0, then we can find γ such that f(γ) = γ.

Let γ = φ(γ) ∪ α(φ(γ)). Then f(γ) = γ. That means that g has afixed point in S1. Therefore, ρ(g) = 0. If θ = p/q such that p 6= 0 and(p, q) = 1, then we can find such a curve γ such that fnq(γ) = γ. Letγi = f i(γ) for i = 0, 1, · · · , nq − 1. Then f(γi) = γi+p (mod q). Thus ghas a periodic point s in S1 such that for si = gi(s), i = 0, 1, · · · , nq−1,g(si) = si+p (mod q). Therefore, ρ(g) = p/q.

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Now suppose θ ∈ [0, 1] \ Q. Let θn be a sequence of rationalnumbers such that θn → θ as n → ∞. Let fn = e2πi(θn−θ)f . Thenfn is defined on U0 and has an invariant set Kr,n in ∆r. Let gn bethe corresponding circle diffeomorphism. Then ρ(gn) = θn. Supposeφn : C\Kr,n → ∆ is the Riemann map with φn(∞) = 0 and φ′n(∞) > 0.

Suppose fn = φnfnφ−1n be the corresponding map defined on annulus

An = φn(U \ Kr,n). Let S±τ = z ∈ C | |z| = 1 ± τ where τ is anumber close to 0. Since Kn,r tends to Kr in the Hausdorff topology

as n goes to infinity, fn|S±τ → f as n → ∞. The maximal principleimplies that gn|S1 → g|S1 as n→∞. So ρ(gn)→ ρ(g) as n→∞.

Let G = f(z) = λz +O(z2) | |λ| = 1 be the set of all holomorphicgerms at 0. Let A be the set of λ, |λ| = 1, such that f(z) = λz+O(z2)is linearizable. Then Ac is the set of λ, |λ| = 1, such that thereexists a germ f(z) = λz + O(z2) such that f is not linearizable. LetS = g : S1 → S1 be the set of all holomorphic diffeomorphisms ofthe circle. Let B be the set of rotation numbers ρ such that for anyg ∈ S with ρ(g) = ρ is holomorphically conjugate to the rigid rotationwith rotation number ρ. Then Bc is the set of rotation numbers ρ suchthat there exists g ∈ S such that g is not holomorphically conjugateto the rigid rotation with rotation number ρ. Then one can check thatA = exp(2πiB).

Suppose f is univalent on U0 and suppose 0 is an irrational neutralfixed point of f . A domain 0 ∈ S ⊂ U is called a Siegel disk of f ifthere is a conformal map φ : S → ∆1 such that φ f φ−1(z) = λz forz ∈ ∆.

Theorem 44. Suppose S is a Siegel disk of f and S ⊂ U0. Then thereis no periodic point of f on ∂S.

Proof. Suppose f has a periodic orbit C = x1, · · · , xn on ∂S. SinceS is a Siegel disk of f , there is a conformal map φ : S → ∆ suchthat φ f φ−1(z) = λz. Let rn = z ∈ C | |z| = 1 − 1/n is acircle in ∆. Let Rn = φ−1(rn) be a curve in S. Let Un be the domainbounded by Rn and Kn be the maximal f -invariant set in Un. ThenKn tends to K = S in the Hausdorff topology as n goes to infinity. Let

Vn = C \Kn and V = C \ S. Let φn : Vn → ∆ be the Riemann mapwith φn(∞) = 0 and φ′n(∞) > 0 and let φ : V → ∆ be the Riemann

map with φ(∞) = 0 and φ′(∞) > 0. Denote fn = φn f φ−1n and

f = φ f φ−1. Let Cn = φn(C). Then fn (or f) can be extendedto a circle diffeomorphism gn (or g) of S1 (by applying the Schwarzreflection principle). Since gn(Cn) = Cn, gn has a periodic orbit. Since

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the number of Cn is less than or equal to n, so there is a subsequenceCnk tends to C = φ(C) ⊂ S1 in the Hausdorff topology. The number

of C is less than or equal to n too. Since g(C) = C, g has a periodicorbit in S1. This contradicts with the fact that the rotation numberρ(g) is irrational.

The following lemma used to be useful in the proof of the non-wandering domain theorem (Theorem 48) for rational maps. Sincewe will use infinitesimal deformations to prove Theorem 48 so we willnot use this lemma in the proof of Theorem 48. But it is still fan toknow it.

Lemma 11 (Snail Lemma). Suppose f 6≡ id is an analytic map definedon a domain U . Suppose 0 ∈ U is a fixed point of f such that |f ′(0)| =1. If there is a curve γ in U ending at 0 and a point in ∂U such thatf(γ) ⊂ γ, then f ′(0) = 1.

Proof. The map f can have only finite many fixed points in γ becausef is not identity. By cutting γ short, we can assume that f has nofixed point in γ except for 0. Under this assumption, f is monotone onγ. This means that there is a natural order on γ from 0 to the otherend of γ, for z1 < z2 in γ, f(z1) < f(z2) (decreasing) or f(z1) > f(z2)(increasing).

Since |f ′(0)| = 1, there is a domain U ′, U ′ ⊂ U , such that f |U ′ isunivalent. Assume γ connects 0 and a point z in ∂U ′. Let K be themaximal completely f -invariant set in U ′. Then K ∩ γ = 0 sincef−1(γ) 6= γ if f is decreasing and f(γ) 6= γ if f is increasing. (Notethat f−1(z) 6∈ U ′ if f is decreasing and f(z) 6∈ U ′ if f is increasing.) Let

φ : C \K → ∆ be the Riemann map with φ(∞) = 0 and φ′(∞) > 0.Then φ(U \K) is an annulus in ∆ and η = φ(γ) is a curve connectingφ(∂U ′) and S1. Let g = φf φ−1. By the Schwarz reflection principle,g can be extended to an analytic diffeomorphism of S1. We still denotethis diffeomorphism as g. Since g(η) ⊂ η, g has a fixed point on S1.This implies that the rotation number ρ(g) = 0. Therefore, f ′(0) =1.

Suppose f(z) = λz + a2z2 + · · · , λ = e2πiθ, is a germ at 0. Then

it is clear that λ is an analytic invariant, i.e., if h is a conformal mapdefined on a neighborhood about 0, then hf h−1(z) = λz+b2z

2+· · · .Moreover, we have that

Theorem 45. The multiplier λ is a topological invariant. More pre-cisely, if f1(z) = λ1z+ a2z

2 + · · · , λ1 = e2πiθ1 and f2(z) = λ2z+ b2z2 +

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· · · , λ2 = e2πiθ2, are two germs at 0 and if h is a homeomorphismdefined on a neighborhood about 0 such that f1 = h f2 h−1, thenλ1 = λ2.

Proof. Suppose f1 and f2 are both univalent on a domain 0 ∈ U0 andh is defined on U0. Let ∆r = z ∈ C | |z| < r be a disk suchthat ∆r ⊂ U0. Let K1 and K2 be the maximal completely f1 and f2

invariant sets in ∆r. Let φ1 : C \ K1 → ∆ and let φ2 : C \ K2 → ∆be the Riemann maps with φ1(∞) = φ2(∞) = 0 and φ1(∞) > 0 andφ2(∞) > 0. Then A1 = φ1(U \K1) and A2 = φ2(U \K2) are annuli in∆. Define

f1 = φ1 f1 φ−11 : A1 → ∆, and f2 = φ2 f2 φ−1

2 : A2 → ∆.

By the Schwarz reflection principle, both f1 and f2 can be extendedto analytic diffeomorphisms of S1. We use g1 and g2 to denote thesediffeomorphisms. Let H = φ2 h φ−1

1 : A1 → A2. Then H can beextended to a homeomorphism of S1. We still denote this homeomor-phism as H. Then g1 = H g2 H−1 on S1. This implies that therotation numbers ρ(g1) = ρ(g2). So λ1 = λ2.

Next we are going to have a study of the Fatou set F . Suppose fis a rational map of degree d > 1. Suppose O = zi = f i(z)n−1

i=0 isan attractive or super-attractive or rational neutral periodic point ofperiod n. Define

B = BO = z ∈ C | fn(z)→ O, as n→∞the basin of O. Then B ⊂ F . On the other hand, for any domain Usuch that U ∩ ∂B 6= ∅, then fn(z) → O as n → ∞ for z ∈ U ∩ Bbut fn(z) 6→ O as n → ∞ for z ∈ (C \ B) ∩ U . So fn|U∞n=0 isnot normal. Therefore, the boundary of the basin of any attractive orsuper-attractive periodic orbit of f is contained in the Julia set. Sincef(B) = f−1(B) = B, ∂B ⊂ J is completely invariant and closed. SoJ = ∂B from Theorem 30. It is clear that BO ⊂ F .

Denote B∗i,O be the component containing zi. Then B∗O = ∪n−1i=0 B

∗i,O

is called the immediate basin of O.

Theorem 46. If O is attractive or super-attractive, then the immediatebasin B∗O contains at least one critical point of f .

Proof. If O is super-attractive, then (fn)′(z0) = 0. So one of the pointsin O is a critical point. Suppose O is attractive. Since fn(z0) = z0 and0 < |λ| = |(fn)′(z0)| < 1, there is a neighborhood U1 about z0 such thatfn|U1 : U1 → U0 ⊂ U1 is homeomorphic. Let g : U0 → U1 be its inverse.

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Then |g′(z0)| = 1/|λ| > 1. Take U1 simply connected. If U1 containsno critical point from fn, then g can be extended holomorphicallyto U1, we still denote this extension as g and let U2 = g(U1), whichis still simply connected. Inductively, suppose we have defined Unsuch that Un is simply connected and g : Un−1 → Un is an analyticdiffeomorphism. If Un contains no critical point from fn, then g canbe extended to Un as an analytic diffeomorphism and we can defineUn+1 = g(Un). Thus, if B∗(0, O) contains no critical point from fn,then we can define a sequence of analytic maps gn : U0 → Un forall n > 0. This is a normal family because it omits the Julia set off which contains more than three points. So there is a subsequencegni : U0 → Uni converging uniformly to a holomorphic map g0 definedon U0. The Cauchy formula implies that (gni)′(z0) closes to g′0(z0) fori large. But since |g′(z0)| > 1, |gn(z0)| → ∞ as n → ∞. This isimpossible. Therefore, there is n such that Un contains a critical pointfrom fn. So B∗O contains a critical point from f .

Theorem 47. If O is rational neutral, then the immediate basin B∗Ocontains at least one critical point of f .

Proof. Without loss of generality, we may assume f(z0) = z0 andf ′(z0) = 1. Then the immediate basin B∗O contains an attracting petalS at z0. Let φ be the conformal map on S such that φ(f(z)) = φ(z)+1.Since f(B∗O) = B∗O, φ can be extended to B∗O, where we still haveφ((f(z)) = φ(z) + 1. Since φ(S) contains a half-plane, φ(B∗O) ⊃ C.If φ has no critical point, then there is a branch ψ of φ−1 such thatψ : C → B∗O is an entire function. Since B∗O ∩ J = ∅, ψ can be only aconstant function. It is impossible. So φ has a critical point c in B∗O,i.e., φ′(c) = 0. Since fn(c)→ z0 as n→∞, we have an integer m > 0such that fm(c) is in the attracting petal where φ is conformal. Sinceφ(fm(c)) = φ(c) + m, φ′(fm(c))(fm)′(c) = φ′(c) = 0. So (fm)′(c) = 0.Therefore, f has a critical point in B∗O.

Corollary 6. Suppose the degree of f is d. Then the number of at-tractive and super-attractive periodic orbits plus the number of rationalneutral periodic orbits is at most 2d− 2.

Proof. A degree d rational function f has at most 2d−2 critical points.

Remark 11. The total number of attractive period orbits and neutralperiodic orbits (including rational and irrational) is less than or equal

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2d−2. For a polynomial of degree d > 1, the total number of attractiveand super-attractive period orbits and neutral periodic orbits (includingrational and irrational) in the complex plane is less than or equal d−1.Refer to

• M. Shishikura, On the quasiconformal surgery of rational func-tions. Ann. Sci. Ecole Norm. Sup., 20, 1987, 1-29.

Suppose R = P/Q is a rational function whose degree d > 1.

Definition 20 (Hyperbolicity by Critical Points). We say that R(z)is hyperbolic if every critical orbit R tends to an attractive or super-attractive periodic orbit.

Recall that we used to define the hyperbolicity as

Definition 21 (Hyperbolicity by Euclidean Metric). We say thatR(z) is hyperbolic if there are constants C > 0 and λ > 1 such that|(Rn)′(z)| ≥ Cλn for all z ∈ J and all n ≥ 1.

Then the above two definitions are equivalent follows by hyperbolicgeometry and the non-wandering domain theorem and the classificationof Fatou components. Let us have these now.

Let F be the Fatou set for R. A connect component of F is called aFatou component of R. If U is a Fatou component, then R(U) is also aFatou component and R−1(U) consists of at most d Fatou components.If R(U) = U , we call U a fixed Fatou component. If Rn(U) = U forsome n ≥ 1, we call U a periodic Fatou component. The minimal n iscalled the period. If Rm(U) is periodic but U is not periodic, we callU a preperiodic Fatou component. If all Rn(U) are distinct, we callU a wandering domain.

Theorem 48 (Sullivan). A rational map has no wandering domains.

We will prove it in Lecture 8. An important conclusion of this theo-rem is a complete classification of Fatou components.

Theorem 49. Let U be a Fatou domain. Then there is a positiveinteger m such that Rm(U) is periodic. Suppose U itself is periodic ofperiod n, then U can be only following cases:

(1): Attracting domain: U contains an attracting fixed point z ofRn and any point w ∈ U tends to z under iterates of Rn.

(2): Super-attracting domain: U contains a super-attracting fixedpoint z of Rn and any point w ∈ U tends to z under iterates ofRn.

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(3): Parabolic domain: ∂U contains a rational neutral fixed pointz of Rn with multiplier 1 and any point w ∈ U tends to z underiterates of Rn.

(4): Siegel disk: U is simply connected and contains an irrationalneutral fixed point z of Rn such that there is a conformal mapφ : U → D1 with φ(z) = 0 and φ′(z) > 0, φ Rn φ−1(w) = λwfor w ∈ U .

(5): Herman ring: U is an annulus such that there is a conformalmap φ : U → D1\Dr, 0 < r < 1, such that φRnφ−1(w) = λw(or φRnφ−1(w) = r(λw) where r is an inversion) for w ∈ U .

Let C be the (finite) set of all critical points of R. Let CO =∪∞i=0R

i(C) be the set of all critical orbits of R. The set CO is importantbecause it is the complement all branches of R−n, n ≥ 1, are locallydefined and analytic. As we have proved every attracting domain orsuper-attracting domain or parabolic domain contains a critical point.A Herman ring or Siegel disk contains no critical point. But we havethat

Theorem 50. If U is a Siegel disk or Herman ring, then ∂U is con-tained in CO.

Proof. Let U be a Siegel disk or Herman ring. Suppose the period of Uis n. Let g = Rn. Then g(U) = U . Suppose CO does not contain ∂U .Let D be an open disk disjoint from CO which meets ∂U . Since U is aSiegel disk or Herman ring, we can pick D such that D is disjoint fromsome open invariant non-empty set V of U . Define fn to be any branchof g−1 on D. Since all the fn omit V , they form a normal family onD. Since g is injective on D, there are other components of g−1(U).Since ∪∞n=0R

−n(z) is dense in the Julia set J for any z ∈ J , there is aninteger m > 0 such that some component W of g−m(U) distinct fromU meets D. If w ∈ W ∩ D, then fj(w) and fk(w) belong to differentcomponents of the Fatou set F for j 6= k, or else they belong to aperiodic component, which could not be iterated eventually to W thenU . Hence fk(w) tends to J for w ∈ D∩W , and since J has no interior,any limit of the normal family fk is constant on D∩W . On the otherhand, since fk are rotations of U , any limit of fk is non-constant onW ∩D. This is a contradiction.

We callR a critical finite rational function if CO is finite. In this case,every point c ∈ CO is periodic or preperiodic, that is, Rm(c) is periodicfor some m > 0. If c is periodic, it must be an super-attracting periodicpoint. If R is critical finite and has no super-attracting periodic points,

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then every c ∈ CO is preperiodic. Thus we call such a rational mappreperiodic.

Theorem 51. If R is preperiodic, J = C.

Proof. If R has Siegel disks or Herman rings, then CO will be infinitefrom the previous theorem. So R has no Siegel disk and Herman ring.Also R has no attracting, super-attracting, and parabolic domain be-cause, otherwise, either R has a super-attracting periodic point or COis inifinity. From the no wandering domain theorem, the Fatou set F

is empty. So the Julia set J = C.

Following this theorem, we have an example of a rational functionwhose Julia set is the whole Riemann sphere.

Example 15 (Lattes Map). Let

R(w) = 1− 2

w2.

Then the Julia set J = C.

Proof. The critical points of R are 0 and ∞. One can check that

R(0) =∞, R(∞) = 1, R(1) = −1, R(−1) = −1,

and −1 is an expanding fixed point of R. So R is preperiodic.

A Fatou component U is called completely invariant if R(U) =R−1(U) = U . For example, if P is a polynomial, then the basin B∞ ofthe infinity is a fixed Fatou component and completely invariant. ForP (w) = w2−1, P (0) = −1 and P (−1) = 0, the immediate basin B∗(0)of 0 is a periodic Fatou component of period 2.

Theorem 52. If U is a completely invariant Fatou component, thenJ = ∂U and every other Fatou component is simply connected. Thereare at most two completely invariant Fatou components.

Proof. If U is completely invariant, then ∂U = R−1(∂U) = R(∂U) ⊆ J .So ∂U = J .

Consider V = C\U . Since Rn|V omits U , Rn|V is normal and V ⊂F . Since U is connected, every component of V is simply connected.

If U further is also simply connected, then R : U → U is a degree dmap and it must contain d − 1 critical points of R. Since R can hasonly 2d− 2 critical points, R can has at most two completely invariantFatou components.

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Theorem 53. The number of Fatou components of a rational functioncan be 0, 1, 2, or ∞ and all this number occurs.

Proof. Suppose F has only finitely many Fatou component and let U0

be one of them. Consider a chain of inverse images

R(U−1) = U0, R(U−2) = U−1, · · · , R(U−n) = U−n+1.

There is an integer n > 0 such that R(U−n) = U−k+1 = R(U−k) for0 ≤ k < n. Then U0 = Rn(U−n) = Rn(U−k) = Rn−k(U0). Thus eachFatou component U0 is periodic. Since there are only finite number ofFatou components, there is an integer N > 0 such that RN(U) = U forevery Fatou component. Hence each component is completely invariantfor RN . There are at most two such Fatou components. For the Lattesexample which we will lecture later, it has 0 Fatou component; forP (w) = w2−2, it has one Fatou component; for P (w) = w2, it has twoFatou component. Most of rational maps have ∞ Fatou components.

Now we prove that

Theorem 54. Definition 21 and Definition 20 are equivalent.

Proof. Suppose Definition 21 holds. Then there is a neighborhood Ucontains J such that |(Rm)′(z)| > 1 for all z ∈ U . This implies that

U ⊂ Rm(U) and all critical point is outside of U . Let V = C\U . Then

V ⊂ F and Rn(V ) ⊂ V . From Theorem 48, every component W of

V is eventually periodic. So we suppose Rm(W ) ⊂ W . Since W is ahyperbolic Riemann surface with the hyperbolic metric dH,W , we havea real number 0 < τ < 1 such that

dH,W (Rm(z), Rm(w)) ≤ τdH,W (z, w), z, w ∈ W1 = Rm(W ).

From the contracting mapping theorem, Rm has a fixed point p ∈ W .Then p is a super-attracting or attracting fixed point of Rm and allcritical points in components of V which are eventually mapped to Ware convergent to p under iteration of R. This implies Definition 20.

Now suppose Definition 20 holds. Let p1, · · · , pk are all attractingand super-attracting periodic points. Let B∗(pi) be the immediatebasin of pi. Let B = ∪ki=1B∗(pi). Then we have an integer m > 0such that B0 = R−m(B) contains all critical points. Thus CO is aclosed set disjoint with J . Thus we have a neighborhood U ⊃ J suchthat R(U) ⊃ U and R(U) ∩ CO = ∅. Every component of U is ahyperbolic Riemann surface. Thus R strictly expands the hyperbolicmetric dH,U , that is, dH,U(R(z), R(w)) ≥ λdH,U(z, w) for some λ > 1

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and all z, w ∈ U . The hyperbolic metric dH,U and the Eucildean metricare equivalent on J (by assume∞ 6∈ J). This implies Definition 21.

Finally, we mention a theorem which is useful in the study of complexdynamics.

Theorem 55 (Denjoy-Wolf). Let f : ∆→ ∆ be an analytic function.Suppose f is not an elliptic Mobius transformation. Then there is apoint α ∈ ∆ such that for any z ∈ ∆, fn(z)→ α as n→∞.

Proof. Suppose f is a Mobius transformation. Then either M has onefixed point or two fixed points. If f has one fixed point a. Thenlet M(z) = 1/(z − a). then g(z) = M f M−1(z) fixes ∞. Thusg(z) = z + b, b 6= 0. Thus fn(z) → a as n → ∞. If f has two fixedpoints a 6= b, then a, b ∈ ∂∆. Let M(z) = (z − a)/(z − b). Theng(z) = M f M−1(z) fixes 0 and ∞. Thus g(z) = αz. If |α| = 1,then f is elliptic, so by the assumption, |α| < 1. Thus fn(z) → a asn→∞.

Now suppose f is not a Mobius transformation. It contracts thehyperbolic metric dH , that is, dH(f(z), f(w)) < dH(z, w) for z, w ∈ ∆.

Consider fε(z) = (1 − ε)f(z). The image fε(∆) is contained in acompact set of ∆. Thus it contracts dH strictly on fε(∆), that is, thereis a real number 0 < τε < 1 such that

dH(fε(z), fε(w)) ≤ τεdH(z, w), z, w ∈ fε(∆).

The contracting mapping theorem says fε has a unique fixed pointαε ∈ ∆. Let Dε be the hyperbolic disk of radius dH(0, αε) centered atαε. Then we have fε(Dε) ⊂ Dε.

The we have two cases. The first one is that there is a limit point αof αε as ε→ 0+ in ∆. In this case, Let D be the corresponding limitinghyperbolic disk of Dε, then f(D) ⊂ D. Since 0 ∈ ∂D, we have thatfn(0) → α as n → ∞. This implies that α is the unique limit pointof αε as ε → 0+ and f(α) = α. Now we can use either the Schwarzlemma or hyperbolic geometry to prove that for any z ∈ ∆, fn(z)→ αas n→∞.

The other case is that all limit points of αε are on ∂∆. Any limitinghyperbolic disk D of Dε has 0 on its boundary and is tangent with ∂∆at a point α. This implies that f(D) ⊂ D and fn(0) → α as n → ∞.Since fn(0)∞n=1 ⊂ D, α is the unique limiting point of αε. For anyz ∈ ∆, we have an integer k > 0 such that fkε (z) ∈ Dε. This impliesthat fk(z) ∈ D. So fn(z)→ α as n→∞.

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Lecture 7. Lebesgue density and areas of Julia setsrevisited.

Suppose A is a measurable subset of C. For any point p ∈ A. let∆ε(p) = z ∈ C | |z− p| < ε be the disk of radius r > 0 centered at p.Let Area(S) mean the Lebesgue measure of a measurable set S ⊂ C.We say p is a density point of A if the following limit exists and

limε→0

Area(A ∩∆ε(p))

Area(∆ε(p))= 1.

Theorem 56 (Lebesgue Density Theorem). If AreaA > 0, then al-most all points p ∈ A are Lebesgue density points.

Using the Koebe distortion and the Lebesgue density theorem andproperty of the Julia set, we have another proof of Corollary 1.

Theorem 57. The Julia set J of a hyperbolic rational map f has zeroLebesgue measure, that is, Area(J) = 0.

Proof. Since a hyperbolic rational map has an attractive or super-attractive periodic point, its Fatou set is not empty. So its Julia set

is not the whole Riemann sphere C. Let C be the set of all criti-cal points of f and let P = ∪∞n=1f

n(C) be the post-critical set of f .Then P ∩ J = ∅. Let ε0 > 0 be a real number such that the disk∆ε0(p) ∩ P = ∅ for any p ∈ J . Let ε = ε0/4. Then we claim that wehave a number a > 0 such that Area(∆ε(p) \ J) ≥ a for any p ∈ J .Let us prove it by contradiction. Suppose we have a sequence pn ∈ Jsuch that Area(∆ε(pn) \ J) < 1/n for all n > 1. Since J is compact,we have convergent subsequence, which we still denote as pn, such thatpn → p ∈ J . Then Area(∆ε(p) \ J) = 0. This implies that J is dense

in ∆ε(p). Thus ∆ε(p) ⊂ J . But if J has an interior point, then J = Cwhich is impossible.

Now take arbitrary point p ∈ J . Let pn = fn(p). Consider the in-verse branch gn of fn mapping pn to p. Then gn can be analyticallyextended to ∆ε0(pn) as a conformal map gn : ∆ε0(pn) → gn(∆ε0(pn)).The Koebe’s 1/4-theorem (Theorem 22) implies that gn(∆ε0(pn)) con-tains a round disk g′n(pn) ·∆ε(p) which is a disk of radius εn = |g′n(pn)|εcentered at p. Since f |J is expanding and g′n(pn) = 1/(fn)′(p), wehave that εn → 0 as n→∞. Now from the Koebe distortion theorem(Theorem 23), we have a constant C > 0 such that

C−1|g′n(pn)||z − p| ≤ |gn(z)− gn(pn)| ≤ C|g′n(pn)||z − p|

76

and

C−1 ≤ |g′n(ξ)||g′n(ξ)|

< C

for all n ≥ 1, p ∈ J , ξ, η ∈ ∆ε(pn). This implies that

Area(∆εn(p) \ J)

Area(∆εn(p))≥ C

∫∆ε(p)\J |g

′n(ξ)|dξ∫

∆ε(p)|g′n(η)|dη

.

Furthermore,

Area(∆εn(p) \ J)

Area(∆εn(p))≥ C2Area(∆ε(p) \ J)

Area(∆ε(p))≥ C2a = b > 0.

So we have thatArea(J ∩∆εn(p))

Area(∆εn(p)< 1− b.

Thus p is not a Lebesgue density point. This implies that Area(J) =0.

Similarly, replacing the Koebe distortion by the naive distortion(Lemma 1), we have a proof of Theorem 5 by using the Lebesgue den-sity theorem.

77

Lecture 8. Sullivan’s no wandering domain theo-rem.

In this lecture, we will prove Theorem 48.Let Rat(d) be the space of all degree d > 1 rational maps. Then

it can be identified with an open set of the 2d + 1 projective spacePC2d+1 = C2d+2/complex lines, that is, we can correspond a rationalmap

g(z) =adz

d + · · · a0

bdzd + · · ·+ b0

∈ Rat(d)

to

ι(g) = (ad, ad−1, . . . , a0, bd, bd−1, . . . , b0)/complex lines ∈ PC2d+1.

Thus, for g in Rat(d), the tangent space TfRat(d) = C2d+1. Recallthat the definition of TgRat(d) is the space of equivalent classes [gt] ofall C1 paths gt in Rat(d) passing g at t = 0, where two paths gt and gtare equivalent if

dgtdt

∣∣∣t=0

=dgtdt

∣∣∣t=0.

Let T C be the continuous tangent bundle over C. By the definition,

the continuous tangent bundle T C is the space of all equivalent classes

[ht] of C1 paths ht in the space of all homeomorphisms Hoem(C) pass-

ing the identity, where two paths ht and ht are equivalent if

v =dhtdt

∣∣∣t=0

=dhtdt

∣∣∣t=0.

Here v can be viewed as a continuous vector field on C. We normalizeall paths fixing 0 and 1 and∞. Then the only holomorphic vector fieldis the zero vector field.

Let f be the given rational map in Theorem 48. For any v = [ht] ∈T C, we have a C1 path

ft = h−1t f ht

of degree d branched coverings of C. The tangent vector, that is theequivalent class of this C1 path, is

Dv =dftdt

∣∣∣t=0

= f ′(z)v(z)− v(f(z)).

Thus we can define a linear operator D from T C to the tangent space ofthe space of all degree d branched coverings at f . Note that TfRat(d)is a subspace of the later tangent space. Define

H = D−1(TfRat(d)).

78

It is a subspace of T C. Thus we have a linear operator, which we stilldenote as D, from H onto TfRat(d). Let

K = Ker(D) = v ∈ H | Dv = 0

be the kernel of D. Then we have the induced operator

D : QH = H/K → TfRat(d)

from D. It is surjective (1-1 and onto). Since TfRat(d) is a 2d + 1dimensional vector space, the key point in the proof of Theorem 48 isto show that QH is an infinite dimensional vector field space if f hasa wandering domain. This will give us a contradiction.

An infinite dimensional vector space V∆ on the open unit disk ∆ caneasily constructed as the space of all vector fields

v∆(z) =

∑nk=1 akz

k, |z| ≤ 12∑n

k=1ak4kz−k, 1

2≤ |z| < 1

where ak are complex numbers. In this vector space, only holomorphicvector field is the zero vector field. Moreover, since v∆ has an analyticextension in a neighborhood of ∂∆, v∆|∂∆ = 0 if and only if v∆ ≡ 0on ∆.

Now let us give a characterization of an element v ∈ H which isequivalent to say that Dv is holomorphic. From the complex analysis,Dv is holomorphic if and only if

∂Dv(z) = f ′(z)∂v − ∂v(f(z))f ′(z) = 0.

Let µ = ∂v. Then v ∈ H if and only if µ is f -invariant, that means

(2) µ(z) = µ(f(z))f ′(z)

f ′(z), a.e. z ∈ C.

The key point in this calculation is that we only need µ(z) is defined

almost everywhere on C and is measurable and in L∞(C), that is,‖µ‖∞ <∞. (As long as v is a Zyhmund continuous function, we havethe above calculation. Thus we have the above calculation for v∆.)

Conversely, if µ(z) ∈ L∞(C) has a compact support in C, define

(3) v(z) = Pµ(z) =1

π

∫ ∫C

µ(ζ)

z − ζdξdη, ζ = ξ + iη.

Then v is a continuous vector field on C (actually Zgymund continuous

vector field on C) and one can check that ∂v = µ. Thus if µ is f -invariant (that is, it satisfies Equation (2)), then v ∈ H.

79

The following lemma characterizes v when it is in K = Ker(D).Recall J is the Julia set of f .

Lemma 12. If v ∈ K, then v|J = 0.

Proof. Since Dv = 0, if z is a repelling periodic point of period n > 0,then

(fn)′(z)v(z) = v(fn(z)) = v(z).

This implies that v(z) = 0 since |(fn)′(z)| > 1. This implies thatv|J = 0 since the set of all repelling periodic points is dense in J andv is continuous.

We also need the following Lemma to embed V∆ into QH.

Lemma 13. If f has a wandering domain, then it must have a simplyconnected wandering domain.

Proof. Suppose U is a wandering domain. Let Un = fn(U) for n > 0.Then Un ∩ Um = ∅ for all n 6= m. Since f has only finite number ofcritical points, there is an integer N ≥ 0 such that all Un contains nocritical points for n ≥ N . This implies that

UNf−→ UN+1

f−→ · · · f−→ Unf−→ Un+1

f−→ · · ·is a chain of covering maps and fn|UN is a normal family. First Uncan not be a punched disk since ∂Un ⊂ J and J has no isolated point.Assume that∞ is in a Fatou component V which is not one of Un∞n=N .

Since∑∞

n=N Area(Un) < ∞, we have (fn)′(z) → 0 as n → ∞;otherwise, by 1/4-Kobe Theorem, we have infinitely many Un containsdisks of a definite radius, the sum of the areas of these disks is infinite,it is impossible. If Um is not simply connected for all m. Let γ0

be a simply closed curve in Um enclosing some points from the Juliaset. Then the diameter of γn = fn(γ0) tends to zero as n → ∞.The bounded component Dn of C \ γn contains no pole of f sincefn(Um) 6= V . Since f is a Lipschitz map, the bounded component Dn

of C\γn is mapped to the bounded component Dn+1 of C\γn+1. Thusthe diameter of Dn tends to zero as n goes to∞. But since D0 containspoints in the Julia set, the image Dn = fn(D0) should be bigger andbigger. This is a contradiction. The contradiction implies that one ofUn must be simply connected.

Proof of Theorem 48. The proof is to embed V∆ into QH if f has awandering domain.

Suppose U is a simply connected wandering domain. Since ∂U ⊂ J ,the limit set (∂U)′ 6= ∅ (thus there are at least two points on the

80

boundary), we have a Riemann mapping φ : U → ∆. Without loss ofgenerality, we assuming ∞ in U such that 1/2 < |φ(∞)| < 1.

For any v∆ ∈ V∆, we have a pullback vector field

vU(z) = φ∗v∆(z) = v∆(φ(z))

on U . Let µU = ∂vU . Then µU(z) = 0 if 1/2 < |φ(z)| < 1. Moreover,µU is continuous except on the analytic curve

γ = φ−1(z | |z| = 1

2

).

Let U0 = U and Un = fn(U) for all integers n ≥ 1. Note thatUn ∩ Un′ = ∅ for any n 6= n′ ≥ 0. Then let Unm = f−m(Un) for allm ≥ 0 and n ≥ 0. The grand orbit of U is, by definition,

U = ∪∞m=0 ∪∞n=0 Unm,

which is the union of pair-wise disjoint domains Unm. Pullback vUagain by using all inverse branches gn of fn, n ≥ 0, and fm, m ≥ 1, we

define a vector field vU on U , that is, define

vUn(z) = vU(gn(z))

on Un for n ≥ 0 and define

vUmn(z) = vUn(fm(z))

on Umn for all m ≥ 1 and n ≥ 0. Then vU is a continuous vector field

on U . Let µU = ∂vU . Then µU extends µU and is continuous except onγ = ∪m≥1 ∪n≥0 f

−m(fn(γ)), a union of countable number of analyticcurves. It is clear that γ has zero Lebesgue measure. Note that µU isf -invariant from our definition due to the fact that all Unm are pairwisedisjoint. Now define

µ(z) =

µU(z), z ∈ U ;

0, z ∈ C \ U .

From our definition, µ is continuous on C except on γ and has a com-pact support in C. µ is also f -invariant. Thus we have a continuous

vector field v = Pµ on C from Equation (3) such that ∂v = µ. Since

∂(v − vU) = 0, a. e. on U , h = v − vU is a holomorphic vector field on

U . Since µ is f -invariant, v ∈ H and Dv ∈ TfRat(d). This defines alinear map v = Lv∆ from V∆ into H.

If v = Lv∆ ∈ K, Lemma 12 implies that v|J = 0 and thus v|∂U =0. Since the limit set (∂U)′ 6= ∅ and v = h + vU is analytic on

81

φ−1(z | 1/2 < |z| < 1), we have v = 0 on φ−1(z | 1/2 < |z| < 1).This implies that

h(φ−1(z)) = −vU(φ−1(z)) = −v∆(z) = −n∑k=1

ak4kz−k, 1/2 < |z| < 1.

So we have that

h(φ−1(z)) = −n∑k=1

ak4kz−k, |z| < 1.

This implies that ak = 0 for all 1 ≤ k ≤ n. Thus h ≡ 0 on U , v∆ ≡ 0

on ∆, and v ≡ 0 on C.Concluding from the above, L embeds V∆ into QH. More precisely,

D L : V∆ → TfRat(d) = C2d+1

is an injective linear operator. But V∆ is an infinite dimensional vec-tor space. It is impossible. The contradiction completes the proof ofTheorem 48.

82

Lecture 9. Hausdorff dimension

Let (X, d) be a metric space. If U is a subset of X, we use diam(U)to denote the diameter of U . Suppose E is a subset of X. We say Uiis a δ-cover of E if every diam(Ui) ≤ δ and E ⊂ ∪iUi.

For δ > 0 and s > 0, we define

Hsδ(E) = inf

∞∑i=1

(diam(Ui))s

where the infimum is over all countable δ-covers Ui of E. It is clearthat for fixed s, Hs

δ(E) is increasing with respect to δ. Now let δ → 0,we define

Hs(E) = limδ→0Hsδ(E).

We will show that Hs(·) defines an s-Hausdorff (outer) measure on X.Recall a function µ(·) : 2X → [0,+∞] is called an outer measure on Xif

1) µ(∅) = 0 (Nullity);2) µ(A) ≤ µ(B) if A ⊂ B ⊂ X (Monotonicity);3) For any sequence (An)n≥1 in X we have

µ(∞⋃n=1

An) ≤∞∑n=1

µ(An).

(σ-subadditivity)

Proposition 3. Hs(·) is an outer measure.

Proof. The nullity and the monotonicity of Hs are obvious. Only theσ-subadditivity needs to be explained. For any ε > 0, δ > 0, and n ≥ 1one can find a countable δ-cover Rn for An such that

|Hsδ(An)−

∑U∈Rn

(diam(U))s| ≤ ε

2n.

Take R =⋃nRn which is a countable δ-cover of A =

⋃nAn. Then

Hsδ(A) ≤

∑U∈R

(diam(U))s =∞∑n=1

∑U∈Rn

(diam(U))s ≤∞∑n=1

Hsδ(An) + ε.

Let δ → 0 then ε→ 0.

For any set E ⊂ X, Hs(E) is non-decreasing as a function of s.Furthermore, if s < t, then

Hs(E) ≥ δ−(t−s)Ht(E), ∀δ > 0,

83

which implies that if Ht(E) > 0, then Hs(E) = ∞. We define theHausdorff dimension of E, denoted by dimH(E) or simply by dimE,as follows

dimE = infs > 0 : Hs(E) =∞ = sups > 0 : Hs(E) = 0.

As a consequence of the preceding proposition we have the followingproperties of Hausdorff dimension.

Proposition 4. We have

(1) dim∅ = 0.(2) If E ⊂ F , then dimE ≤ dimF .(3) For a countable family Ei of subsets of X we have

dim(⋃

Ei) = supi

dimEi.

Proof. Here (1) and (2) are obvious. We only prove (3). For (3), weonly need to prove dim(

⋃Ei) ≤ supi dimEi because the other direction

is also obvious. Assume dimEi < s for all i. Then Hs(Ei) = 0. By theσ-subadditivity of the Hs, we get Hs(

⋃iEi) = 0 so that dim(

⋃iEi) ≤

s.

Let µ(·) be an outer measure on X. A subset M ⊂ X is said to beµ-measurable if for any subset S ⊂ X we have

µ(S) = µ(S ∩M) + µ(S \M).

It is well known that the subfamily Mµ ⊂ 2X consisting of all µ-measurable sets is a σ-algebra and that the restriction µ(·) : Mµ →[0,∞] is a measure. Actually (X,Mµ, µ) is a complete measure space.If every Borel set is µ-measurable (i.e. B(X) ⊂ Mµ), we say µ (con-sidered as an outer measure on 2X or a measure on Mµ) is a Borelmeasure. A Borel outer measure µ is said to be (outer) Borel regularif for each subset A ⊂ X there exists a Borel set B ⊃ A such thatµ(B) = µ(A). We will see that the Hausdorff measure Hs is Borelregular.

Proposition 5. Hs is a Borel measure.

Proof. There is a criterion due to Caratheodory for an outer measureµ to be a Borel measure:

(4) µ(A ∪B) ≥ µ(A) + µ(B)

for all subsets A ⊂ X,B ⊂ X such that d(A,B) > 0, where d(A,B) isthe distance between A and B.

84

Let us first prove the proposition from the criterion. We have onlyto show the µ-measurability of an open set O, that is,

µ(S) = µ(S ∩O) + µ(S \O) (∀S ⊂ X).

We only need to prove that

µ(S) ≥ µ(S ∩O) + µ(S \O) (∀S ⊂ X).

Let F = Oc. Let Fk = x ∈ X : d(x, F ) ≤ 1/k be the closed 1/k-neighborhood of F . Write Ok = O \ Fk. This is an open set containedin O, which has a distance to F at least 1/k. Thus d(S∩Ok, S \O) > 0and so

(5) µ(S) ≥ µ(S ∩Ok) + µ(S \O).

Assume that µ(S) <∞. Otherwise there is nothing to prove. ConsiderCk = S ∩Ok. The sequence Ck is increasing. Let Rk = Ck \Ck−1 (withC0 = ∅). It is easy to check that d(Rk+2, Rk) ≥ 1

k(k+1). So,

µ(S) ≥ µ

(m⋃j=1

R2j−1

)≥

m∑j=1

µ(R2j−1),

µ(S) ≥ µ

(m⋃j=1

R2j

)≥

m∑j=1

µ(R2j).

It follows that both series∑∞

j=1 µ(R2j−1) and∑∞

j=1 µ(R2j) are conver-gent. Observe

µ(S ∩O) = µ

(∞⋃k=1

Ck

)= µ

(Cm ∪

∞⋃k=m

Rk

)≤ µ(Cm) +

∞∑k=m

µ(Rk).

Since∑∞

k=m µ(Rk)→ 0 as m→∞, the above inequality implies

µ(S ∩O) ≤ limmµ(S ∩Om).

Taking limit in (5), we get

µ(S) ≥ µ(S ∩O) + µ(S \O).

We get the µ-measurability of O.Now let us prove the criterion for Hs. Given two subsets A ⊂ X

and B ⊂ X such that d(A,B) > 0. Take 0 < δ < d(A,B)2

. Let R be anarbitrary δ-cover of A∪B. None of element in R can meet both A andB. Hence∑

E∈R

(diamE)s =∑

E∈R:E∩A 6=∅

(diamE)s +∑

E∈R:E∩B 6=∅

(diamE)s

≥ Hsδ(A) +Hs

δ(B).

85

Taking the infimum over all such covers R and then taking limit asδ → 0 we get

Hs(A ∪B) ≥ Hs(A) +Hs(B).

This is (4) for the Hausdorff measure. We proved the criterion for Hs.This proves the proposition.

Proposition 6. Hs is Borel regular.

Proof. Since diam(U) = diam(U), we may assume that the coveringsets in the definition of Hs are all closed. Let A ⊂ X be a fixed subset.For each n ≥ 1, there exists a 1/n-cover of closed sets Fn,ii≥1 of Asuch that ∑

i

(diam(Fn,i))s ≤ Hs

1/n(A) +1

n.

Take the Borel set B =⋂∞n=1

⋃∞i=1 Fn,i. Then A ⊂ B and

Hs(A) = Hs(B).

In the definition of the Hausdorff dimension of a subset E of a metricspace (X, d), we use all δ-covers of E. However, in applications, it ismore convenient to use δ-covers of E by balls to calculate the Hausdorffdimension. Here we will prove it is possible. A ball is defined as a subsetB(x, r) = y ∈ X | d(y, x) < r, where x is the center and r > 0 is theradius.

Let E be an arbitrary subset of X. For δ > 0 and s > 0, we define

Bsδ(E) = inf∞∑i=1

(diam(B))s

where the infimum is over all countable δ-covers Bi of E by balls. Itis clear that for fixed s, Bsδ(E) is increasing with respect to δ. Now letδ → 0 we get the s-(outer) measure of E:

Bs(E) = limδ→0Bsδ(E).

Theorem 58.

dimE = infs > 0 : Bs(E) =∞ = sups > 0 : Bs(E) = 0.

Proof. For any δ > 0 and s > 0, it is clear that that

Hsδ(E) ≤ Bsδ(E)

since δ-covers of E by balls are also δ-covers of E in the definition ofHsδ(E). Suppose Ui is a δ-cover of E. Let Bi be some ball containing

86

Ui of radius diam(Ui). Then Bi is a δ-cover of E by balls. We havethat ∑

i

diam(Bi)s ≤

∑i

(2diam(Ui))s = 2s

∑i

(diam(Ui))s.

Take infima,

Bs2δ(E) ≤ Hsδ(E).

Letting δ → 0, it follows that

Hs(E) ≤ Bs(E) ≤ 2sHs(E).

It proves the theorem.

One can also prove that the Hausdorff dimension dimE can be de-fined by all δ-covers of E by just closed subsets or just by open subsets.When E is a compact set, one can prove that the Hausdorff dimensiondimE can be defined by all finite δ-covers of E (or by just balls, or justclosed subsets or just by open subsets).

The following the Frostmann lemma is useful in the calculation ofHausdorff dimension.Lemma 14. Let E be a Borel set in Rn. Suppose there is a probabilitymeasure µ on E and positive constants C > 0 and α such that

µ(B(x, r) ∩ E) ≤ Crα

for all x ∈ E, r > 0, where B(x, r) = y ∈ Rn | ‖y − x‖ < r is theball of radius r centered at x. Then dimE ≥ α.

Proof. Let s ≤ α and δ > 0. Let Bi = B(xi, ri) be a δ-cover of Eby balls. There is a universal constant C(n) > 0 such that after weremove unnecessary balls each point x ∈ E will only be covered by atmost C(n) balls. Then∑i

(diam(Bi))s =

∑i

(2ri)s ≥

∑i

1

C2sµ(Bi∩E) =

1

C2s

∫E

(∑i

1Bi

)dµ

≥ 1

C2sC(n)

∫E

1∪iBidµ =1

C2sC(n).

This implies that

Bs(E) ≥ 1

C2sC(n)> 0

and

dimE ≥ α.

87

As an application of the distortion result in Lecture 3, we calculatethe Hausdorff dimension of a Cantor set which is the invariant set of aC1+α-hyperbolic dynamical system.

Theorem 59 (C1+α-hyperbolic Cantor set, Hausdorff dimension). Sup-pose f : I0 ∪ I1 → I is a C1+α expanding map for some 0 < α ≤ 1.Then the Hausdorff dimension of the non-escaping set Λ = Λf underf is bigger than 0 and less than 1, that is,

0 < dimΛ < 1.

Proof. We use the same nation as that in the proof of Theorem 5.Without loss of generality, we assume that

0 < γ ≤ 1

|f ′(x)|≤ µ < 1.

Then we have thatγn+1 ≤ |Iwn| ≤ µn+1

for all wn = i0 . . . in−1 and all n > 0. Thus we can use all partitionsηn = Iwn to calculate the Hausdorff dimension. That is, we consider

βn(s) =∑wn

|Iwn|s.

Since βn(s) ≥ 2n+1γs(n+1), we have that dimΛ ≥ log 2/(− log γ) > 0.For the other direction, we use the inequality

as + bs

2≤(a+ b

2

)sto get

|Iwn−10|s + |Iwn−11|s ≤ 21−s(|Iwn−10|+ |Iwn−11|)s ≤ 21−s(1− c)|Iwn−1|s.Thus we have that

βn(s) ≤ 2(n+1)(1−s)(1− c)n+1

and that

dimΛ ≤ log 2 + log(1− c)log 2

.

88

Lecture 10. Hausdorff dimension and pressure func-tion.

Now let come back to study the Hausdorff dimension of an expandingholomorphic dynamical system (or C1+α expanding one dimensionaldynamical system).

Suppose Ω is a bounded simply connected region. Suppose Ω1, . . .,Ωd are bounded simply connected regions such that

Ωi ∩ Ωj = ∅ ∀ 1 ≤ j 6= j ≤ d

and

∪di=1Ωi ⊂ Ω.

Let f : ∪di=1Ωi → Ω be a holomorphic map, that means that f isholomorphic on a neighborhood of ∪di=1Ωi. Suppose f |Ωi : Ωi → Ω isconformal for every 1 ≤ i ≤ d. Let

J = ∩nn=0f−n(Ω)

be the non-escaping set of f (we will call it the Julia set later).(In one-dimensional dynamical systems, we have a similar setting as

follows: Let I = [0, 1] and Ii = [ai, bi], i = 1, . . . , d, a1 = 0 < b1 < a2 <b2 < · · · < ad < bd = 1. Suppose f : K0 = ∪di Ii → I is a C1+α mapsuch that each f : Ii → I is a homeomorphism. Let

J = ∩∞n=1f−n(I)

be the non-escaping set. Suppose f is expanding, that is, |(fn)′(x)| ≥Cλn for two constants C > 0 and λ > 1 and all n > 0 and all x suchthat f i(x) ∈ K0, 0 ≤ i ≤ n − 1. Then all arguments in the followingin this section work for this setting just replacing the Koebe distortionby the naive distortion.)

Then from the hyperbolic geometry we know that f : J → J is ahyperbolic Markovian map with the initial partition

η0 = J0, . . . , Jd

where Ji = J ∩ Ωi. Then we have a bijective map

π : Σd =∞∏n=0

1, . . . , d → J.

Thus J is a Cantor set.Without loss of generality, we assume that Ω = ∆ = z ∈ C | |z| <

1. Let 0 < r < 1 be a such number such that

J ⊂ ∆r = z ∈ C | |z| < r.

89

Let

gi = (f |Ωi)−1 : ∆→ Ωi, 1 ≤ i ≤ d.

For any wn = i0 . . . in−1 of a sequence of 1, . . . , d,let

gwn = gi0 . . . gin−1 : ∆→ Ωwn = gwn(∆).

Then gwn is conformal. From the Koebe distortion theorem (Theo-rem 23), we have that(1− r

1 + r

)4

≤|g′wn(z1)||g′wn(z2)|

≤(1 + r

1− r

)4

, ∀|z1|, |z2| ∈ ∆r,

and

|g′wn(0)| |r|(1 + |r|)2

≤ |gwn(z)− gwn(0)| ≤ |g′wn(0)| |r|(1− |r|)2

, ∀z ∈ ∆r.

Therefore, we have a constant C > 0 such that

(6)|g′wn(z)|C

≤ diam(Ωwn) ≤ C|g′wn(z)|, ∀z ∈ ∆r.

In other words,

(7) C−1|(fn+1)′(z)|−1 ≤ diam(Ωwn) ≤ C|(fn+1)′(z)|−1, ∀z ∈ Ωwn .

Since gwn : Ω → Ωwn is a contracting map, from the contractingfixed point theorem, we have a unique fixed point pwn ∈ Ωwn of gwn . Itis the unique fixed point pwn ∈ Ωwn of fn+1, that is, fn+1(pwn) = pwn .Let λwn = (fn+1)′(pwn) be the multiplier of fn+1 at pwn . Then we havethat

(8) C−1|λwn|−1 ≤ diam(Ωwn) ≤ C|λwn|−1, ∀wn.

(In the C1+α expanding one-dimensional case, we can use the naivedistortion (Lemma 1) to replace the Koebe distortion to get the in-equalities (6), (7), and (8).)

Now consider the function φt(z) = −t log |f ′(z)|. For any z ∈ J ,let Sn,t(z) =

∑ni=0 φt(f

i(z)). The topological pressure for φt is, bydefinition,

P (φt) = limn→∞

1

nlog(∑

wn

eSn,t(pwn ))

= limn→∞

1

nlog(∑

wn

|λwn|−t)

It is clearly that

P (φt) = limn→∞

1

nlog( ∑z∈Ωwn

eSn,t(z))

= limn→∞

1

nlog( ∑z∈Ωwn

|(fn)′(z)|−t)

90

because from the Koebe distortion (or the naive distortion in one-dimensional dynamical systems)), we have a constant C > 0 such that

C−1|λwn| ≤ |(fn+1)′(z)| ≤ C|λwn|, ∀z ∈ Ωwn , ∀wn.

Lemma 15. Suppose an∞n=1 is a sequence of nonnegative real num-bers. Suppose an+m ≤ an + am. Then

limn→∞

ann

exists.

Proof. Let l = infan/n | n ≥ 1. For any ε > 0, there is n0 > 0 suchthat

l ≤ an0

n0

≤ l + ε/2.

Let n1 > n0 be a number such that maxak/n1, k = 1, · · · , n0 − 1 ≤ε/2. Then any n > n1, let n = pn0 + k, where 0 ≤ k < n0. Definea0 = 0. Then

l ≤ ann≤ pan0 + ak

pn0 + k≤ an0

n0

+akn≤ l + ε.

This says that

limn→∞

ann

= l.

Theorem 60. The limit

P (φt) = limn→∞

1

nlog(∑

wn

eSn,t(pwn ))

= limn→∞

1

nlog(∑

wn

|λwn|−t)

exists.

Proof. Since

P (φt) = limn→∞

1

nlog( ∑z∈Ωwn

eSn,t(z)),

let bn =∑

z∈ΩwneSn,t(z)), we have that

bn+p =∑

z∈Ωwn+p

eSn+p,t(z) =∑

z∈Ωwn+p

eSn,t(z)eSp,t(fn(z))

≤∑z∈Ωwn

eSn,t(z)∑w∈Ωwp

eSp,t(w) ≤ bnbp.

Let an = log bn. Then we have that an+p ≤ an + ap. The previouslemma implies that the limit exists.

91

Remark 12. The topological pressure can be defined for any continuousfunction φ. In particular, when φ ≡ 0, the P (0) = h(f) is just thegrowth rate we studied before (or called the topological entropy). As wealready know, in our case studied above, P (0) = log d.

Consider the function α(t) = P (φt). It is differentiable with α′(t) < 0and α′′(t) > 0. Thus α(t) is strictly decreasing and concave upwardfunction of t. Since α(0) = h(f) > 0 and α(t) → −∞ as → ∞. Wehave a unique number t0 such that α(t0) = 0. The following theoremis due to Bowen and Ruelle originally, refer to

• R. Bowen. Hausdorff dimension of quasi-circles. Publ. of IHES,Volume 50 (1979).• D. Ruelle, Repellers for real analytic maps. Ergodic Theory

Dynamical Systems 2 (1982), 99107.

Theorem 61. The Hausdorff dimension dimJ = t0.

Combing with Theorem 19, we got

Corollary 7. 0 < dimJ < 2.

Proof of Theorem 61. The collection of set Ωwn is a δ-cover of J forany large n. From the Koebe distortion, there is a constant C > 0 suchthat

C−1|λwn|−1 ≤ diam(Ωwn) ≤ C|λwn|−1, ∀n > 0,∀wn.Thus

C−1 = C−1∑wn

|λwn|−t0 ≤∑wn

(diam(Ωwn)

)−t≤ C

∑wn

|λwn|−t0 = C.

This implies that diamJ ≤ t0.To prove diamJ ≥ t0. We need the Gibbs measure for φt0 (see

Theorem 62), that is, we have a measure µ on J such that there isconstant D > 0,

D−1|(fn+1)′(z)|−t0 = D−1eSn,t(z) ≤ µ(Jwn) ≤ DeSn,t(z) = D|(fn+1)′(z)|−t0

for any wn and z ∈ Jwn (since P (φt0) = 0).We also need the Koebe distortion, that is, we have two constants

c1, c2 > 0 such that

B(gwn(0), c1|g′wn(0)|) ⊂ Ωwn ⊂ B(gwn(0), c2|g′wn(0)|).for any n > 0, wn. This implies that there are two constants d1, d2 > 0such that

B(gwn(0), d1|(fn+1)′(z)|−1) ⊂ Ωwn ⊂ B(gwn(0), d2|(fn+1)′(z)|−1)

92

for any n > 0, wn, and z ∈ Ωwn .Suppose B(a, r) = z ∈ C | |z − a| < r is any ball such that

B(a, r) ∩ J 6= ∅. For any z ∈ B(a, r) ∩ J , B(a, r) ⊂ B(z, 2r). There isa unique integer n1 > 0 such that

|(fn1+2)′(z)|−1 ≤ 2rd−11 ≤ |(fn1+1)′(z)|−1.

Thus

B(a, r) ⊂ B(gwn1 (0), 2r) ⊂ B(gwn1 (0), d1|(fn1+1)′(z)|−1) ⊂ Ωwn1.

Let m = infz∈J |f ′(z)|−1. Then

rt0 ≥(d1m

2

)t0|(fn1+1)′(z)|−t0 ≥ D−1

(d1m

2

)t0µ(Jwn1 ) ≥ Cµ(B(a, r)∩J)

where

C = D−1(d1m

2

)t0is a constant. From Lemma 14, we have that

dimJ ≥ t0.

In the proof of Theorem 61, we need the Gibbs theory, let us stateit here and we will give a full study later, for example, refer to

• Y. Jiang, Nanjing Lecture Notes In Dynamical Systems. PartOne: Transfer Operators in Thermodynamical Formalism. FIMPreprint Series, ETH-Zurich, June 2000.• Y. Jiang, A proof of existence and simplicity of a maximal eigen-

value for Ruelle-Perron-Frobenius operators. Letters in Math.Phys. 48 (1999), no. 3, 211-219.• A. Fan and Y. Jiang, On Ruelle-Perron-Frobenius Operators. I.

Ruelle Theorem. Comm. in Math. Phys., 223 (2001), 125-141.

Let A = (aij)d×d be a 0− 1 matrix, that is, aij = 0 or 1. A (n+ 1)-sequence wn = i0i1 . . . ik−1ik . . . in−1in of 1, . . . , d is called admissibleif aik−1ik = 1 for all 1 ≤ k ≤ n. Let

ΣA = w = i0i1 . . . ik−1ik . . .be the space of all ∞-admissible sequence of 1, . . . , d. For a givenwn = i0i1 . . . ik−1ik . . . in−1in,

[wn] = w′ = j0j1 . . . jk−1jk . . . | jk = ik, 0 ≤ k ≤ nis a cylinder. All cylinders form a topological basis for ΣA. With thistopology, ΣA is a compact topological space. Let

σA : w = i0i1 . . .→ σA(w) = i1 . . .

93

be the shift.Consider the space C(ΣA) of all continuous functions defined on ΣA.

Let|φ|∞ = max

w∈ΣA|φ(w)|

be the maximal norm of φ. A function φ ∈ C(ΣA) is called a Holdercontinuous if there are constants C > 0 and 0 < β < 1 such that

|φ(w)− φ(w′)| ≤ Cβn

for any w,w′ ∈ [wn]. Let Ch(ΣA) be the space of all Holder continuousfunctions.

Theorem 62. For any φ ∈ Ch(ΣA), there is a real number P = P (φ)and a probability measure µ = µφ on ΣA such that

C1 ≤ µ([wn])

exp(−nP +∑n

i=0 log φ(σiA(w))≤ C

for any cylinder [wn] and and w ∈ [wn], where C > 0 is a constant.

Remark 13. The function φ is called a potential. The number P iscalled the pressure for the potential φ. The measure µ is called theGibbs measure associate to the potential φ. Usually, we write

Sn(w) = Sn,φ(w) =n∑i=0

log φ(σiA(w)).

Using the Ruelle’s Perron-Frobenius theorem and the study of trans-fer operators (which we will fully develop with the Gibbs theory), weknew that

P (φ) : Ch(ΣA)→ Ris a real analytic function. Refer to

• Y. Jiang, Nanjing Lecture Notes In Dynamical Systems. PartOne: Transfer Operators in Thermodynamical Formalism. FIMPreprint Series, ETH-Zurich, June 2000.

Now consider a holomorphic family of hyperbolic holomorphic dy-namical systems

fc : ∪di=1Ωi(c)→ Ω, c ∈ ∆

where

(1) all Ωi(c) and Ω are simply connected regions in C,(2) Ωi(c) ∩ Ωj(c) = for any 1 ≤ i, j ≤ d,(3) fc : Ωk(c)→ Ω is conformal for every 1 ≤ k ≤ d,(4) ∪di=1Ωk(c) ⊂ Ω,

94

(5) for any z ∈ Jc, fc(z) depends on c ∈ ∆ holomorphically.

For c = 0, φ0,t(w) = −t log |f ′0(π(w))| defined on Σd is a Holdercontinuous function. For any wn, there exists a unique pwn(c) ∈ Ωwn(c)such that fn+1

c (pwn(c)) = pwn(c). The multiplier

|λwn(c)| = |(fn+1c )′(pwn(c))| > 1.

Thus, implicitly function theorem implies that pwn(c) defines a holo-morphic function on ∆. Let Per be the set of all periodic points of f0.Then

(9) h(c, pwn) = pwn(c) : ∆× Per → Cis a holomorphic motion.

In general, let E be a subset in the Riemann sphere C, a map h(c, z) :

∆× E → C is called a holomorphic motion if

• h(0, z) = z for every z ∈ E,• for each c ∈ ∆, hc(z) = h(c, z) defines a bijective map from E

to hc(E),• for each z ∈ E, hz(c) = h(c, z) defines a holomorphic map from

∆ to C.

The following lemma is first proved by Mane, Sad, and Sullivan in

• Mane, R. M., Sad, P. and Sullivan, D. On the dynamics ofrational maps. Ann. c. Norm. Sup. 96 (1983), 193217.

Lemma 16 (λ-Lemma). The map h can be extended to a holomorphic

motion h(c, z) : ∆× E → C.

Furthermore, we have

Theorem 63 (Slodkowski’s Theorem). The map h can be extended to

a holomorphic motion H(c, z) : ∆× C→ C.

And

Theorem 64 (Bers-Royden’s Theorem). Suppose H(c, z) : ∆×C→ Cis a holomorphic motion. Then for each c ∈ ∆, Hc(z) = H(c, z) : C→C defines a quasiconformal homeomorphism with whose quasiconformaldilatation K(Hc) ≤ (1 + |c|)/(1− |c|).

We will lecture these two theorems in details later following papers.

• F. Gardiner, Y. Jiang, and Z. Wang, Holomorphic motions andrelated topics, Geometry of Riemann Surfaces, London Mathe-matical Society Lecture Note Series, No. 368, 2010, 166-193.

95

• Y. Jiang, S. Mitra, and Z. Wang, Liftings of holomorphic mapsinto Teichmller spaces, Kodai Mathematical Journal, 32 (2009),No. 3, 544-560.• F. Gardiner and Y. Jiang, Guiding isotopies and Slodkowskis

theorem, Preprint.

Theorem 65. Suppose H is a quasiconformal homeomorphism on asimply connected region Ω with quasiconformal dilatation K > 1. LetJ ⊂ Ω be a compact subset. Then H|J is an α-Holder continuousfunction for some 0 < α = α(K) < 1, that is,

|H(z)−H(w)| ≤ C|z − w|α, ∀z, w ∈ J,

where C = C(K) > 0 is a constant.

We will also lecture this theorem later, refer to

• L. V. Ahlfors. Lectures on Quasiconformal Mapping,, volume38 of University Lecture Series. Amer. Math. Soc., 2006.

Applying holomorphic motions, we can extend the holomorphic mo-tion in (9) to a holomorphic motion, which we still denote as

h(c, z) : ∆× J → C

since J = Per. From the definition and the continuity of hc(z) = h(c, z)on J , we have hc : J → Jc and

hc(f0(z)) = fc(hc(z)), z ∈ J.

Since hc is the restriction of a quasiconformal homeomorphism Hc, sohc and h−1

c are both α-Holder continuous. Define

φc,t(w) = −t log |f ′c(h−1c (π(w)))|

on Σd. It is a family of Holder continuous functions which depends onc ∈ ∆ real analytically. Thus the pressure function P (φc,t) is a functiondepends on c and t both real analytic. The unique solution tc such thatP (φc,tc) = 0 is a real analytic function defined on c ∈ ∆. Since tc isthe Hausdorff dimension diamJc, we conclude that, originally due toRuelle, refer to

• D. Ruelle, Repellers for real analytic maps. Ergodic TheoryDynamical Systems 2 (1082), 99107.

Theorem 66. The Hausdorff dimension diamJc is real analyticallydepends on c ∈ ∆.

96

In the proof of this theorem, I still owes you two big theories, theGibbs theory and the holomorphic motion theory. We will lecture themlater. Using the above theorem, Ruelle showed that

Corollary 8. Suppose qc(z) = z2 + c is the quadratic polynomial andsuppose Jc is its Julia set. The Hausdorff dimension diamJc is analyticfunction on the main cardioid M0 and

diamJc = 1 +|c|2

4 log 2+ o(|c|2).

97

Lecture 11. Quasiconformal Mappings andApplications

11.1. Grotzsch argument.

Recall that the Riemann Mapping Theorem (Theorem 10). A simpleclosed curve in C is called a Jordan curve if it is an image of the unitcircle under a homeomorphism.

Theorem 67 (Caratheodory Theorem). Suppose Ω is a simply con-nected domain. Suppose ∂Ω is a Jordan curve. Then H can be extendedto the closure Ω as a homeomorphism from Ω onto ∆.

We say ∂Ω is locally connected if for any point z ∈ ∂Ω and anyneighborhood U about z, U ∩ ∂Ω is connected. More generally, wehave

Theorem 68 (Caratheodory Theorem). Suppose Ω is a simply con-nected domain and suppose ∂Ω is locally connected. Then the inverseH−1 : ∆→ Ω can be extended to a continuous map from ∆→ Ω.

Now let us consider the following problem. Suppose R = [0, a, a +ib, ib] is a rectangle with vertices 0, a, a + ib, and ib in the complexplane. Let R′ = [0, a′, a′ + ib′, ib′] be another rectangle with vertices 0,a′, a′+ ib′, and ib′ in the complex plane. According to Theorem 10, wehave a conformal map H : R → R′ and according to Theorem 67, Hcan be extended as a homeomorphism H : R → R′. How we add onemore condition that H(0) = 0, H(a) = a′, H(a + ib) = a′ + ib′, andH(ib) = ib′.

Problem 6. When can we find a conformal map H from R to R′

satisfying the above condition?

Before answering the above problem, let us have some basic calcula-tion in complex analysis.

Let z = x+ iy and z = x− iy be the coordinate system in C. Thenwe have the change of coordinate

x =z + z

2and y =

z − z2i

.

We calculate differentials

dz = dx+ idy and dz = dx− idy.

98

Since ∂/∂x and ∂/∂y are two differential operators, we have otherdifferential operators

∂

∂z=

1

2

( ∂∂x− i ∂

∂y

)and

∂

∂z=

1

2

( ∂∂x

+ i∂

∂y

).

So for any orientation-preserving C1 map f : Ω→ C, we have

df(z) =∂f(z)

∂xdx+

∂f(z)

∂ydy =

∂f(z)

∂zdz +

∂f(z)

∂zdz.

We use the following notation

∂f(z) =∂f(z)

∂zand ∂f(z) =

∂f(z)

∂z.

So we have

df(z) = ∂f(z)dz + ∂f(z)dz

is a linear map from the tangent plane at z ∈ Ω to the tangent planeat f(z). Consider f(z) = f(z). Then

∂f(z) = ∂f(z) and ∂f(z) = ∂f(z).

The Jacobian J(f)(z) is

J(f)(z) =

∥∥∥∥ ∂f(z) ∂f(z)∂f(z) ∂f(z)

∥∥∥∥ = |∂f(z)|2 − |∂f(z)|2 > 0.

Consider the unit circle |dz| = 1 on the tangent plane at z, then theimage of the unit circle is df(z) with

0 < |∂f(z)| − |∂f(z)| ≤ |df(z)| ≤ |∂f(z)|+ |∂f(z)|.

Let

µ(z) =∂f(z)

∂f(z)= re2πiθ.

Then we have 0 ≤ |µ(z)| < 1 and

df(z) = (1 + µ(z)dz

dz)∂f(z)dz

and

|∂f(z)|(1− |µ(z)|) ≤ df(z) ≤ |∂f(z)|(1 + |µ(z)|).Suppose arg µ = θ. Then the maximum can be reached at arg dz = θ/2and the minimum can be reached at arg dz = (θ + π)/2. Thus df(z) isan ellipse with the major axis l(z) = |∂f(z)|(1 + |µ(z)|) and the minoraxis s(z) = |∂f(z)|(1−|µ(z)|). The ratio K(f)(z) = l(z)/s(z) is calledthe quasiconformal dilatation of f at z. A conformal map preserves

99

circles. That is df(z) is a circle. In other words, f is conformal at z ifand only if

K(z) = 1⇐⇒ µ(z) = 0⇐⇒ ∂f(z) = 0.

Now suppose f : R→ R′ is a C1 diffeomorphism such that f(0) = 0,f(a) = a′, f(a+ib) = a′+ib′, and f(ib) = ib′. Let cy(t) = t+iy, 0 ≤ t ≤a be a straight line in R. Then f(cy(t)) is a C1 curve in R′ connectingf(iy) and f(a+ iy). Then

a′ ≤ |∫cy

df(z)| ≤∫cy

|df(z)|

≤∫cy

(|∂f(z)|+ |∂f(z)|)|dz| =∫cy

J(f)(z)1/2K(f)(z)1/2|dz|.

So

a′b ≤ |∫ ∫

R

df(z)| ≤∫ ∫

R

J(f)(z)1/2K(f)(z)1/2|dz ∧ dz|

≤ (

∫ ∫R

J(f)(z)|dz ∧ dz|)1/2(

∫ ∫R

K(f)(z)|dz ∧ dz|)1/2

= (a′b′)1/2(ab)1/2(maxz∈R

K(f)(z))1/2.

This implies that

(a′/b′) : (a/b) ≤ maxz∈R

K(f)(z).

Thus the answer to Problem 6 is that there is a conformal map from Ronto R′ preserving the vertices if and only if a′/b′ = a/b. The numbera/b is called the modulus of R is denoted as mod(R). It is a conformalinvariant.

In general, for a C1 diffeomorphism f : Ω→ C, let

K(f) = maxz∈Ω

K(f)(z).

Then we call it a K-quasiconformal diffeomorphism. Furthermore, wehave that

Theorem 69. Suppose R and R′ are two rectangles. Then there is a K-quasiconformal C1-diffeomorphism preserving the vertices f : R → R′

if and only if

1

Kmod(R′) ≤ mod(R) ≤ Kmod(R′).

100

Suppose Ω is a double connected region. Then we have a conformalmap f : Ω → Ar = z ∈ C | 1 < |z| < r, where r is unique. Themodulus mod(Ω) of Ω is defined as

mod(Ω) =1

2πlog r.

Then it is a conformal invariant. Furthermore, we also have by theabove argument that

Theorem 70. Suppose Ω and Ω′ are two double connected regions.Then there is a K-quasiconformal C1-diffeomorphism preserving thevertices f : Ω→ Ω′ if and only if

1

Kmod(Ω) ≤ mod(Ω′) ≤ Kmod(Ω).

11.2. Extremal length.

Suppose Ω is a domain in the complex plane. Let ρ ≥ 0 be a measur-able function on Ω. Then ρ is called a conformal metric on Ω. SupposeF = γ(t) be a family of piecewise C1 curves in Ω. The ρ-length is

lρ(γ) =

∫γ

ρ|dz| =∫ b

a

ρ(γ(t))γ′(t)dt

if t→ ρ(γ(t)) is measurable; otherwise, lρ(γ) =∞. The ρ-length of Fis

Lρ(F) = inflρ(γ) | γ ∈ F.The extremal length of F is, by definition,

λ(F) = supρ

(Lρ(F))2

Aρ,

where supremum is taken over all conformal metrics ρ ≥ 0 on Ω suchthat areas Aρ =

∫ ∫Ωρ2dxdy <∞. If the supremum can be reached by

a conformal metric ρ ≥ 0 on Ω. Then such a ρ is called an extremalconformal metric of F .

Theorem 71. Suppose Ω and Ω′ are two domains. If f : Ω→ Ω′ is aK-quasiconformal C1-diffeomorphism, then

1

Kλ(F) ≤ λ(f(F)) ≤ Kλ(F)

for any family F of piecewise C1 curves in Ω (then f(F) is a fam-ily of piecewise C1 curves in Ω′. In particular, the extremal length isconformal invariant.

101

Proof. Consider a conformal metric ρ(z)dz on Ω. Then

ρ(w) =ρ(f−1(w))

|∂f(f−1(w))| − |∂f(f−1(w))|defines a conformal metric on Ω′. For γ(t) ∈ F , δ(t) = f(γ(t)) ∈ f(F).We have that

γ′(t) = ∂f(γ(t))γ′(t) + ∂f(γ(t))γ′(t).

This implies that lρ(γ) ≤ lρ(γ). Thus Lρ(F) ≤ Lρ(f(F)). On the otherhand, for w = u+ iv and z = x+ iy,

Aρ =

∫ ∫Ω′

(ρ(w))2dudv =

∫ ∫Ω

( ρ(z)

|∂f(z)| − |∂f(z)|

)2

J(f)(z)dxdy

=

∫ ∫Ω

(ρ(z))2 |∂f(z)|+ |∂f(z)||∂f(z)| − |∂f(z)|

dxdy ≤ KAρ.

This implies that

λρ(f(F)) ≥ 1

Kλρ(F).

The other side inequality can be obtained similarly.

Let R = R(0, a, a+ ib, ib) be the rectangle. Let F be the family of allpiecewise C1-curves connecting two points on the both vertical sides ofR. Let ρ0 = 1. Then |dz| is the Lebesgue measure on R. It is easy tosee Lρ0(F) = a and Aρ0 = ab. Thus

λ(F) ≥ a2

ab=a

b.

Suppose ρ ≥ 0 is any conformal metric on R. Suppose Lρ(F) 6= 0.By multiplying a constant on ρ, we can assume that Lρ(F) = a. Thisimplies that ∫ a

0

(ρ(z)− 1)dx ≥ 0.

Then we get

0 ≤∫ ∫

R

(ρ− 1)2dxdy =

∫ ∫R

ρ2dxdy − 2

∫ ∫ρdxdy +

∫ ∫R

1dxdy

≤∫ ∫

R

ρ2dxdy −∫ ∫

1dxdy.

This says thatAρ ≥ ab.

So(Lρ(F))2

Aρ≤ a2

ab=a

b.

102

Thus λ(F) = a/b = mod(R) and the Lebesgue metric is the extremalconformal metric.Exercise 3. Let Ar = z ∈ C | 1 ≤ |z| ≤ r. Let F be the family ofpiecewise C1 curves connecting two points in the unit circle |z| = 1 andthe circle |z| = r. Find the extremal length λ(F) and compare with themodulus mod(Ar) = (1/2π) log r. Find the extremal conformal metric.

11.3. Quasiconformal mappings.

We now are ready to give a general definition of a quasiconformalhomeomorphism. A quadrilteral B = B(z1, z2, z3, z4) is a (closed) Jor-dan domain with four distinct points z1, z2, z3, z4 marked on the bound-ary of B in the counterclockwise sense. Then two arcs [z1, z2] and [z3, z4]are called horizontal sides and two arcs [z2, z3] and [z4, z2] are calledvertical sides. Let F be the family of all C1 piecewise curves connectingtwo points on two different vertical sides. The modulus of B is definedas the extremal length of F , that is,

mod(B) = λ(F).

We have a conformal map f from B onto a unique R = R(0, a, a+ i, i)such that f(z1) = 0, f(z2) = a, f(z3) = a + i, and f(z4) = i. Thenumber mod(R) = a is uniquely determinant. It is just the modulusmod(B) of B. That is,

mod(B) = λ(F) = mod(R).

The quadrilateral B′ = B(z2, z3, z4, z1) has the modulus mod(B′) =1/mod(B).

Theorem 72. Suppose B = B(z1, z2, z3, z4) = B1(z1, w2, w3, z4) ∪B2(w2, z2, z3, w4) is a quadrilateral which is a union of two quadrilat-erals B1 = B1(z1, w2, w3, z4) and B2 = B2(w2, z2, z3, w4). Then

mod(B) ≥ mod(B1) + mod(B2).

The equality holds if and only if there is a conformal map f : B →R(0, a, a+i, i) such that f(B1) = R(0, a1, a1+i, i) and f(B2) = R(0, a2, a2+i, i), where a = a1 + a2.

Corollary 9. Suppose B is a quadrilateral and n−1 curves connectingpoints on its horizontal sides cut it into n sub-quadrilaterals B1, · · · ,Bn. Then

mod(B) ≥n∑i=1

mod(Bi).

103

Definition 22. Suppose Ω is a domain in the complex plane. Supposef : Ω→ f(Ω) is an orientation-preserving homeomorphism. We say fis a K-quasiconformal if

1

Kmod(B) ≤ mod(f(B)) ≤ Kmod(B)

for any quadrilateral B ⊂ Ω.

Clearly, f−1 is also a K-quasiconformal homeomorphism if f is a K-quasiconformal. If f1 isK1-quasiconformal and f2 isK2-qausiconformal,then f1 f2 is K1K2-quasiconformal.Theorem 73. Suppose f : Ω → Ω′ is 1-quasiconformal. Then f isconformal.

Proof. Let B be a quadrilateral in Ω. Then B′ = f(B) is a quadrilateralin Ω′ and a = mod(B′) = mod(B). Let g1 : B → R = R(0, a, a + i, i)and g2 : B′ → R = R(0, a, a + i, i) be two conformal map. Then

f = g2 f g−11 : R → R is 1-quasiconformal. For any z ∈ R, let l be

the vertical line passing z. Then l cuts R into two rectangles R1 andR2 such that a1 = mod(R1) and a2 = mod(R2) and a = a1 + a2. Since

f is 1-quasiconformal, mod(f(R1)) = a1 and mod(f(R2)) = a2. Thisimplies that

mod(R) = mod(f(R1)) + mod(f(R2)).

This implies that f(l) = l. Similarly , if l is a horizontal line in R, we

have that f(l) = l. This implies that f(z) = z and f(z) = g−12 g1(z)

is conformal.

Instead to using quadrilaterals in the definition of a quasiconformalmapping, one can also use topological annuli. A topological annulusA is a (closed) double connected domain whose boundary consists oftwo Jordan curves. Then we have a conformal mapping mapping f :A → Ar = z ∈ C | 1 ≤ |z| ≤ r where r is uniquely determinant.The modulus of A is mod(A) = (1/2π) log r. As Exercise 3, mod(A) =λ(F), the extremal length of the the family of piecewise C1 curvesconnecting two points in two boundary Jordan curves.

Definition 23. Suppose Ω is a domain in the complex plane. Supposef : Ω→ f(Ω) is an orientation-preserving homeomorphism. We say fis a K-quasiconformal if

1

Kmod(A) ≤ mod(f(A)) ≤ Kmod(A)

for any topological annulus A ⊂ Ω.

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Using the extremal length technique, we have that

Corollary 10. Suppose A is a topological annulus and A1, · · · , An ⊂ Aare n pairwise disjoint annulus separate two boundary components ofA. Then

mod(A) ≥n∑i=1

mod(Ai).

Suppose f : Ω → f(Ω) is a homeomorphism. We say it is locallyK-quasiconformal if for every point z ∈ Ω, there is a neighborhood Uof z such that f : U → f(U) is K-quasiconformal.

Theorem 74. The homeomorphism f : Ω → Ω′ is K-quasiconformalif and only if it is locally K-quasiconformal.

Proof. We need only to prove the “if” part. Let B ⊂ Ω be a quadri-lateral. Since B is compact, we can cut it into small quadrilateralsQij such that each Qij falls in the definition neighborhood of localquasiconformalilty and such that

mod(Q) =∑i

mod(Qi)

where Qi =∑

j Qij, and

1

mod(Qj)=∑i

1

mod(Qij)

Thus1

Kmod(Qij) ≤ mod(f(Qij)) ≤ Kmod(Qij).

Using the property that

mod(f(Q)) ≥∑i

mod(f(Qi))

and1

mod(f(Qi))≤∑j

1

mod(f(Qij)),

we get1

Kmod(Q) ≤ mod(f(Q)) ≤ Kmod(Q).

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Suppose A1 and A2 are two doubly connected domains. Supposef : A1 → A2 is a holomorphic degree two covering map. Then we have

mod(A2) = 2mod(A1).

The reason is that consider conformal maps g1 : A1 → Ar1 = z ∈C | 1 ≤ |z| ≤ r1 and g2 : A2 → Ar2 = z ∈ C | 1 ≤ |z| ≤ r2.Then f = g2 f g−1

1 : Ar1 → Ar2 is a holomorphic degree coveringmap preserving inner circles and outer circles. Now by the Schwarzreflection theorem, f can be extended to a holomorphic degree twomap from C into itself and fixes 0 and preserves the unit circle. Thisimplies that f(z) = λz2 with |λ| = 1. Thus we have that r2 = r2

1.Now consider the hyperbolic disk (∆, d(·, ·)). For any two points

z, w ∈ ∆, let x = d(z, w) be its hyperbolic distance. Now considerthe disk ∆r(x) = z ∈ C | |z| < r(x) with the hyperbolic distancedr(x)(·, ·) such that dr(x)(0, 1) = d(z, w). We have a conformal mapfrom ∆r(x) \ [0, 1] onto ∆ \ lzw where lzw is the hyperbolic geodesicconnecting z and w. Here r(x) is uniquely determined by x and realanalytic depending on x. It is also strictly decreasing with r(x)→∞ asx→ 0+ and r(x)→ 1+ as x→∞. Now consider the doubly connecteddomain Dr(x) = ∆r(x) \ [0, 1]. The modulus m(x) = mod(Dr) is also ananalytic strictly increasing function with m(x) → ∞ as x → ∞ andm(x)→ 0 as x→ 1+. Let Ae2πm(x) = z ∈ C | 1 < |z| < e2πm(x) be theround annulus. Then we have a conformal map g : Ae2πm(x) → Dr(x)

which can be extended as a continuous function on the boundary. Theboundary mapping maps the unit circle into the interval [0, 1]. TheSchwarz reflection theorem implies that we have a degree two branchedholomorphic cover g : A = z ∈ C | −e2πm(x) ≤ |z| ≤ e2πm(x) →(∆, z, w) with two branched points−1 and 1 and two branched valuesz and w. The modulus mod(A) = 2m(x).

Theorem 75. Suppose f : ∆→ ∆ is a K-quasiconformal homeomor-phism. Then we have that

1

Km(d(z, w)) ≤ m(d(f(z), f(w)) ≤ Km(d(z, w))

for any z, w ∈ ∆.

Proof. Let g1 : A1 \ −1, 1 → ∆ \ z, w and g2 : A2 \ −1, 1 → ∆ \f(z), f(w) be two degree two holomorphic covers constructed above.Then the K-quasiconformal map f : ∆ \ z, w → ∆ \ f(z), f(w)can be lift to a K-quasiconformal homeomorphism f : A1 → A2. This

106

implies that

1

Kmod(A1) ≤ mod(A2) ≤ Kmod(A1).

Thus1

Km(d(z, w)) ≤ m(d(f(z), f(w)) ≤ Km(d(z, w)).

We have the following important corollaries.

Corollary 11. Suppose f : ∆→ ∆ is a K-quasiconformal homeomor-phism with f(0) = 0. For any 0 < r < 1, there is a constant C(r) > 0such that

|f(z)− f(w)| ≤ C(r)|z − w|1/K

for any |z|, |w| ≤ r.

Remark 14. Furthermore, by using the Grozsch extremal modulus, wehave a more strong result:

|f(z)− f(w)| ≤ 16|z − w|1/K

for any |z|, |w| < 1. Refer to

• Ahlfors, L. V. Lectures on Quasiconformal Mapping, UniversityLecture Series. 38, Amer. Math. Soc., 2006.

Corollary 12. Any K-quasiconformal homoemorphism f : ∆ → ∆can be extended to a homeomorphism f : ∆→ ∆.

Corollary 13. The family K of all K-quasiconformal homeomorphismsf : ∆ → ∆ with f(0) = 0 is a normal family, that is, any sequence inK has a convergent subsequence on any compact subset of ∆ and thelimiting function is again in K.

11.4. Analytic definition and the measurable Riemann map-ping theorem.

Suppose I is an interval in the real line R. A function f : I → R isabsolutely continuous on I if for every positive number ε > 0, there is apositive number δ > 0 such that whenever a finite sequence of pairwisedisjoint sub-intervals (xk, yk) of I satisfies∑

k

|yk − xk| < δ,

107

then ∑k

|f(yk)− f(xk)| < ε.

From the real analysis, the statement that f is absolutely continuouson I = [a, b] is equivalent to the following statements,

i. f has a derivative f ′ almost everywhere, the derivative is Lebesgueintegrable, and

f(x) = f(a) +

∫ x

a

f ′(x)dx ∀x ∈ I;

ii. there exists a Lebesgue integrable function g on I such that

f(x) = f(a) +

∫ x

a

g(x)dx ∀x ∈ I.

Thus we have the following properties for an absolutely continuousfunction f : I = [a, b]→ R:

1) f is of bounded variation;2) f maps every zero Lebesgue measure subset E of I to a zero

Lebesgue measure set f(E).

Therefore, we have another equivalent statement: f : I → R is abso-lutely continuous if and only if f is continuous, is of bounded variation,and maps every zero Lebesgue measure subset E of I to a zero Lebesguemeasure set f(E).

Definition 24. Suppose U is a domain in the complex plane. Acontinuous map f = u+ iv : U → C is called absolutely continuous onlines (ACL) if u and v when restricted on almost all horizontal linesand almost all vertical lines are absolutely continuous.

If f is absolutely continuous on lines, then it has partial derivativesfx and fy for almost all z = x + iy ∈ U . From Egoroff’s theorem, fis differentiable at almost all z ∈ U . Thus we have fz(z) and fz(z) foralmost all z ∈ U such that

f(w)− f(z) = fz(z)(w − z) + fz(z)(w − z) + o(|w − z|).We then define a measurable function

µ(z) =fz(z)

fz(z), z ∈ U.

Moreover. fz and fz are partial derivatives in the distribution sense,that is, ∫ ∫

fz(z)h(z)dzdz = −∫ ∫

f(z)hz(z)dzdz

108

and ∫ ∫fz(z)h(z)dzdz = −

∫ ∫f(z)hz(z)dzdz

hold for all C∞ functions with compact support.

Definition 25. Suppose f : U → V is an orientation-preservinghomeomorphism from the domain U to the domain V . The we say fis K-quasiconformal for some K > 1 if

a) f is ACL;b) ‖µ‖∞ ≤ k = (K − 1)/(K + 1).

Suppose f is K-qiasiconformal in Definition 25. Then for any quadri-lateral B = B(z1, z2, z3, z4) ⊂ U , we have a conformal map φ fromB onto the rectangle R = R(0, a, a + bi, bi) and a conformal mapψ from f(B) = B′(f(z1), f(z2), f(z3), f(z4)) to the rectangle R =

R(0, a′, a′ + b′i, b′i). Then f = φ f ψ−1 : R → R′ is a homeomor-phism, ACL, maps vertices to vertices. Following Grotzsch argumentin Lecture 10.1, we can show that

(a′/b′) : (a/b) ≤ K.

This says m(B′) ≤ Km(B). Thus f is K-quasiconformal in Defini-tion 23. If f is a quasiconfornmal mapping in Definition 23, then it isACL. So it has partial derivatives fz and fz a.e. and differentiable a.e.This implies b) in Definition 25. For more detailed arguments, refer to

• L. V. Ahlfors, Quasiconformol Mappings. Van Nostrand, 1966• O. Lehto and K. Virtanen, Quasiconformal Mappings in the

Plane. Springer-Verlag, 1973.

Thus Definition 25 and Definition 23 are equivalent.Checking f is ACL in Definition 23 is important even if f has partial

derivatives almost everywhere. For example, let Λ be a Cantor set as weconstructed in Lecture 3 as the non-escaping set set for a dynamicalsystem f : I0 ∪ I1 → I with zero Lebesgue measure. Then we have[0, 1] = G ∪ ∪wkGwk ∪ Λ. Define f(x) = 1/2 on G and

β(x) =1

22+

k∑j=0

ij2k+1

x ∈ Gwk .

Then β : [0, 1]→ [0, 1] is a onto map with β′(x) = 0 for all x ∈ ∪wkGwk

but is not absolutely continuous and maps Λ to [0, 1]. It is called adevil staircase. Now consider the homeomorphism

f(z) = z + β(z + z

2)i

109

on the unit square D = [0, 1]× [0, 1]. It is clear f has partial derivatives

fz = 1

and

fz = 0

for almost all z ∈ D. But it is not quasiconformal because, otherwise, itmust be conformal (refer to Theoren 73), but clearly it is not conformal.The reason is that it is not ACL.

From Definition 25, for a K-quasiconformal map f on U , we definea measurable function µf on U such that k = ‖µf‖∞ < 1, whereK = (1 + k)/(1− k). The measurable function µf is called a Beltramicoefficient. In general, let

M(C) = µ ∈ L∞ | ‖µ‖∞ < 1

is the unit open ball of L∞ functions on C. Consider the Beltramiequation,

(10) wz = µwz, a.e.

where µ ∈ M(C) is called a Beltrami coefficient. If w is a quasicon-formal homeomorphism, then it is a solution of the Betrami equationwith the Betrami coefficient µw = wz/wz. A quasiconformal homeo-morphism is called normalized if it fixes 0, 1,∞. The following theoremplays an important role in Teichmuller theory and theory of Riemannsurfaces.

Theorem 76 (Measurable Riemann Mapping Theorem). For anyµ ∈ M(C), the Beltrami equation has a quasiconformal solution. Andthe normalized solution, denoted as wµ, is unique and holomorphicallydepends on µ.

To prove this theorem, we first need to study two operators: Letζ = ξ + ηi,

Tµ(z) = − 1

π

∫ ∫C

µ(ζ)

(ζ − z)2dξdη = lim

ε→0+− 1

π

∫ ∫|z−ζ|>ε

µ(ζ)

(ζ − z)2dξdη

where µ ∈ C20 (C2 with compact support in C) and

Pµ(z) = − 1

π

∫ ∫Cµ(ζ)

( 1

ζ − z− 1

ζ

)dξdη

where µ ∈ Lp(C) for p > 2.

110

Lemma 17. Pµ is continuous and satisfies the uniform Holder con-dition with exponent 1− 2/p, that is,

|Pµ(z)− Pµ(w)| ≤ Kp‖µ‖p|z − w|1−2/p, z, w ∈ C

where Kp > 0 is a constant.

Lemma 18. For any µ ∈ C20 , Tµ exists and is of class C1. Further-

more,

(Pµ)z = µ and (Pµ)z = Tµ

and ∫ ∫C|Tµ(ζ)|2dξdη =

∫ ∫C|µ(ζ)|2dξdη.

In the above lemma, Tµ is an isometry on L2(C) ∩ C20 . Since C2

0 isdense in L2, T can be extended to an isometry on L2. As we noticethat Tµ also acts on Lp for p > 2. We have the following importantinequality.

Lemma 19 (Zygmund-Calderon). For any µ ∈ C20 , we have

‖Tµ‖p ≤ Cp‖µ‖pfor any p > 1 and Cp → 1 as p→ 2.

This lemma implies that Tµ can be extended as a bounded operatordefined on Lp(C) for p > 1. Thus we two operators Tµ and Pµ definedon Lp(C) for p > 2 and generalize Lemma 18 as follows.

Lemma 20. For any µ ∈ Lp(C), p > 2, we have

(Pµ)z = µ and (Pµ)z = Tµ

in the distribution sense.

Theorem 77 (Special Case). For any µ ∈ M(C) with compact sup-port, the Beltrami equation (10) has a unique solution such that f(0) =0 and fz − 1 ∈ Lp(C).

Proof. Let us prove the uniqueness first. Suppose f is a solution in thetheorem. Then we have fz = µfz. Since µ has a compact support andfz − 1 ∈ Lp(C), fz ∈ Lp(C). Then we have Pfz such that (Pfz)z = fz.Consider the function F = f − Pfz. We have Fz = 0 a.e.. Therefore,F is analytic from Weyl’s lemma. F ′ = fz − (Pfz)z = fz − Tfz. SinceTfz ∈ Lp(C) and fz − 1 ∈ Lp(C), we get F ′ − 1 ∈ Lp(C). This ispossible only if F ′ = 1. Thus F = z + a for some constant a. Thisimplies that f = z + a+ Pfz. But f(0) = 0 implies that f = z + Pfz.

111

Suppose g is another solution. Then we also have g = z + Pgz. Thisimplies that

f − g = P (fz − gz) = P (µ(fz − gz)).This implies that

fz = gz = T (µ(fz − gz)).Lemma 19 implies that

‖fz − gz‖p ≤ kCp‖fz − gz‖pwhere k = ‖µ‖∞ < 1. Let p→ 2. Then ‖µ‖∞Cp → k, this implies that

fz = gz and fz = gz a.e. Thus both f − g and f − g are analytic. Thusf − g = const. This constant must be 0. We get f = g.

The proof of the uniqueness also suggests a proof of the existence.Consider the equation

(11) h = T (µh) + Tµ.

We first solve this equation in Lp(C) for a number p > 2 such thatλ = kCp < 1. Let µ0 = µ, µn = Tµn−1 for n ≥ 1. Lemma 19 impliesthat

‖µn‖p ≤ λ‖µn−1p ≤ · · · ≤ λn‖µ0‖p.Then we have a convergent series

(12) h =∞∑n=1

µn

in Lp(C). Then h is a solution of Equation (11). We claim

f = z + P (µ(h+ 1))

is a solution of Equation (10). The reason is that first, µ(h+ 1) ∈ Lp,so P (µ(h+ 1)) is well-defined and continuous function on C (actually,a Holder continuous function on C). Secondly,

(13) fz = µ(h+ 1), fz = T (µ(h+ 1)) + 1 = h+ 1.

This implies that fz − 1 ∈ Lp(C). Clearly f(0) = 0.

The solution in the theorem is called a normalized solution of (10).Next we claim that the solution f is a homeomorphism. From thisclaim that f is a quasiconformal homeomorphism since it satisfies theequation (10).

Now let us prove the claim. First we prove it under assumption thaµ ∈ C1

0 . Then we have µz. We would like first solve the equation

(14) λz = µλz + µz.

112

Similar to find a solution of the Beltrami equation (10), we first findthe solution

q = T (µq) + Tµz

by convergent power series

q = Tµz + TµTµz + . . .+ TµTµTµz + . . .

in Lp(C) by choosing p > 2 such that kCp < 1. Set

σ = P (µq + µz) + const

such that σ → 0 as z →∞. Thus σ is continuous and

σz = µq + µz, σz = T (µq + µz) = q.

Define λ = eσ. Then it is the solution of Equation (14). Let p = λand q = µλ. We have that pz = qz are continuous. Green’s Theoremimplies that

f(z) =

∫ z

0

pdz + qdz

defines a C2 function on C. It is the unique solution of Equation (10).Since σ → 0 as z → ∞, fz = λ → 1 as z → ∞. Thus f(z) → ∞ asz →∞. The Jacobian

J(f)(z) = |fz|2 − |fz|2 = (1− |µ|2)e2σ > 0, ∀z ∈ C.This implies that f is locally 1-1. This combining with f(z) → ∞ asz →∞, we get that both f and f−1 are 1-1 and onto and continuous,therefore, f : C→ C is a homeomorphism.

From Lemma 19 and (11), we have

‖h‖p ≤ kCp‖h‖p + Cp‖h‖p.So

(15) ‖h‖p ≤Cp

1− kCp‖µ‖p.

From (6),

‖fz‖p ≤1

1− kCp‖µ‖p.

Lemma 17 implies that

(16) |f(z)− f(w)| ≤ Kp

1− kCp‖µ‖p|z − w|1−2/p + |z − w|.

Applying on f−1, we get

(17) |z − w| ≤ Kp

1− kCp‖µ‖p|f(z)− f(w)|1−2/p + |f(z)− f(w)|.

113

Now, for any µ ∈ L∞(C) with compact support, let µn ∈ C10 such

that µn → µ a.e. and ‖µn‖∞ ≤ k. Then the normalized solutions fnare K-quasiconformal homeomoprhisms for K = (1 + k)/(1− k) for alln > 0. Then (17) implies that f is 1-1 since fn → f and ‖µn‖p → ‖µ‖pas n→∞. Similarly, f−1 is also 1-1. Both f and f−1 are continuous.So f is K-quasiconformal homeomorphism.

Proof of Theorem 76. If µ has a compact support, we only need tonormalize f by using a Mobis transformation. The unique solutionfixing 0, 1, ∞ is denoted as fµ.

If µ has no compact support, we decompose µ = µ1 + µ2 such thatµ1 has a compact support and µ2 = 0 in a neighborhood of 0. Thenwe fµ1 . Define

µ3(z) = µ2(1/z)z2

z2 .

Then µ3 has a compact support. So we have fµ3 . Then 1/fµ3(1/z)defines a homeomorphism of C which fixes 0, 1,∞. We call this home-omorphism as fµ2 = 1/fµ3(1/z). It is the unique solution of the Bel-trami equation (10) with the Beltrami coefficient µ2. Now define

µ4 =(( µ− µ2

1− µµ2

)(fµ2zfµ2z

)) (fµ2)−1.

Then µ4 has a compact support. So we have fµ4 . Now define

fµ = fµ4 fµ2 .We have that fµ is the unique solution of Beltrami equation (10) withthe Beltrami coefficient µ.

From Equation (12), if we consider tµ for ‖µ‖∞ = 1 and t ∈ ∆, h isconvergent power series of t. Thus f tµ2 and f tµ4 are holomorphicallydependent on t ∈ ∆. This implies f tµ is also holomorphically dependenton t ∈ ∆.

114

Lecture 12. Local dynamics of a holomorphic map,irrational neutral case.

Suppose f is a holomorphic map defined on a domain U containing0. Suppose 0 ∈ U is an irrational neutral fixed point, that is, f(0) = 0andf ′(0) = e2πiθ for θ ∈ [0, 1] \ Q. Let λ = e2πiθ. Consider the Taylorexpansion

f(w) = λw + a2w2 + a3w

3 + · · · .If all ai = 0, then f(w) = λw is a linear map. In general, we would liketo know that it is linear under an appropriate coordinate. We still use

∆r = w ∈ C | |w| < r

to denote the disk centered 0 with radius r > 0. If there is a conformalmap φ : U ′ ⊂ U → ∆r such that φ f = λφ on U ′, then we call fis linearizable. By considering h = φ−1, this is equivalent to that theSchroder equation

f(h(w)) = h(λw), h(0) = 0, h′(0) = 1,

has a solution as we studied before.A real number θ ∈ [0, 1] \ Q is called Diophantine if there are con-

stants C > 0 and 0 < µ <∞ such that

|θ − p

q| ≥ C

qµ

for all p and q 6= 0. This condition is equivalent to

|λn − 1| ≥ Cn1−µ

for all n ≥ 1. For a fixed µ > 2, |θ − pq| ≥ C/qµ holds for almost all

real number θ ∈ [0, 1] \Q. Indeed if E is the set of θ ∈ [0, 1] \Q suchthat |θ − p/q| < q−µ infinitely many often, then

m(E) ≤∞∑q=n

2q−µ+1 = O(n2−µ)→ 0.

In particular, almost all real numbers in [0, 1] \Q are Diophantine. Apositive answer for the linearization problem is given in 1942 by Siegelby using the Diophantine condition.

Theorem 78. If θ is Diophantine and if f(w) = λw + · · · has a fixedpoint 0 with multiplier λ = e2πiθ, then there is a solution to the Schroderequation, that is, f can be conjugated near 0 to the linear map w 7→ λw.

115

Proof. Let C > 0 and µ > 0 be two constants such that |λn − 1| ≥Cn1−µ for all n ≥ 1. We want to find a conformal solution h defined ona disk ∆r such that h(λw) = f(h(w)). If we define f(w) = λw + f(w)

and h(w) = w + h(w). Then the equation can be written as

h(λw)− λh(w) = f(h(w)), w ∈ ∆r.

Instead to solve h directly, we consider a coordinate change ψ(w) =

w + ψ(w) with ψ(w) = O(w2). Then

ψ−1 f ψ(w) = λw + g(w).

If g ≡ 0, then h = ψ is a solution for the Schroder equation. If g 6= 0,we want |g(w)| < |f(w)|. Suppose

f(w) =∞∑j=2

ajwj, w ∈ ∆r, r < 1.

Let

ψ(w) =∞∑j=2

bjwj.

Solving the equation

ψ(λw)− λψ(w) = f(w),

we get bj = aj/(λj − λ). By the Diophantine condition, the series

ψ(w) =∞∑j=2

ajλj − λ

wj

converges and defines an analytic function on ∆r.For a δ > 0, let r > 0 small so that |f ′(w)| < δ for w ∈ ∆r. Now let

us estimate g in

g(w) = ψ−1 f ψ = λw + g(w).

Since

f (j)(0) =(j − 1)!

2πi

∫∂∆r

f ′(ξ)

ξjdξ,

|aj| = |f (j)(0)/j!| ≤ δ

jrj−1.

116

Let ∆r(1−η) = w ∈ C | |w| < r(1 − η), where 0 < η < 1/5. Forw ∈ ∆r(1−η),

|ψ′(w)| ≤∞∑j=2

jaj|λj−1 − 1|

rj−1(1− η)j−1

≤ Cδ

∞∑j=2

jµ−1(1− η)j−1

≤ Cδ∞∑j=2

j[µ](1− η)j−1

≤ C0δ∞∑j=2

j(j − 1) · · · (j − [µ] + 1)(1− η)j−1 =C0δ

η[µ]+1.

Take 0 < δ < η small, we also assume C0δ < η[µ]+2. Then

|ψ′(w)| < η, w ∈ ∆r(1−η).

So we haveψ(∆r(1−4η)) ⊂ ∆r(1−3η).

Furthermore, ψ takes every value in ∆r(1−2η) precisely once because|ψ(w)| ≥ r(1 − 2η) for w ∈ ∂∆r(1−η), the argument principle, andψ(w) = 0 has only one solution 0 in Dr(1−η). Now consider g = ψ−1 f ψ and Ui = ∆r(1−iη). Since ψ(U4) ⊂ U3 and δ < η, f(U3) ⊂ U2.Moreover, since ψ−1(U2) ⊂ U3, g(U4) ⊂ U1.

Let us further estimate g on U4. Since

g(w) + ψ(λw + g(w)) = λψ(w) + f(w + ψ(w)), w ∈ U4,

g(w) = ψ(λw)− ψ(λw + g(w)) + f(w + ψ(w))− f(w), w ∈ U4.

Now we conclude our discussion above. Take f(w) = λw + f(w) such

that |f ′| ≤ δ < η < 1/5 on U0 = ∆r and such that C0δ < η[µ]+2. Then

we can find a conjugacy ψ(w) = w + ψ(w) such that for

g(w) = ψ−1 f ψ(w) = λw + g(w),

|g′(w)| ≤ C0δ2rη−([µ]+2)(1− η)−1, w ∈ U4.

Let us label δ0 = δ and η0 = η. Define

rn+1 = rn(1− 5ηn), ηn+1 =ηn2, δn+1 = C0δ

2nη−([µ]+2)n 2−([µ]+2).

The require condition C0δn ≤ η[µ]+2n is now easy to verify by the induc-

tion because if it holds for n, then

C0δn+1 = C20δ

2nη−([µ]+2)n 2−([µ]+2) ≤ η2([µ]+2)

n η−([µ]+2)n 2−([µ]+2) = η

[µ]+2n+1 .

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So it also holds for n+ 1. Thus we can construct inductively sequencesof maps ψn and gn with g0 = f and

gn = ψ−1n gn−1 ψn = ψ−1

n ψ−1n−1 · · · ψ−1

1 f ψ1 · · · ψn.Let τ = r

∏∞i=1(1− 5ηn) > 0. Then we have that

g′n(w) ≤ δnrn/(1− ηn), w ∈ ∆τ = w ∈ C; |w| < τ, n > 0.

Thus ψ1· · ·ψn and gn tend to a conformal map h and the linear mapλw, respectively, as n goes to infinity. So h conjugates f to λw.

The above Theorems view the linearization problem from the com-binatorical point of views. There are many new developments in thisdirection. For example, a more precise combinatorical condition suchthat f is linearizable is so called the Brjuno condition as follows. Sup-pose θ ∈ [0, 1] \Q. Let

θ = [a1, a2, · · · , ] =1

a1 + 1

a2+...

be the continuous fractional expansion of θ. Let

pnqn

= [a1, a2, · · · , an] =1

a1 + 1

a2+...+ 1

an

.

Then θ is called a Brjuno number if∞∑n=1

log qn+1

qn<∞.

The sufficiency of this condition was proved by Brjuno in 1965 forf(w) = e2πiθw + w2 + · · · and necessity condition was established byYoccoz in 1988.

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We will discuss the following topics in the future semesters.

Lecture 13. Holomorphic motions and ap-plications

13.1. λ-Lemma, from Mane-Sad-Sullivan’s paper

13.2. Bers-Royden theorem, from Bers-Royden’s paper.

13.3. Slodkowski theorem, Chirka’s proof from Gardiner-Jiang-Wang’s paper.

13.4. Proof of the lifting theorem, an equivalent statement toSlodkowski’s theorem, from Jiang-Mitra-Wang’s paper

13.5. Canonical holomorphic replacement, a new proof ofSlodkowski’s theorem, from Gardiner-Jiang’s paper

13.6. Infinitesimal holomorphic motions, from Gardiner-Jiang-Wang’s paper

13.7. Structural stability theorem, from Mane-Sad-Sullivan’spaper

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Lecture 14. Gibbs Theory

14.1. Expanding and mixing dynamical systems.

14.2. Maximal eigenvalue of a Ruelle-Perron-Frobenius op-erators.

14.3. Smooth invariant measures for expanding dynamicalsystems.

14.4. Chains of Markovian projections.

14.5. Gibbs distributions.

14.6. Hilbert metric, second eigenvalue, and convergencespeed.

14.7. Spectra of Ruelle-Perron-Frobenius operators.

14.8. Essential spectrum radius of a Ruelle-Perron-Frobeniusoperator.

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Lecture 15. Characterization of rationalmaps

15.1. Branched coverings.

15.2. Orbifold structures.

15.3. Rational maps

15.4. The Teichmuller space of a closed subset

15.5. Pull-back operator.

15.6. Multi-curves

15.7. Characterization of post-critically finite rational maps

15.8. Geometrically finite branched coverings

15.9. Characterization of sub-hyperbolic rational maps

15.10. Characterization of some entire and meromorphic func-tions