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Geometric Gibbs theory

Jiang Yunping

Citation: SCIENCE CHINA Mathematics; doi: 10.1007/s11425-019-1638-6

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Geometric Gibbs theoryIn Memory of Professor Shantao Liao

Yunping Jiang1,2

1Department of Mathematics, Queens College of the City University of New York,New York, NY 11367-1597, USA;

2Department of Mathematics, The CUNY Graduate Center, New York, NY 10016, USA

Email: [email protected]

Received July 19, 2019; accepted November 19, 2019

Abstract We extend the classical Gibbs theory for smooth potentials to the geometric Gibbs theory for certain

continuous potentials. We study the existence and uniqueness and the compatibility of geometric Gibbs measures

associated with these continuous potentials. We introduce a complex Banach manifold structure on the space

of these continuous potentials as well as on the space of all geometric Gibbs measures. We prove that with this

complex Banach manifold structure, the space is complete and, moreover, is the completion of the space of all

smooth potentials as well as the space of all classical Gibbs measures. There is a maximum metric on the space,

which is incomplete. We prove that the topology induced by the newly introduced complex Banach manifold

structure and the topology induced by the maximal metric are the same. We prove that a geometric Gibbs

measure is an equilibrium state, and the infimum of the metric entropy function on the space is zero.

Keywords geometric Gibbs measure, continuous potential, smooth potential, Teichmuller’s metric, maximum

metric, Kobayashi’s metric, symmetric rigidity, complex Banach manifold

MSC(2010) 37D35, 37F30, 37E10, 37A05

Citation: Jiang Y P. Geometric Gibbs theory. Sci China Math, 2020, 63, https://doi.org/10.1007/s11425-019-

1638-6

1 Introduction

The mathematical theory of Gibbs measures, an important idea originating in physics, is a beautiful

mathematical theory from the celebrated work of Sinai [28, 29] and Ruelle [25, 26]. It leads to the study

of Sinai-Ruelle-Bowen (SRB)-measures in Anosov dynamical systems and, more generally, Axiom A

dynamical systems due to the further work of Sinai, Ruelle, Bowen and many others (see [2]). An essential

feature of a Gibbs measure is that it is an equilibrium state. This equilibrium state plays a significant

role in mathematics, as well as many other areas such as physics, chemistry, biology, economics, and

game theory.

An important topic in the current study of Gibbs measures is to study the deformation of a Gibbs

measure and its density (see, for example, [20, 27]). To have a nice deformation theory, an appropriate

metric on the space of all Gibbs measures is helpful. In this paper, we introduce a complex Banach

manifold structure on the space of all Gibbs measures and, then study a generalized Gibbs measure

which we call a geometric Gibbs measure.

A classical Gibbs measure is for a smooth potential (i.e., at least Holder smoothness). In general,

for a continuous potential, one may not have an appropriate Gibbs theory. We prove that for a specific


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2 Jiang Y P Sci China Math

continuous potential defined in this paper, there exists a geometric Gibbs measure. In addition, we study

the uniqueness and the compatibility of this geometric Gibbs measure.

We introduce a complex Banach manifold structure on the space of all continuous potentials defined

in this paper, and then on the space of all geometric Gibbs measures. We prove that under this complex

Banach manifold structure, the space is complete and, moreover, is the completion of the space of all

smooth potentials. This further says that the complex Banach manifold structure on the space of all

geometric Gibbs measures is complete and is the completion of the space of all classical Gibbs measures.

There is a maximum metric on this space. It is incomplete. We prove that the topology induced from

the newly introduced complex Banach manifold structure and the topology induced by the maximum

metric are the same. We prove that a geometric Gibbs measure is an equilibrium state, study the metric

entropy function defined on the space, and prove that the infimum of the metric entropy function is zero.

We organize the rest of this paper as follows. In Section 2, we review the classical Gibbs theory for

smooth potentials and the classical g-measure theory for continuous potentials with a detailed proof of

Ledrappier’s theorem (see Theorem 2.2) on g-measures. In Section 3, we study circle homeomorphisms

and circle endomorphisms. We review a quasisymmetric circle homeomorphism (see Definition 3.1) and a

symmetric circle homeomorphism (see Definition 3.3). We then review a uniformly quasisymmetric circle

endomorphism (see Definition 3.8) and a uniformly symmetric circle endomorphism (see Definition 3.9).

We prove two examples of uniformly symmetric circle endomorphisms. One is a C1+α expanding cir-

cle endomorphism (see Example 3.10) and the other is a C1+Dini expanding circle endomorphism (see

Example 3.12). We mention the complex analytic characterization of a uniformly symmetric circle endo-

morphism (see Definition 3.14 and Lemma 3.15). In Section 4, we review the symbolic representation of

the unit circle and the sequence of Markov partitions induced by a circle endomorphism. In Section 5, we

define the bounded nearby geometry (see Definition 5.1) on the sequence of Markov partitions induced

by a circle endomorphism and prove that the bounded nearby geometry is an equivalent condition to

the uniform quasisymmetry condition on a circle endomorphism (see Theorem 5.3). In Section 6, we

define those continuous potentials which we study in this paper (see Definition 6.1 and Theorem 6.4).

Furthermore, we prove that these continuous potentials are complete symmetric conjugacy invariants

(see Theorem 6.7). In Section 10, we define geometric Gibbs measures associated with the continuous

potentials in this paper (see Definition 10.1). In the same section, we state and prove our existence

and uniqueness and compatibility results (see Theorem 10.2). We prove the equilibrium property for a

geometric Gibbs measure in the same section (see Theorem 10.3). We also prove our related symmetric

rigidity results and smooth geometric Gibbs measure results and state our conjectures in the same section

(see Theorems 10.6, 10.7 and 10.13, Corollaries 10.8 and 10.9, and Conjectures 10.10–10.12). In Sections 7

and 11, we introduce and prove a complex Banach manifold structure on the space of these continuous

potentials, as well as the space of all geometric Gibbs measures (see Theorem 11.1). In Section 8, we prove

that under this complex Banach manifold structure, the space is complete and, moreover, the completion

of the space of all smooth potentials (see Theorem 8.1). In Section 9 we show that there is an incomplete

maximum metric on the space and prove that the topology induced by the newly introduced complex

Banach structure and the topology induced by the maximum metric are the same (see Theorem 9.1 and

Corollary 9.2). For the complete complex Banach manifold which we introduced in Sections 7 and 11,

we can define Kobayashi’s metric on it. In Section 12, we conjecture that Kobayashi’s metric and Teich-

muller’s metric on this complete complex Banach manifold coincide (see Conjecture 12.1). In Section 13,

we define the metric entropy function on the space of all geometric Gibbs measures and prove that the

infimum of the metric entropy function is zero (see Theorem 13.1).

2 Classical Gibbs theory

In this section, we give a brief review of the classical Gibbs theory. Suppose d > 2 is a positive integer.


Jiang Y P Sci China Math 3

Consider the symbolic dynamical system σ : Σ → Σ, where

Σ = w = · · · in−1 · · · i1i0 | in−1 ∈ 0, . . . , d− 1, n = 1, 2, . . . =0∏∞

0, . . . , d− 1

and the shift map σ : w = · · · in−1 · · · i1i0 → σ(w) = · · · in−1 · · · i1. The space Σ =∏0

∞0, 1, . . . , d − 1is a compact topological space with the product topology. We purposely write w = · · · in−1 · · · i1i0 from

the right to the left because we will use v = j0j1 · · · jn−1 · · · to represent a point on the unit circle and

treat σ+ : Σ+ = v = j0j1 · · · jn−1 · · · → Σ+ as the symbolic representation of the dynamics of a circle

endomorphism of degree d in Section 4. We will treat σ : Σ → Σ as the dual symbolic representation for

the dynamics of a circle endomorphism of degree d later.

An n-cylinder [w]n containing w = · · · in−1 · · · i1i0 is the subset of all elements

[w]n = w′ = · · · i′n+m · · · i′nin−1 · · · i0 | i′n+m ∈ 0, . . . , d− 1, m = 0, 1, . . ..

All cylinders [w]n are open sets and form a topological basis for Σ. Let B be the standard Borel σ-field

on Σ generated by all cylinders [w]n.

Let R be the real line and let ϕ : Σ → R be a real-valued function. Recall that a continuous function ϕ

is called Holder continuous if there are two constants C > 0 and 0 < τ < 1 such that |ϕ(w)−ϕ(w′)| 6 Cτn

as long as w and w′ are in the same n-cylinder.

We use C = C(Σ) to denote the space of all continuous functions ϕ and CH = CH(Σ) to denote the

space of all Holder continuous functions ϕ. A positive function in C is called a continuous potential and

a positive function in CH is called a smooth potential.

Let M be the space of all finite Borel measures on Σ, which is the dual space of C and let M(σ) be

the space of all σ-invariant probability measures in M, i.e., the space of measures with total measure 1

and satisfying

µ(σ−1(A)) = µ(A), ∀A ∈ B.

The classical Gibbs theory (see, for example, [2, 14, 25, 29]) ensures that given a smooth potential ψ,

there are a number P = P (logψ), called the pressure, and a unique σ-invariant probability measure

µ = µψ, called the Gibbs measure associated with ψ, such that

C−1 6 µ([w]n)

exp(−Pn+∑n−1i=0 logψ(σi(w)))

6 C (2.1)

for any n-cylinder [w]n, where C is a fixed constant.

Two smooth potentials ψ and ψ are said to be cohomologous, denoted as ϕ ∼C ψ, if there is a

continuous function u such that

log ψ = logψ + u σ − u. (2.2)

If ψ and ψ are cohomologous, then Gibbs measures associated with them are the same. Let

[ψ]C = ψ ∼C ψ | ψ ∈ CH

be the cohomologous equivalence class of ψ and CCH be the space of all cohomologous equivalence classes.

Gibbs measures depend only on cohomologous equivalence classes, i.e., ϕ ∼C ψ if and only if µψ = µψ.

Then we have a bijective map between CCH and the space of all Gibbs measures.

A Gibbs measure is an equilibrium state in the sense that

P (logψ) = entµ(σ) +

∫Σ

logψ dµ = entµ(σ) +

∫Σ

logψ dµ = supν∈M(σ)

entν(σ) +

∫Σ

logψdν

,

where entν(σ) is the measure-theoretical entropy of σ with respect to ν. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


We can pick a normalized smooth potential in each cohomologous equivalence class as follows. Given

a smooth potential ψ, consider the positive transfer operator (or called the Ruelle operator) associated

with ψ:

Lψϕ(w) =∑

σ(w′)=w

ψ(w′)ϕ(w′) : CH → CH . (2.3)

For any ϕ ∈ C and ν ∈ M, define

⟨ϕ, ν⟩ =∫Σ

ϕdν.

Then we can define the dual operator L∗ψ : M → M as

⟨ϕ,L∗ψν⟩ = ⟨Lψϕ, ν⟩, ϕ ∈ C, ν ∈ M. (2.4)

We have a positive real number λ = eP and a positive function ρ ∈ CH such that Lψρ = λρ, where λ

is the unique, maximal, positive, simple eigenvalue of Lψ (see, for example, [13]). We have also a unique

νψ ∈ M(σ) such that L∗ψνψ = λνψ. Take ρ such that ⟨ρ, νψ⟩ = 1. Then µ = ρνψ is the Gibbs measure

associated with ψ. Take

g =ψ · ρ

λ · ρ σ. (2.5)

It is in [ψ]C and uniquely characterized by a normalization condition

d−1∑i=0

g(wi) = 1. (2.6)

One can check that the Gibbs measure µ associated with [ψ]C satisfies that L∗gµ = µ. We have a bijective

map between the space of all normalized smooth potentials and the space of all Gibbs measures. In

general, a continuous potential satisfying (2.6) is called a g-function (see, for example, [21]).

For any continuous function g, we can define the transfer operator Lg associated with it similarly to

Lgϕ(w) =∑

σ(w′)=w

g(w′)ϕ(w′) : C → C. (2.7)

Let L∗g be its dual operator. For any ν ∈ M, we define a new measure ν = L∗

1ν.

Lemma 2.1. Any ν ∈ M(σ) is absolutely continuous with respect to ν.

Proof. For any A ∈ B,

ν(A) = ⟨1A,L∗1ν⟩ = ⟨L11A, ν⟩ =

d−1∑j=0

ν(σ(A ∩ [j])).

Since ν is σ-invariant, we have that

ν(A) =d−1∑j=0

ν(σ(A ∩ [j])) =d−1∑j=0

ν(σ−1(σ(A ∩ [j]))) >d−1∑j=0

ν(A ∩ [j]) = ν(A).

Therefore, ν(A) = 0 whenever ν(A) = 0. So ν is absolutely continuous with respect to ν.

From this lemma, we have a Radon-Nikodym derivative for any ν ∈ M(σ),

RNDν,ν(w) =dν

dν(w), ν-a.e. w ∈ Σ.

The following theorem is due to Ledrappier [22] (see also [30]). We give a complete proof in this section

in order to have a self-contained paper. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Theorem 2.2 (Ledrappier’s theorem). Suppose g is a g-function. The following statements are

equivalent:

(i) L∗gµ = µ.

(ii) µ ∈ M(σ) and RNDµ,µ(w) = g(w) for µ-a.e. w ∈ Σ.

(iii) µ ∈ M(σ) and

E[ϕ | σ−1(B)](w) = Lgϕ(σ(w)) =∑

σ(w′)=σ(w)

g(w′)ϕ(w′), for µ-a.e. w ∈ Σ,

where E[ϕ | σ−1(B)] is the conditional expectation of ϕ with respect to σ−1(B).(iv) µ ∈ M(σ) and is an equilibrium state for g in the sense that

0 = entµ(σ) +

∫Σ

log gdµ = supν∈M(σ)

entν(σ) +

∫Σ

log gdν

.

Proof. Note that since C is dense in the space L1(µ) of all µ-measurable and integrable functions (as

well as in the space L1(µ)), ⟨·, ·⟩ can be extended to L1(µ) (as well as to L1(µ)).

We prove (i) implies (ii). Let us show that any Gibbs measure µ is σ-invariant as follows: For any

A ∈ B,µ(σ−1(A)) = ⟨1σ−1(A), µ⟩ = ⟨1A σ,L∗

gµ⟩ = ⟨Lg1A σ, µ⟩ = ⟨1A, µ⟩ = µ(A).

So µ ∈ M(σ).

For any ϕ ∈ C,

⟨ϕg, µ⟩ = ⟨ϕg,L∗1µ⟩ = ⟨L1(ϕg), µ⟩ = ⟨Lgϕ, µ⟩ = ⟨ϕ,L∗

gµ⟩= ⟨ϕ, µ⟩ = ⟨ϕRNDµ,µ, µ⟩.

Thus RNDµ,µ(w) = g(w) for µ-a.e. w ∈ Σ.

We prove that (ii) implies (i). For any ϕ ∈ C,

⟨ϕ, µ⟩ = ⟨ϕRNDµ,µ, µ⟩ = ⟨ϕg,L∗1µ⟩

= ⟨L1(ϕg), µ⟩ = ⟨Lgϕ, µ⟩ = ⟨ϕ,L∗gµ⟩.

It implies L∗gµ = µ. This is (i).

We prove (i) implies (iii). For any A ∈ B,

⟨(Lgϕ) σ · 1σ−1(A), µ⟩ = ⟨(L1(ϕg)) σ · 1A σ, µ⟩ = ⟨L1(ϕg) · 1A, µ⟩= ⟨L1(ϕg1A σ), µ⟩ = ⟨Lg(ϕ1A σ), µ⟩ = ⟨ϕ1A σ, µ⟩ = ⟨ϕ1σ−1(A), µ⟩,

i.e.,

E[ϕ | σ−1(B)] = (Lgϕ) σ, µ-a.e. w ∈ Σ.

Note that

((Lgϕ) σ)(w) =∑

v∈σ−1(σ(w))

g(v)ϕ(v).

We now prove that (iii) implies (i). Since, for any ϕ ∈ C,

E[ϕ | σ−1(B)] = Lgϕ(σ(w)), µ-a.e. w ∈ Σ,

we have ⟨ϕ, µ⟩ = ⟨(Lgϕ) σ, µ⟩ = ⟨Lgϕ, µ⟩ = ⟨ϕ,L∗gµ⟩. Thus L∗

gµ = µ. This is (i).

We prove that (ii) implies (iv). For any ν ∈ M(σ), let

RNDν,ν =dν

dν, µ-a.e. w ∈ Σ



be the Radon-Nikodym derivative. We claim that

hν(σ) = −∫Σ

log RNDν,νdν.

We prove this claim. Since − log RNDν,ν is a non-negative ν-measurable function and since ν is absolutely

continuous with respect to ν, it is also a ν-measurable function. Thus∫Σ

− log RNDν,νdν =

∫Σ

−RNDν,ν log RNDν,νdν.

By definition,

RNDν,ν(w) = limn→∞

ν([w]n)

ν([σ(w)]n−1), ν-a.e. w ∈ Σ.

Denote

RNDn,ν,ν(x) =ν([w]n)

ν([σ(w)]n−1).

Then

−RNDν,ν(w) log RNDν,ν(w) = limn→∞

(−RNDn,ν,ν(w) log RNDn,ν,ν(w)), ν-a.e. w ∈ Σ.

Since −t log t is a positive bounded function on [0, 1], by the Lebesgue control convergence theorem,

we have∫Σ

limn→∞

(−RNDn,ν,ν(w) log RNDn,ν,ν(w)) dν = limn→∞

∫Σ

−RNDn,ν,ν(w) log RNDn,ν,ν(w)dν.

Because ν([w]n) = ν([σ(w)]n−1),∫Σ

−RNDn,ν,ν(w) log RNDn,ν,ν(w)dν =∑[w]n

− ν([w]n)

ν([σ(w)]n−1)log

(ν([w]n)

ν([σ(w)]n−1)

)ν([w]n)

=∑[w]n

−ν([w]n) log(

ν([w]n)

ν([σ(w)]n−1)

).

We know that

hν(σ) = limn→∞

1

n

∑[w]n

−ν([w]n) log ν([w]n)

= limn→∞

∑[w]n

−ν([w]n) log(

ν([w]n)

ν([σ(w)]n−1)

).

We proved the claim.

The claim says that hν(σ) = −⟨log RNDν,ν , ν⟩ for any ν ∈ M(σ). By using the fact that log t 6 t− 1,

we get

hν(σ) + ⟨log g, ν⟩ =⟨log

g

RNDν,ν, ν

⟩6

⟨g

RNDν,ν− 1, ν

⟩=

⟨g

RNDν,ν, ν

⟩− 1

= ⟨g, ν⟩ − 1 = ⟨L1g, ν⟩ − 1 = ⟨1, ν⟩ − 1 = 1− 1 = 0.

The assumption in (ii) is that RNDµ,µ(w) = g(w), µ-a.e. w ∈ Σ. But for µ ≪ µ, RNDµ,µ(w) = g(w),

µ-a.e. w ∈ Σ too. Since log t = t − 1 if and only if t = 1 and since g/RNDµ,µ(w) = 1 µ-a.e. w ∈ Σ, we

can improve the above inequality to get hµ(σ) +∫Σlog g dµ = 0. So µ is an equilibrium state for g.

Finally, we prove that (iv) implies (i). Suppose µ ∈ M(σ) is an equilibrium state for g. We have that

hµ(σ) + ⟨log g, µ⟩ = 0. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


We already know that hµ(σ) + ⟨log RNDµ,µ, µ⟩ = 0. So we have that

hµ(σ) + ⟨log g, µ⟩ = hµ(fσ + ⟨log RNDµ,µ, µ⟩).

Therefore,

0 = ⟨log g − log RNDµ,µ, µ⟩ =⟨log

g

RNDµ,µ, µ

⟩6

⟨g

RNDµ,µ− 1, µ

⟩=

⟨g

RNDµ,µ, µ

⟩− 1 = ⟨g, µ⟩ − 1

= ⟨g,L∗1µ⟩ = ⟨L1g, µ⟩ − 1 = ⟨1, µ⟩ − 1 = 1− 1 = 0.

Again by using the fact that log t = t− 1 if and only if t = 1, we get

g(w)

RNDµ,µ(w)= 1, µ-a.e. w ∈ Σ. (2.8)

For any ϕ ∈ C,

⟨ϕ,L∗gµ⟩ = ⟨Lgϕ, µ⟩ = ⟨L1(gϕ), µ⟩

= ⟨gϕ,L∗1µ⟩ = ⟨ϕg, µ⟩ =

⟨ϕ

g

RNDµ,µ, µ

⟩= ⟨ϕ, µ⟩.

This says that L∗gµ = µ. We proved (i).

Remark 2.3. The equality (2.8) cannot imply that

g(w)

RNDµ,µ(w)= 1, µ-a.e. w ∈ Σ,

since µ may not be absolutely continuous with respect to µ. So, it will not imply (ii). However, if

g(w) > 0 for all w ∈ Σ, then

RNDµ,µ(w) =dµ

dµ(w) =

1

RNDµ,µ(w)=

1

g(w), µ-a.e. w ∈ Σ.

This implies that µ is absolutely continuous with respect to µ. Then Equation (2.8) implies (ii). Fur-

thermore, if g is a smooth potential, then RNDµ,µ is actually a Holder continuous function and

RNDµ,µ(w) ≡ g(w) (2.9)

on Σ. However, this fact may not be true in general for an only continuous function g. One of our goals

is to generalize (2.9) for the certain continuous potentials g. The associated measure µ will be called

a geometric Gibbs measure. One of the main purposes of this work is to study the existence and the

uniqueness of geometric Gibbs measures associated with these continuous potentials defined in Section 6.

These continuous functions will satisfy (2.6). This suggests that they relate to circle endomorphisms of

degree d, which we will discuss in the next section.

3 Circle endomorphisms

We use C = z = x+ yi | x, y ∈ R to denote the complex plane with the norm |z| =√x2 + y2. Let

H = z ∈ C | y > 0

be the upper-half plane and D = z ∈ C | |z| < 1 be the open unit disk. Let

T = ∂D = z ∈ C | |z| = 1 http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


be the unit circle. Then R is the universal cover of T. Let π(x) = e2πix : R → T be the universal covering

map with π(0) = 1.

Consider an orientation-preserving covering map f : T → T. Let d = df be the degree of f . Then

every f can be lifted to an orientation-preserving homeomorphism

F : R → R, F (x+ 1) = F (x) + d, ∀x ∈ R.

We will assume throughout this paper that F (0) = 0, i.e., f(1) = 1. Then there is a one-to-one corre-

spondence between f and F . If d > 1, we call f as well as F a circle endomorphism. If d = 1, we call it

a circle homeomorphism and use h and H to denote it and its lift. Note that

H : R → R, H(x+ 1) = H(x) + 1, ∀x ∈ R and H(0) = 0.

Definition 3.1 (Quasisymmetric circle homeomorphism). A circle homeomorphism h is called qua-

sisymmetric if there is a constant M > 1 such that

M−1 6 |H(x+ t)−H(x)||H(x)−H(x− t)|

6M, ∀x ∈ R, ∀ t > 0. (3.1)

We also call h an M -quasisymmetric circle homeomorphism.

We use QS to denote the space of all quasisymmetric circle homeomorphisms h with h(1) = 1.

Lemma 3.2 (Compactness). Suppose M > 1 is a fixed real number. Then the set QS(M) of all

M -quasisymmetric circle homeomorphisms h ∈ QS is compact under the uniform convergence topology.

Proof. Let H be the lift of h on R such that H(0) = 0. Then H(1) = 1. From Condition (3.1), we have

1

M + 16 H

(1

2

)6 M

M + 1.

Inductively, we have (1

M + 1

)n6 H

(1

2n

)6

(M

M + 1

)n.

For any x0 ∈ [0, 1], consider H(x) = H(x0 + x) − H(x0). It is also a circle homeomorphism since

H(x+ 1) = H(x) + 1 and M0-quasisymmetric. So we have also

H(x0 + x)−H(x0) 6 H

(x0 +

1

2n

)−H(x0) 6

(M

M + 1

)nfor any 0 6 x 6 1/2n. This says QS(M) is equicontinuous on T. Clearly, the set of all lifts of QS(M)

on any compact set of R is uniformly bounded. This implies that QS(M) is compact under the uniform

convergence topology.

An important feature about a quasisymmetric homeomorphism h is that it can be extended to a

quasiconformal homeomorphism h of D. Here, we can use Buerling-Ahlfors’ extension on the lift H.

Suppose

H(z) = u(x, y) + iv(x, y), ∀ z = x+ yi ∈ H,

where

u(x, y) =1

2y

∫ y

−yH(x+ t)dt and v(x, y) =

1

2y

∫ y

0

(H(x+ t)−H(x− t))dt.

Then we have u(x, 0) = H(x) and v(x, y) > 0 and v(x, y) → 0 as y → 0. Thus H |R = H(x). Since H is

a homeomorphism of H satisfying H(z + 1) = H(z) + 1, it is a lift of a homeomorphism h of D \ 0 and

h |T = h. Considering the Beltrami coefficient for H,

µH =Hz

Hz

∈ L∞(H),



we have that µH(z + 1) = µH(z) on H and

k = ∥µH∥∞ 6 2M(1 +M)− 1

2M(1 +M) + 1< 1

(refer to [1]). Since k is finite, h is quasiconformal and 0 is a removable point, thus h is actually a

quasiconformal homeomorphism of D and µH is its complex quasiconformal dilatation on D called a

Beltrami coefficient.

In general, a Beltrami coefficient on D is a periodic complex function µ of period 1 in L∞(H) such that

k = ∥µ∥∞ < 1. Let K = (1+ k)/(1− k) > 1. A homeomorphism h of D is called K-quasiconformal if its

lift H is a solution of the Beltrami equation

Hz = µ(z) · Hz.

So another equivalent complex analytic definition of an M -quasisymmetric circle homeomorphism h is

that it is the restriction of a K-quasiconformal homeomorphism h of D. We have the following relation

between M and K (refer to [1]):

K 6 2M(1 +M), M 6 16eπK .

Definition 3.3 (Symmetric circle homeomorphism). A circle homeomorphism h ∈ QS is called sym-

metric if there is a bounded real function ε(t) > 0 for t > 0 such that ε(t) → 0+ as t → 0+ and such

that1

1 + ε(t)6 |H(x+ t)−H(x)|

|H(x)−H(x− t)|6 1 + ε(t), ∀x ∈ R, ∀ t > 0.

We use S to denote the space of all symmetric circle homeomorphisms h ∈ QS. Let C1 be the space

of all C1-circle diffeomorphisms fixing 1.

Example 3.4. It holds that C1 ⊂ S.However, S is a much larger space than C1. For example, h ∈ S may not be absolutely continuous and

it is a typical phenomenon in S.From [8], we know that a circle homeomorphism H is symmetric if and only if it has a quasiconformal

homeomorphism extension H on the upper-half plane H such that its Beltrami coefficient

µH(z) =Hz(z)

Hz(z)→ 0, as z = x+ yi, y → 0 uniformly on x. (3.2)

Thus we have the following complex analytic characterization of a symmetric circle homeomorphism.

Definition 3.5 (Asymptotically conformal circle homeomorphism). A quasiconformal homeomorphism

h of the unit disk D is said to be asymptotically conformal if its lift H satisfies (3.2).

Lemma 3.6 (Analytic characterization for symmetry). A circle homeomorphism h is symmetric if and

only if it has an asymptotically conformal extension h on D.We will not provide a proof of this lemma. The reader who is interested in this lemma can find a

detailed proof in [8].

The following composition formula for Beltrami coefficients is useful for us. Suppose h1 and h2 are two

quasiconformal homeomorphisms of D. Then

µH1H−12

H2 =(H2)z

(H2)z·µH1

− µH2

1− µH2· µH2

. (3.3)

Then Lemma 3.6 and Equation (3.3) imply the following lemma.

Lemma 3.7 (Composition). Suppose h1 and h2 are two symmetric circle homeomorphisms. Then

h1 h2 is also a symmetric circle homeomorphism.



Proof. Let h3 = h1 h2. Then h1 = h3 h−12 . Let h1 and h2 be asymptotical conformal extensions

of h1 and h2, respectively. Then h3 = h1 h2 is an extension of h3 on D. Since there is a constant

0 6 k < 1 such that ∥µH1∥∞, ∥µH2

∥∞ 6 k, the equation (3.3) implies that ∥µH1∥∞ < 1. Thus h3 is

quasisymmetric. Moreover, the equation (3.3) implies that µH3satisfies (3.2) too. This says that h3 is

an asymptotically conformal extension of h3. Therefore, h3 is symmetric.

Now we pay attention to circle endomorphisms.

Definition 3.8 (Uniformly quasisymmetric circle endomorphism). A circle endomorphism f is called

uniformly quasisymmetric if there is a constant M > 1 such that

1

M6 |F−n(x+ t)− F−n(x)|

|F−n(x)− F−n(x− t)|6M, ∀x ∈ R, ∀ t > 0, ∀n > 1.

Furthermore, we have the following definition.

Definition 3.9 (Uniformly symmetric circle endomorphism). A circle endomorphism f is called uni-

formly symmetric if there is a bounded real function ε(t) > 0 for t > 0 such that ε(t) → 0+ as t → 0+

and such that

1

1 + ε(t)6 |F−n(x+ t)− F−n(x)|

|F−n(x)− F−n(x− t)|6 1 + ε(t), ∀x ∈ R, ∀ t > 0, ∀n > 1.

Let UQS = UQS(d) and US = US(d) be the space of all uniformly quasisymmetric circle endomor-

phisms of degree d > 2 and the space of uniformly symmetric circle endomorphisms of degree d > 2. It

is clear that US ⊂ UQS.A circle endomorphism f is C1 if its derivative F ′ exists and is continuous. Furthermore, it is called

C1+α for some 0 < α 6 1 if F ′ is α-Holder continuous, i.e.,

supx =y∈R

|F ′(x)− F ′(y)||x− y|α

= supx =y∈[0,1]

|F ′(x)− F ′(y)||x− y|α

<∞.

A C1 circle endomorphism f is called expanding if there are constants 0 < C 6 1 and λ > 1 such that

(Fn)′(x) > Cλn, x ∈ R, ∀n > 1.

Example 3.10. A C1+α circle expanding endomorphism f is uniformly symmetric. Furthermore, we

have ε(t) 6 Dtα for some constant D > 0 and 0 6 t 6 1.

Proof. Since F (x+ 1) = F (x) + d, F ′(x+ 1) = F ′(x) is a periodic function. Since F is C1+α, we have

a constant C1 > 0 such that

|F ′(x)− F ′(y)| 6 C1|x− y|α, ∀x, y ∈ R.

Since F is expanding, we have a constant C2 > 0 and λ > 1 such that

(Fn)′(x) > C2λn, ∀x ∈ R, n > 0.

For any x, y ∈ R and n > 0, let xk = F−k(x) and yk = F−k(y), 0 6 k 6 n. Then∣∣∣∣ log (F−n)′(x)

(F−n)′(y)

∣∣∣∣ = ∣∣∣∣ log (Fn)′(yn)

(Fn)′(xn)

∣∣∣∣ 6 n∑k=1

| logF ′(xk)− logF ′(yk)|

6 1

C2λ

n∑k=1

|F ′(xk)− F ′(yk)| 6C1

C2λ

n∑k=1

|xk − yy|α 6 C1

C1+α2 λ

n∑k=1

λ−αk|x− y|α.

Let

C =C1λ

α

C1+α2 (λα − 1)λ

.



Then we have the following Holder distortion property:

e−C|x−y|α 6 (F−n)′(x)

(F−n)′(y)6 eC|x−y|α , ∀x, y ∈ R, ∀n > 0. (3.4)

Furthermore, let

ε(t) =

eCt

α − 1, 0 < t 6 1,

eC − 1, t > 1.

Then ε(t) > 0 is a bounded function such that ε(t) → 0 as t→ 0+ and such that

1

1 + ε(t)6 (F−n)′(ξ)

(F−n)′(η)=

|F−n(x+ t)− F−n(x)||F−n(x)− F−n(x− t)|

6 1 + ε(t), ∀x ∈ R, ∀ t > 0,

where ξ and η are two numbers in [0, 1]. Thus F is uniformly symmetric. Furthermore, one can see that

ε(t) 6 Dtα for some constant D > 0 and 0 6 t 6 1. We proved the example.

Remark 3.11. The uniformly symmetric condition is a much weaker condition than the C1+α ex-

panding condition. For example, a uniformly symmetric circle endomorphism could be totally singular,

i.e., it could map a set with positive Lebesgue measure to a set with zero Lebesgue measure. But we will

see in the rest of the paper, from the dual geometric point of view, many dynamical aspects of a C1+α

expanding circle endomorphism will be preserved by a uniformly symmetric circle endomorphism.

Another example of a uniformly symmetric circle endomorphism is a C1+Dini expanding circle endo-

morphism as follows. Suppose f is a C1 circle endomorphism. The function

ω(t) = sup|x−y|6t

|F ′(x)− F ′(y)|, t > 0 (3.5)

is called the modulus of continuity of F ′. Then f is called C1+Dini if∫ 1

0

ω(t)

tdt <∞. (3.6)

Consider a C1+Dini expanding circle endomorphism f . Let 0 < C 6 1 and λ > 1 be two constants such

that

(Fn)′(x) > Cλn, x ∈ R, n > 1.

Without loss of generality, we assume that C = 1. Define

ω(t) =∞∑n=1

ω(λ−nt). (3.7)

Then

ω(t) 6∫ ∞

0

ω(λ−xt)dx =1

log λ

∫ λ−1t

0

ω(y)

ydy <∞

for all 0 6 t 6 1 and ω(t) → 0 as t→ 0.

Example 3.12. A C1+Dini circle expanding endomorphism f is uniformly symmetric. Furthermore,

ε(t) 6 Dω(t), 0 6 t 6 1, for some constant D > 0.

Proof. Since F (x+ 1) = F (x) + d, F ′(x+ 1) = F ′(x) is a periodic function. Suppose

(Fn)′(x) > λn, ∀x ∈ R, n > 0,

where λ > 1 is the expanding constant.

For any x, y ∈ R and n > 0, let xk = F−k(x) and yk = F−k(y), 0 6 k 6 n. Then∣∣∣∣ log (F−n)′(x)

(F−n)′(y)

∣∣∣∣ = ∣∣∣∣ log (Fn)′(yn)

(Fn)′(xn)

∣∣∣∣ 6 n∑k=1

| logF ′(xk)− logF ′(yk)|



6 1

λ

n∑k=1

|F ′(xk)− F ′(yk)| 61

λ

n∑k=1

ω(λ−k|x− y|).

Let C = 1/λ. Then we have the following Dini distortion property:

e−Cω(|x−y|) 6 (F−n)′(x)

(F−n)′(y)6 eCω(|x−y|), ∀x, y ∈ R, ∀n > 0. (3.8)

Furthermore, let

ε(t) =

eCω(t) − 1, 0 < t 6 1,

eCω(1) − 1, t > 1.

Then ε(t) > 0 is a bounded function such that ε(t) → 0 as t→ 0+ and such that

1

1 + ε(t)6 (F−n)′(ξ)

(F−n)′(η)=

|F−n(x+ t)− F−n(x)||F−n(x)− F−n(x− t)|

6 1 + ε(t), ∀x ∈ R, ∀ t > 0,

where ξ and η are two numbers in [0, 1]. Thus F is uniformly symmetric. Furthermore, we have a constant

D > 0 such that ε(t) 6 Dω(t) for all 0 6 t 6 1. We proved the example.

Remark 3.13. We note that ε(t) in Example 3.12 may not be a Dini modulus of continuity anymore.

This fact will affect our study of symmetric rigidity later.

The complex analytic characterization of a uniformly symmetric circle endomorphism is given in [7].

Suppose f is a circle endomorphism. Then its lift F is a homeomorphism of R. Let F be a quasiconformal

extension of F to H. Then we can calculate the Beltrami coefficient

µF−n =F−nz

F−nz

∈ L∞(H)

for every n > 0. The quasiconformal homeomorphism F gives a quasi-regular map f of D of degree d with

one branched point 0 such that F is its lift to H. We have the following characterization of a uniformly

symmetric circle endomorphism as given and proved in [7].

Definition 3.14 (Uniformly asymptotically conformal circle endomorphism). A quasi-regular map f

of D is said to be uniformly asymptotically conformal if

µF−n(z) =F−nz (z)

F−nz (z)

→ 0, as z = x+ yi, y → 0 uniformly on x and n > 0. (3.9)

Theorem 3.15 (Analytic characterization for uniform symmetry). A circle endomorphism f is uni-

formly symmetric if and only if it has a uniformly asymptotically conformal quasi-regular extension f

on D.Since its proof is quite lengthy, we will not provide it here. The reader who is interested in this

characterization can find a detailed proof of this theorem in [7].

4 Symbolic representation of T and dynamics of circle endomorphisms

For any interval I = [a, b] in [0, 1], we use |I| = b−a to denote its Lebesgue length. We also use |J | = b−ato mean the Lebesgue length of the interval J ⊂ T with π(I) = J . We use m0 to denote the Lebesgue

probability measure on T. The measure m0 on any interval J ⊂ T is m0(J) = |J | = |I|.Consider the space

Σ+ =∞∏0

0, 1, . . . , d− 1

with the product topology. Then

Σ+ = v = j0j1 · · · jn−1 · · · | jn−1 ∈ 0, 1, . . . , d− 1, n = 1, . . .. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


An n-cylinder containing v = j0j1 · · · jn−1 · · · is

[v]+n = v′ = j0j1 · · · jn−1j′nj

′n+1 · · · | j′n+k ∈ 0, 1, . . . , d− 1, k = 0, 1, . . ..

All cylinders [v]+n are open sets and form a topological basis for Σ+. The space Σ+ with this topology

is a compact space and can be thought of as the symbolic representation of the unit circle T as follows:

Any z = e2πix ∈ T can be represented by a point v = j0j1 · · · jn−1 · · · ∈ Σ+ for

x = x(v) =∞∑k=0

jkdk+1

. (4.1)

Define the length on Σ+ as |v − v′| = |x(v)− x(v′)| = |z − z′|. The Lebesgue measure on T can be lifted

to Σ+ as the Lebesgue probability measure which we still denote as m0 such that m0([v]+n ) = 1/dn for

any n-cylinder in Σ+. Let B+ be the standard Borel σ-field generated by all cylinders [v]+n in Σ+.

For v = j0j1 · · · jn−1 · · · ∈ Σ+, let

σ+(v) = j1 · · · jn−1jn · · ·

be the shift map. Then

σ+ : Σ+ → Σ+

is called a symbolic dynamical system. Let M+ be the space of all Borel finite measures on Σ+ and

M+(σ+) be the space of all σ+-invariant probability measures in M+, i.e., all probability µ+ ∈ M+

such that

µ+((σ+)−1(A)) = µ+(A)

for all Borel subsets A ∈ B+.

Now consider a circle endomorphism f of degree d > 2 with f(1) = 1. Consider the preimage f−1(1).

Then f−1(1) cuts T into d closed intervals J0, J1, . . . , Jd−1, ordered by the counter-clockwise order of T.Suppose J0 has an endpoint 1. Then Jd−1 also has an endpoint 1. Let

ϖ0 = ϖ0,f = J0, J1, . . . , Jd−1.

Then it is a Markov partition, i.e.,

(1) T =∪d−1j=0 Jj ,

(2) the restriction of f to the interior of Jj is injective for every 0 6 j 6 d− 1,

(3) f(Jj) = T for every 0 6 j 6 d− 1.

Let I0, I1, . . . , Id−1 be the lifts of J0, J1, . . . , Jd−1 in [0, 1]. Then we have

(i) [0, 1] = ∪d−1j=0Ij ,

(ii) F (Ij) = [j, j + 1] for every 0 6 j 6 d− 1.

Let

η0 = η0,f = I0, I1, . . . , Id−1.

Then it is a Markov partition of [0, 1].

Consider the pull-back partition ϖn = f−nϖ0 of ϖ0 by fn. It contains dn intervals and is also a

Markov partition of T. Intervals J in ϖn can be labeled as follows. Let vn = j0j1 · · · jn−1 be a word of

length n of 0’s, 1’s, . . . , and (d−1)’s. Then J = Jvn if fk(J) ⊂ Jjk for 0 6 k 6 n−2 and fn−1(J) = Jjn−1 .

We have

ϖn = ϖn,f = Jvn | vn = j0j1 · · · jn−1, jk ∈ 0, 1, . . . , d− 1, k = 0, 1, . . . , d− 1.

Let ηn be the corresponding lift partition of ϖn in [0, 1] with the same labelings. Then

ηn = ηn,f = Ivn | vn = j0j1 · · · jn−1, jk ∈ 0, 1, . . . , d− 1, k = 0, 1, . . . , d− 1.

We call

ϖn∞n=0 as well as ηn∞n=0 (4.2) http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


the sequence of nested Markov partitions by f .

For a point v = j0 · · · jn−1 · · · ∈ Σ+, let vn = j0 · · · jn−1. Then

· · · ⊂ Jvn ⊂ Jvn−1 ⊂ · · · ⊂ Jv1 ⊂ T.

Since each Jvn is compact,

Jv =∞∩n=1

Jvn = ∅.

If every Jv = xv contains only one point, then we define the projection πf from Σ+ onto T as

πf (v) = xv.

The projection πf is one-to-one except for a countable set B which are mapped to endpoints of all

intervals Jvn . On B, πf is two to one. From our construction, one can check that

πf σ+(v) = f πf (v), v ∈ Σ+, (4.3)

i.e., f is always semi-conjugate to σ+.

Let

ιn,f = maxvn

|Jvn |,

where vn runs over all words of 0, 1, . . . , d − 1 of length n. Then every Jv contains only one point,

which is equivalent to say that ιn,f → 0 as n → ∞. We will always have this assumption in the rest

of the paper. Thus we call σ+ : Σ+ → Σ+ the topological representation for all circle endomorphisms

f : T → T.Two circle endomorphisms f and g are said to be topologically conjugate if there is a circle homeo-

morphism h of T such that

f h = h g.

From (4.3), one can easily check the following lemma.

Lemma 4.1. Let f and g be two circle endomorphisms such that both ιn,f and ιn,g tend to zero as

n→ ∞. Then f and g are topologically conjugate if and only if their topological degrees are the same.

Therefore, for a fixed degree d > 1, we can think σ+ : Σ+ → Σ+ as the symbolic representation of the

dynamics of all circle endomorphisms of degree d with ιn → 0 as n→ ∞.

5 Bounded nearby geometry

In this section, we study the geometry of a circle endomorphism. Most of the proofs are adapted from

our previous work (see also [12]).

Definition 5.1 (Bounded nearby geometry). Suppose f is a circle endomorphism of degree d > 1.

The sequence ϖn∞n=0 (as well as ηn∞n=0) of nested Markov partitions by f is said to have bounded

nearby geometry if there is a constant C > 0 such that for any n > 0 and any two adjacent intervals

J, J ′ ∈ ϖn (i.e., they have the same endpoint),

C−1 6 |J ′||J |

6 C.

Remark 5.2. The sequence ϖn∞n=0 of nested Markov partitions by f is said to have bounded ge-

ometry if there is a constant C > 0 such that

|L||J |

> C, ∀L ⊂ J, L ∈ ϖn+1, J ∈ ϖn, ∀n > 0.

The bounded nearby geometry implies the bounded geometry since each interval I ∈ ηn is divided into d

subintervals in ηn+1. But it is not true for the other direction. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Theorem 5.3. The sequence ϖn∞n=0 of nested Markov partitions by f has bounded nearby geometry

if any only if f is a uniformly quasisymmetric circle endomorphism.

We prove this theorem through several lemmas.

Lemma 5.4. Suppose f is a uniformly quasisymmetric circle endomorphism. Then the sequence

ϖn∞n=0 of nested Markov partitions by f has bounded nearby geometry.

Proof. Let F with F (0) = 0 be the lift of f . Define

Gj(x) = F−1(j + k) : [0, 1] → [0, 1], for j = 0, 1, . . . , d− 1.

For any word vn = j0j1 · · · jn−1, define

Gvn = Gj0 Gj1 · · · Gjn−1 .

Suppose Jvn and J ′vn are two adjacent intervals in ϖn. Let Ivn and I ′vn be their lifts in [0, 1]. Then

Ivn = Gvn([0, 1]) = F−n([m,m+ 1]),

where m = jn−1 + jn−2d+ · · ·+ j0dn−1 and

I ′vn = F−n([m+ 1,m+ 2]) or F−n([m− 1,m]).

Thus

M−1 6 |Ivn ||I ′vn |

=|Jvn ||J ′vn |

6M.

This means that ϖn∞n=0 has the bounded nearby geometry. This proves the lemma.

From this lemma, we have a constant 0 < τ < 1 such that

ιn,f = maxvn

|Jvn,f | 6 τn, ∀n = 1, 2, . . . , (5.1)

for a uniformly quasisymmetric circle endomorphism f .

Let q(z) = zd and Q(x) = dx be its lift. From Lemma 4.1, we have a homeomorphism hf of T such

that

f hf = hf q.

Remember that we use Hf to denote the lift of hf on R.Lemma 5.5. Suppose the sequence ϖn∞n=0 of nested Markov partitions by f has bounded nearby

geometry. Then hf is a quasisymmetric circle homeomorphism.

Proof. Suppose

ηn,f = Ivn,f and ηn,q = Ivn,q, n = 1, 2, . . .

are two sequences of nested Markov partitions by f and by q lift to [0, 1], respectively. Then

Hf (Ivn,q) = Ivn,f .

One can check that

Ivn,q =

[m

dn,m+ 1

dn

],

where vn = j0 · · · jn−1 and 0 6 m = jn−1 + jn−2d + · · · + j0dn−1 < dn. For x = m/dn + k, where k is

an integer, and t = 1/dn, [x, x + t] = Ivn,q + k and [x − t, x] = Iv′n,q + k are two adjacent intervals and

[Hf (x), Hf (x+ t)] = Ivn,f + k and [Hf (x− t),Hf (x)] = Iv′n,f + k. From Lemma 5.4, since the sequence

ηn,f has bounded nearby geometry, we have a constant M > 0 such that

M−1 6 |Hf (x+ t)−Hf (x)||Hf (x)−Hf (x− t)|

=|Ivn,f ||Iv′n,f |

6M.



LL′ L′′ L′′′

x − t x x + t

I I ′

Figure 1 Bounded nearby geometry

Now for any x ∈ R and any t > 0, consider two adjacent intervals [x− t, x] and [x, x+ t]. Without loss

of generality, let us assume that [x − t, x + t] ⊂ [0, 1]. Let n be the smallest integer such that there is

an interval I ∈ ηn contained completely in [x− t, x]. Let L be the interval in ηn−1 containing I. Either

[x − t, x] ⊆ L or we have an interval L′ ∈ ηn−1 adjacent to L such that L ∪ L′ ⊃ [x − t, x]. From the

relations between I, L, L′ and [x − t, x], we see 1/dn 6 t 6 2/dn−1. Assume L′ < L and let L′′ be the

other adjacent interval of L. If L′′ ⊂ [x, x + t], then let L′′′ be the next adjacent interval of L′′, and we

must have [x, x + t] ⊂ L ∪ L′′ ∪ L′′′; otherwise, we must have [x, x + t] ⊂ L ∪ L′′. Let L = L ∪ L′′ or

L ∪ L′′ ∪ L′′′ depending on the above situations. We have a constant integer n0 > 0 independent of x, t,

and n such that there is an interval I ′ ∈ ηn+n0 satisfying I ′ ⊂ [x, x+ t] (see Figure 1).

Because of the bounded nearby geometry for f , we have a constant M > 1 independent of x, t and n

such that1

M6 |Hf (I)|

|Hf (L)|6 |Hf (L

′ ∪ L)||Hf (I ′)|

6M.

This implies that1

M6 |Hf (I)|

|Hf (L)|6 Hf (x− t)−Hf (x)

Hf (x)−Hf (x+ t)6 |Hf (L

′ ∪ L)||Hf (I ′)|

6M.

This implies that Hf is an M -quasisymmetric circle homeomorphism. This completes the proof.

Proof of Theorem 5.3. The “if” part is Lemma 5.4. The proof of the “only if” part follows Lemma 5.5

as follows: Since f = hf q h−1f , fn = hf qn h−1

f . This implies that

F−n(x) = Hf Q−nd H−1

f (x) mod 1 on R,

where Qnd (x) = dnx. Since Qnd (z) = dnz are conformal maps for all n > 1 on the Riemann sphere

C = C ∪ ∞, from the composition formula (3.3) and Buerling-Ahlfors’ extension, F−n is an M2-

quasisymmetric homeomorphism on R for every n > 1. This implies that f is a uniformly quasisymmetric

circle endomorphism. This completes the proof.

Corollary 5.6. The conjugacy h between any two uniformly quasisymmetric circle endomorphisms f

and f of the same degree d > 1 is quasisymmetric.

Proof. Suppose f h = h f . Then h = hf h−1

f. Since both hf and hf are quasisymmetric, from the

composition formula (3.3) and Buerling-Ahlfors’ extension, h = hf h−1

fis quasisymmetric.

Corollary 5.6 says that any two uniformly quasisymmetric circle endomorphisms (therefore, two

uniformly symmetric circle endomorphisms or two C1+Dini expanding circle endomorphisms) are always

quasisymmetrically conjugate. Furthermore, we have the following definition.

Definition 5.7 (Symmetric conjugacy and smooth conjugacy). Two uniformly quasisymmetric circle

endomorphisms f and f are said to be symmetrically conjugate if the conjugacy h between f and f is

a symmetric circle homeomorphism. If, furthermore, h is a C1-diffeomorphism, then we say f and f are

smoothly conjugate.

We have proved the following result in [16].

Theorem 5.8. Suppose f is a C1 circle endomorphism and f is a C1 expanding circle endomorphism.

Suppose h is a Holder continuous circle homeomorphism such that f = h f h−1. Then f itself is

expanding.

Theorem 5.8 and the fact that every quasisymmetric circle homeomorphism is Holder continuous

(see [1]) imply the following corollary. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Corollary 5.9. Any C1 map f ∈ QUS is expanding.

6 Normalized continuous potentials

If f ∈ US is C1+α, then it is expanding from Theorem 5.9. Its derivative ψf = 1/f ′ π is a Holder

continuous function defined on Σ+ with a unique Gibbs measure associated with it. If f and f in US are

both C1+α and also smoothly conjugate by a diffeomorphism h, i.e., f h = h f , then ψf and ψf are

cohomologous, i.e.,

logψf = logψf + u σ+ − u.

However, a general uniformly symmetric circle endomorphism f may not be differentiable. Thus we

cannot use derivatives to study the thermodynamical formalism for a general uniformly symmetric circle

endomorphism.

In this section, we define a normalized continuous potential on Σ associated with a uniformly symmetric

circle endomorphism. We will use this normalized continuous potential to replace a missing derivative in

the study of thermodynamical formalism for a uniformly symmetric circle endomorphisms.

Suppose f is a circle endomorphism of degree d. For any w = · · · in−1 · · · i1i0 ∈ Σ, let

wn = in−1 · · · i1i0 and w′n−1 = σ(wn) = in−1 · · · i1.

If we let j0 = in−1, . . . , jn−2 = i1, jn−1 = i0, then we also have

vn = j0 · · · jn−2jn−1 and v′n−1 = j0 · · · jn−2.

(Note that σ+(vn) = j1 · · · jn−1.) Using the notation in Section 4, we have Jvn and Jv′n−1. For the

convenience of notation, we also use notations Jwn = Jvn and Jw′n−1

= Jv′n−1. Then

Jwn⊂ Jw′

n−1,

where Jwn ∈ ϖn and Jw′n−1

∈ ϖn−1 (see Figure 2).

Define

gn(w) =|Jwn ||Jw′

n−1|. (6.1)

Then we haved−1∑i=1

gn(wi) =d−1∑i=1

|Jwn−1i||Jwn−1 |

= 1. (6.2)

Definition 6.1 (Normalized function). If for every w ∈ Σ,

g(w) = limn→∞

gn(w) (6.3)

exists, then we define a function

g(w) = gf (w) : Σ → R.

It satisfies the equation (2.6) because of (6.2), and thus, is a g-function.

f

I = Iwn vn

I = Iwn−1′ vn−1′ I σ (v )n+

Figure 2 Relationship between intervals



We prove the following quasisymmetric distortion lemma. Suppose [a, b] is an interval and H : [a, b] →[c, d] is a homeomorphism. We say H is M -quasisymmetric on [a, b] if

M−1 6 |H(x+ t)−H(x)||H(x)−H(x− t)|

6M (6.4)

for any x, x+ t, x− t ∈ [a, b] and t > 0.

Lemma 6.2 (Quasisymmetric distortion). Suppose H is an M -quasisymmetric homeomorphism from

[0, 1] onto [0, 1] with H(0) = 0 and H(1) = 1. Then we have

|H(x)− x| 6M − 1, ∀x ∈ [0, 1].

The bound M − 1 is the sharpest estimation.

Proof. Let In,k = [k/2n, (k + 1)/2n] for n > 0 and 0 6 k 6 2n − 1. From (6.4), we have(1

1 +M

)n6 |H(In,k)| 6

(M

1 +M

)n.

This implies that, (1

1 +M

)n+H

(k

2n

)6 H

(k + 1

2n

)6

(M

1 +M

)n+H

(k

2n

).

For any even integer k = 2i,

−gn(M) +

(H

(i

2n−1

)− i

2n−1

)6 H

(k + 1

2n

)− k + 1

2n6

(h

(i

2n−1

)− i

2n−1

)+ fn(M),

where

fn(M) =

(M

M + 1

)n− 1

2nand gn(M) =

1

2n−

(1

M + 1

)n.

Consider hn(M) = fn(M)− gn(M) as a function of M > 1. We have that hn(1) = 0 and

h′n(M) =n(Mn−1 − 1)

(M + 1)n+1> 0, M > 1.

This implies that fn(M) > gn(M) for all M > 1. Thus we get that for any n > 1,

max06k62n

∣∣∣∣H(k

2n

)− k

2n

∣∣∣∣ 6 max06k62n−1

∣∣∣∣H(k

2n−1

)− k

2n−1

∣∣∣∣+ fn(M).

This implies that

max06k62n

∣∣∣∣H(k

2n

)− k

2n

∣∣∣∣ 6 n∑m=1

fm(M) =M − 1 +1

2n−M

(M

1 +M

)n6M − 1.

So we have

supn>0

max06k62n

∣∣∣∣H(k

2n

)− k

2n

∣∣∣∣ 6M − 1, ∀n > 0.

Since the dyadic points k/2nn>0,06k62n are dense in the unit interval [0, 1] and since H is continuous,

we get

|H(x)− x| 6M − 1, ∀x ∈ [0, 1].

For M > 1, let 0 < A < 1− 1/M − 1/M2 and define

H(x) =

Mx, 0 6 x 6 1

M2,

x− 1

M2+

1

M,

1

M26 x 6 1

M2+A,

1−A− 1M

1−A− 1M2

(x−A− 1

M2

)+A+

1

M,

1

M2+A 6 x 6 1.



Since 1/M < (1−A− 1/M)/(1−A− 1/M2) < M , H is M -quasisymmetric. One can check that

|H(1/M2)− 1/M2| = (M − 1)(1/M2).

As M → 1, we have shown that M − 1 is the sharpest possible estimation. This completes the proof.

Corollary 6.3 (Quasisymmetric distortion). Suppose H : [a, b] → [H(a),H(b)] is anM -quasisymmetric

homeomorphism. Then we have ∣∣∣∣H(d)−H(c)

H(b)−H(a)− d− c

b− a

∣∣∣∣ 6 2(M − 1)

for any a 6 c < d 6 b.

Proof. First let us prove it under the assumption that a = c. In this case let α : [a, b] → [0, 1] and

β : [H(a), H(b)] → [0, 1] be orientation-preserving linear maps such that

H = β H α−1 : [0, 1] → [0, 1]

fixes 0 and 1. Since α and β are both linear, H is again an M -quasisymmetric homeomorphism. Let

x = (d− a)/(b− a). Lemma 6.2 implies that∣∣∣∣H(d)−H(a)

H(b)−H(a)− d− a

b− a

∣∣∣∣ = |H(x)− x| 6M − 1.

In general,d− c

b− a=d− a

b− a− c− a

b− a

andH(d)−H(c)

H(b)−H(a)=H(d)−H(a)

H(b)−H(a)− H(c)−H(a)

H(b)−H(a).

Then we have∣∣∣∣H(d)−H(c)

H(b)−H(a)− d− c

b− a

∣∣∣∣ 6 ∣∣∣∣H(d)−H(a)

H(b)−H(a)− d− a

b− a

∣∣∣∣+ ∣∣∣∣H(c)−H(a)

H(b)−H(a)− c− a

b− a

∣∣∣∣ 6 2(M − 1).

This completes the proof.

Theorem 6.4 (Normalized continuous function). Suppose f is a uniformly symmetric circle endomor-

phism in US. Then the normalized function defined in Definition 6.1

g(w) = gf (w) : Σ → R

associated with f exists and is a positive and continuous function. Furthermore, if f is C1+α, 0 < α

6 1, then g(w) is a Holder continuous function. Actually when f is C1+Dini expanding, the modulus of

continuity of g(w) is ω(t) defined in (3.7).

Proof. Suppose w = · · · in−1 · · · i1i0 ∈ Σ. Let

wn = in−1 · · · i1i0 and w′n−1 = in−1 · · · i1.

By definition,

gn(w) =|Jwn ||Jw′

n−1|=

|Iwn ||Iw′

n−1|,

where Iwn ⊂ Iw′n−1

are the lifts of Jwn and Jw′n−1

in [0, 1]. Consider the sequence gn(w)∞n=1. Let

0 < τ < 1 be a constant such that

ιn = maxwn

|Iwn | 6 τn, ∀n > 1.



For any ϵ > 0, there is n0 > 0 such that ϵ(τn−1/2) < ϵ/2 for all n > n0. Now for any m > n > n0,

|fm−n(Jw′m−1

)| = |Jw′n−1

| and |fm−n(Jwm)| = |Jwn |.

Considering the map fm−n : Jw′m−1

→ Jw′n−1

, its inverse

H = f−(m−n) : Jw′n−1

→ Jw′m−1

is an M -quasisymmetric homeomorphism where M = (1 + ϵ(τn−1/2)). Consider [a, b] = Jw′n−1

and

[c, d] = Jwn . Then we have a 6 c < d 6 b. Moreover, [H(a), H(b)] = Jw′m−1

and [h(c),H(d)] = Jwm .

Corollary 6.3 implies that

|gm(w)− gn(w)| 6 2(M − 1) < ϵ.

In other words gn(w)∞n=1 is a Cauchy sequence. Thus the limit

g(w) = limn→∞

gn(w)

exists. Furthermore, consider two points

w = · · · im−1 · · · inin−1 · · · i0 and w = · · · i′m−1 · · · i′nin−1 · · · i0.

Let wm = im−1 · · · inin−1 · · · i0 and wm = im−1 · · · i′nin−1 · · · i0. Then wn = wn. For any m > n > n0,

|gm(w)− gm(w)| 6 |gm(w)− gn(w)|+ |gm(w)− gn(w)| 6 2ϵ.

So by taking a limit, we have

|g(w)− g(w)| 6 2ϵ.

This implies that

g(w) : Σ → R

is a continuous function. From Theorem 5.3 and Remark 5.2, g(w) is a positive function. The modulus

of continuity of g is ϵ(τn−1/2).

When f ∈ US is C1+α, it is expanding from Theorem 5.9. Then from the Holder distortion proper-

ty (3.4), there is a constant C > 0 such that

|g(w)− g(w)| 6 Cτα(n−1).

This implies that g(w) is Holder continuous and thus a smooth potential.

When f is C1+Dini, then there is a constant C > 0 such that |g(w)− g(w)| 6 Cω(τn−1). Thus g(w) is

continuous and its modulus of continuity is ω(τn−1). We complete the proof of the theorem.

Remark 6.5. Given w = · · · in−1 · · · i1i0 ∈ Σ, let wn = in−1 · · · i1i0 and σ(wn) = in−1 · · · i1. Then we

have

D∗f(w) = limn→∞

|Iσ(wn)||Iwn |

=1

g(w).

Thus D∗f = 1/g can be thought of as the dual derivative of f at w. Although a general uniformly

symmetric circle endomorphism f may not have a non-zero derivative at every point on T or even be

totally singular with respect to the Lebesgue measure on T, Theorem 6.4 says that it has always a positive

dual derivative at every point w ∈ Σ. Moreover, the dual derivative defines a continuous function on Σ.

Let

G = G(d) = gf (w) | f ∈ US (6.5)

be the space of all normalized continuous potentials from Theorem 6.4 for all uniformly symmetric circle

endomorphisms of degree d > 2. The constant function g0(w) ≡ 1/d is an example in G which is associated

with qd(z) = zd. However, it is the only one having the local constant property in G (see Subsection 10.2). http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Let C1+ be the space of all C1+α, for some 0 < α 6 1, circle endomorphisms in US. Every map in C1+

is expanding (see Theorem 5.9). Let

HG = HG(d) = gf (w) | f ∈ C1+ (6.6)

denote the space of all smooth potentials in G.Remark 6.6. Not every positive normalized continuous function (i.e., g-function) on Σ is in G. From[3, 4, 15] we know that for d = 2, a normalized continuous (resp. Holder) positive function g(w) ∈ G(2)(resp. ∈ HG(2)) must satisfy another compatibility condition

∞∏n=0

g(w0

n︷︸︸︷1 · · · 1)

g(w1 0 · · · 0︸︷︷︸n

)= const., ∀w ∈ Σ. (6.7)

For example, from Equation (6.7), this function must satisfy

g(· · · 000) = g(· · · 111).

Moreover, (2.6) and (6.7) give a complete characterization of a continuous (resp. Holder) function g(w)

in G(2) (resp. HG(2)).The next theorem shows that elements in G are complete symmetric invariants.

Theorem 6.7 (Complete symmetric invariant). Suppose f and f are two uniformly symmetric circle

endomorphisms in US. Then gf (w) = gf (w) if and only if f and f are symmetrically conjugate.

Proof. Let us first prove the “if” part. Suppose h f = f h and suppose h is symmetric. For any

w = · · · in−1 · · · i1i0 ∈ Σ, let

wn = in−1 · · · i1i0 and w′n−1 = in−1 · · · i1.

From

gn,f (w) =|Jwn ||Jw′

n−1|,

we get that

gn,f (w) =|h(Jwn)||h(Jw′

n−1)|.

Corollary 6.3 and the proof of Theorem 6.4 imply that

|gn,f (w)− gn,f (w)| < 2ϵ(τn−1/2),

where n is large enough such that h | Jw′n−1

is (1 + ϵ(τn−1/2))-quasisymmetric homeomorphism. Let n

tend to infinity. We get

gf (w) = gf (w).

It completes the proof of the “if” part.

Another way to prove the “if” part is to first conjugate F to a linear map. The method of this proof

can be also used to give a proof of the “only if” part. So let me give this proof first. Let

λ =1

g(· · · 000)= limn→∞

F−n(1)

F−(n+1)(1).

(The limit exists because of Lemma 6.2 and the proof is similar to that of Theorem 6.4.) Consider the

linear map

P (x) = λx : R → R. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


For each n > 0, consider the homeomorphism

γn(x) =F−n(x)

F−n(1): R → R.

For any ϵ > 0, there is an n0 > 0 such that F−m | [0, F−n(1)] is (1 + ϵ)-quasisymmetric for any

m > n > n0. Then

H(y) =F−m+n(F−n(1)y)

F−m(1): [0, 1] → [0, 1]

is a (1 + ϵ)-quasisymmetric homeomorphism with H(0) = 0 and H(1) = 1. Lemma 6.2 implies that

|H(y)− y| 6 ϵ, ∀ y ∈ [0, 1].

Thus for y = F−n(x)/F−n(1), x ∈ [0, 1], we have

|γn+m(x)− γn(x)| < ϵ, ∀m > n > n0.

This implies that γn(x)∞n=0 is a uniformly Cauchy sequence on [0, 1]. Thus it converges uniformly to a

function γ(x) on [0, 1]. Similarly the sequence of inverses γ−1n (y) = Fn(F−n(1)y)∞n=0 is also a uniformly

Cauchy sequence on [0, 1], and it converges uniformly to a function on [0, 1] which is the inverse of γ(x).

So γ(x) is a homeomorphism. A direct calculation implies that γ(x) is also symmetric on [0, 1] since the

sequence γn(x) is uniformly symmetric.

For any fixed k > 0, F−k maps [0, dk] onto [0, 1]. Using the relation

γn+k(x) =F−n(1)

F−n−k(1)γn(F

−k(x)), x ∈ [0, dk],

we get that γn(x)∞n=0 is a uniformly Cauchy sequence on [0, dk] and converges to a uniformly symmetric

homeomorphism γ(x) on [0, dk], since F−k∞k=0 is uniformly symmetric. We conclude that γn(x) is

a convergent sequence and converges uniformly on any compact set of R+ = [0,∞). Similarly, we have

that γn(x) is a convergent sequence and converges uniformly on any compact set of R− = (−∞, 0].

One can check that

γ F γ−1(x) = P (x), ∀x ∈ R.

This says that γ conjugates F to the linear dynamical system P (x) = λx.

Let

L(x) = Lf (x) = γ(γ−1(x) + 1). (6.8)

Then we have L(0) = 1. The map L is a symmetric homeomorphism which we call the linear model of f

because it plays a role like the translation x→ x+ 1 for the linear dynamical system P (x) such that

F (x+ 1) = F (x) + d

becomes

P (L(x)) = Ld(P (x)). (6.9)

For any w = · · · in−1 · · · i1i0 ∈ Σ, let wn = in−1 · · · i1i0 and w′n−1 = in−1 · · · i1. Since γ(x) is symmetric

on [0, 1] with γ(0) = 0 and γ(1) = 1, from Corollary 6.3 and the proof of Theorem 6.4, we have

gf (w) = limn→∞

|Jwn ||Jw′

n−1|= limn→∞

|Iwn ||Iw′

n−1|= limn→∞

|γ(Iwn)||γ(Iw′

n−1)|.

Consider the non-negative integers

k = i0 + i1d+ · · ·+ in−1dn−1 and k′ = i1 + i2d+ · · ·+ in−1d

n−2.

Then k = dk′ + i0 and

Iwn = F−n([k, k + 1]) and Iw′n−1

= F−(n−1)([k′, k′ + 1]). http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Since γ(F−n(x)) = λ−nγ(x) and since λγ(k′) = γ(F (k′)) = γ(dk′),

|γ(Iwn)||γ(Iw′

n−1)|

=λ|γ([k, k + 1])||γ([k′, k′ + 1])|

=|γ([k, k + 1])|

|γ([k − i0, k + d− i0])|.

This implies that

|γ([k, k + 1])||γ([k − i0, k + d− i0])|

= gf (· · · 000in−1 · · · i1i0). (6.10)

Since Lk(0) = γ(k), the above equality says that all values

gf (· · · 000in−1 · · · i1i0) | ik = 0, 1, . . . , (d− 1), k = 0, 1, . . . , n− 1

are completely determined by the points Lk(0)∞k=0. Thus if h f = f h and if h is symmetric, then

γ = γ H,

where H is the lift of h with H(x+ 1) = H(x) + 1. But

L(x) = γ(γ−1(x) + 1) = γ(H(H−1(γ−1(x)) + 1)) = γ(γ−1(x) + 1) = L(x).

This implies that

gf (· · · 000in−1 · · · i1i0) = gf (· · · 000in−1 · · · i1i0).

Since gf (w) and gf (w) are both continuous on the compact space Σ and since the set · · · 000in−1 · · · i1i0is dense in Σ, we conclude that gf (w) = gf (w) on Σ. This is another proof of the “if” part.

Now let us prove the “only if” part. Suppose gf (w) = gf (w) on Σ. From Equation (6.10), we get

gf (· · · 000in−1 · · · i1i0) =Lk+1(0)− Lk(0)

Lk+d−i0(0)− Lk−i0(0).

This implies that Lf (0) = 1 and Ldf (0) = gf (· · · 000)−1 = λ. From

gf (· · · 00i) =Li+1f (0)− Lif (0)

λ, 1 6 i 6 d− 1,

we have that the value of Li+1f (0) is uniquely determined by gf (· · · 00i), 1 6 i 6 d − 1. A similar

argument implies that all values of Lkf (0), 1 6 k < ∞ are determined by g(· · · 000in−1 · · · i1i0). Now

from Equation (6.9), we have that λn = Ldn

f (0) and A = λ−nLkf (0)dn

k=0 and B = λ−(n−1)Lkf (0)dn

k=0

give partition points on [0, 1] and on [0, λ], respectively. The map Lf maps partition points in A into

those in B. Inductively, from Equation (6.9), we see that Lf is uniquely determined by gf . This implies

that Lf = Lf and λ = gf (· · · 000)−1 = gf (· · · 000)−1. Since

Lf (x) = γ(γ(x) + 1) = γ(γ(x) + 1) = Lf (x),

H(x) = γ−1 γ(x) is a symmetric homeomorphism satisfying that H(x + 1) = H(x) + 1. This implies

that H is a symmetric circle homeomorphism. Since

F (x) = γ−1(λγ(x)) and F (x) = γ−1(λγ(x)),

we get that

F (x) = H−1 F H(x).

So f and g are symmetrically conjugate. We complete the proof of the theorem.



7 Teichmuller’s metric

Let qd(z) = zd be the basepoint in US. We define the Teichmuller space for US as follows. For any

f ∈ US, let hf be the conjugacy from f to qd, i.e.,

f hf = hf qd.

From Lemma 5.5, hf is quasisymmetric. Thus we can think of US as the space of marking pairs (f, hf ).

We define an equivalence relation ∼T : Two pairs (f, hf ) ∼T (f , hf ) if hf h−1f is symmetric. The

Teichmuller space

T US = [(f, hf )] | (f, hf ) ∈ US, with the basepoint [(qd, id)]

is the space of all ∼T -equivalence classes [(f, hf )]. From Theorem 6.7, we have a one-to-one correspon-

dence between G and T US. Thus we can write

G = T US. (7.1)

By using T US, we can introduce Teichmuller’s metric on G with the basepoint g0(w) ≡ 1/d.

For h ∈ QS, let Eh be the set of all quasiconformal extensions of h into D. For each h ∈ Eh, let

µh = hz/hz be its complex dilatation. Let

kh = ∥µ(z)∥∞ and Kh =1 + kh1− kh

.

Here, Kh is called the quasiconformal dilatation of h. Using quasiconformal dilatation, we can define a

pseudo-distance in QS by

d(h1, h2) =1

2inflogKh2h−1

1| h1 ∈ Eh1 , h2 ∈ E2.

It will induce a distance in the space of QS modulo the space of all Mobius transformations preserving D,which is called the universal Teichmuller space T . It is a complete metric space and a complex manifold

with the complex structure compatible with the Hilbert transform (see Section 11).

Now consider the asymptotical universal Teichmuller space

AQS = QS modulo S.

Given two cosets Sh1 and Sh2 in this factor space, define

d(Sh1,Sh2) = infA,B∈S

d(Ah1, Bh2). (7.2)

It defines a distance on AQS. The asymptotical Teichmuller space (AQS, d(·, ·)) is a complete

metric space and a complex manifold. The topology on (AQS, d) is the finest topology which makes

the projection π : T → AQS continuous, and π is also holomorphic (refer to [8]).

An equivalent topology on the quotient space AQS can be defined as follows. For any h ∈ QS, let hbe a quasiconformal extension of h to a small neighborhood U of T in the complex plane. Let

µh =hz

hz, z ∈ U

and

kh = ∥µ(z)∥∞,U and Bh =1 + kh1− kh

.

Then the boundary dilatation h is defined as

Bh = infU,h

Bh,



where the infimum is taken over all quasiconformal extensions h of h in a neighborhood U of T. It is

known that h is symmetric if and only if Bh = 1. Define

d(h1, h2) =1

2logBh2h−1

1. (7.3)

Then it is a distance on AQS. Two distances d and d on AQS are equal.

There is a natural embedding from

G = T US ∋ [f, hf ]ι→ [hf ] ∈ AQS.

Thus, the restriction of d(·, ·) = d(·, ·) to G gives a distance dG(·, ·) on G, which we call Teichmuller’s

metric on G. We call the space

(G, dG(·, ·)) (7.4)

the Teichmuller space. The basepoint is g0 = [(qd, id)].

We can also think of C1+ as the space of all marking pairs (f, hf ) and define an equivalence relation

∼S : Two pairs (f, hf ) ∼S (f , hf ) if hf h−1f is a C1-diffeomorphism. The Teichmuller space

T C1+ = [(f, hf )] | (f, hf ) ∈ C1+, with the basepoint [(qd, id)]

is defined as the space of all ∼S-equivalence classes [(f, hf )]. For two points Π and Π′ in T C1+, consider

d(Π,Π′) =1

2logBh

fh−1

f,

where Π,Π′ ∈ T C1+ and (f, hf ) ∈ Π and (f , τf ) ∈ Π′. It satisfies the symmetric condition and the

triangle inequality. If we also have that d(Π,Π′) = 0 if and only if Π = Π′, then d(·, ·) is indeed a distance

on T C1+. This is the next theorem.

Theorem 7.1 (Complete smooth invariance). Suppose f, f ∈ C1+. Then (f, hf ) ∼S (f , hf ) if and only

if gf (w) = gf (w). Furthermore, d(Π,Π′) = 0 if and only if Π = Π′.

Proof. Let

h = hf h−1f .

Then

h f = f h.

Let ϖn,f∞n=1 and ϖn,f∞n=1 be the corresponding sequences of nested Markov partitions by f and f ,

respectively. For any interval Jwn ∈ ϖn,f , h(Jwn) ∈ ϖn,f .

Suppose (f, hf ) ∼S (f , hf ). Then h is a C1-diffeomorphism of T. For any w = · · · in−1 · · · i1i0 ∈ Σ, let

wn = in−1 · · · i1i0 and w′n−1 = σ(wn) = in−1 · · · i1. Then

gf (wn) =|h(Jwn)||h(Jw′

n−1)|

=h′(ξ)

h′(ϱ)

|Jwn ||Jw′

n−1|=h′(ξ)

h′(ϱ)gf (wn).

This implies that

gf (w) = gf (w).

Now suppose gf (w) = gf (w). Since f and f are both C1+α expanding for some 0 < α 6 1, the Holder

distortion property (3.4) implies that there are constants C > 0 and 0 < τ < 1 such that

|gf (w)− gn,f (wn)| 6 Cτn and |gf (w)− gn,f (wn)| 6 Cτn.

This implies that there is a constant C ′ > 0 such that

gf (wn)

gf (wn)6 1 + C ′τn, ∀n > 0.



Let C ′′ =∏∞n=0(1 + C ′τ ′). Then

|h(Jwn)||Jwn |

=n−1∏k=0

gf (wn−k)

gf (wn−k)6 C ′′, ∀wn, ∀n > 0.

From the additive formula,

min16i6n

aibi

6 a1 + · · ·+ anb1 + · · ·+ bn

6 max16i6n

aibi

(7.5)

for any positive numbers aini=1 and bini=1, we conclude that h is Lipschitz continuous. Similarly, h−1

is also Lipschitz continuous. A Lipschitz continuous function is absolutely continuous. From Theorem 7.2

in the next section, h is a C1-diffeomorphism. We complete the proof of the first part of the theorem.

From the definition, d(Π,Π′) = 0 if and only if h = hf h−1f is symmetric. If Π = Π′, then h is

a C1-diffeomorphism. So it is symmetric. This implies that d(Π,Π′) = 0. On the other hand, if h is

symmetric, Theorem 6.7 implies that gf (w) = gf (w). This further implies that h is a C1-diffeomorphism.

So Π = Π′. It completes the proof of the theorem.

Thus we have

HG = T C1+. (7.6)

The distance dG(·, ·) on T C1+ is also a distance on HG, and (HG, dG(·, ·)) is a subspace of (G, dG(·, ·)).We call (HG, dG(·, ·)) the smooth Teichmuller space.

To make a self-contained paper, we give a proof of the following theorem which can be used in the

proof of the above theorem. The proof itself is interesting. We also studied the one-point rigidity for

one-dimensional maps with critical points. The reader who is interested in this topic can refer to the

survey article [17].

Suppose f, f ∈ C1+ are conjugate by the homeomorphism h, i.e., f h = h f . If h is differentiable at

a point p with non-zero derivative h′(p), then it is differentiable with non-zero derivatives at all points in

the grand orbit

GO(p) =∞∪m=0

∞∪n=0

f−n(fm(p)).

We say h is differentiable at point p with bound if there are a neighborhood U about p and a constant

C > 0 such that

C−1 6 |h′(q)| 6 C, ∀ q ∈ U ∩GO(p).

Then we have the following theorem.

Theorem 7.2 (One-point rigidity). The conjugacy h is a C1-diffeomorphism if and only if h is dif-

ferentiable at one point with bound.

Proof. Note that h is differentiable if and only if its lift H is differentiable. If H is a C1-diffeomorphism,

then

1 = H(1)−H(0) =

∫ 1

0

H ′(x)dx.

So there is at least one point x0 ∈ [0, 1] such that H ′(x0) = 0. Since H ′(x) is continuous, we have a

neighborhood U about x0 such that H ′(x) = 0 for all x ∈ U . This is the “only if” part.

To prove the “if” part, suppose H is differentiable at x0 with H ′(x0) > 0. Let p0 = π(x0) and

S =∪∞i=0 f

−i(p0) be the set of all preimages of p0 in T. Then S is dense in T. The lift set S of S to [0, 1]

contains all points

xnm = F−n(x0 +m), n = 0, 1, . . . , m = 0, 1, . . . , dn − 1.

Since

H(Fn(xmn)) = Gn(H(xnm)) (mod 1), http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


we have

H ′(xnm) =H ′(x0)(F

n)′(xnm)

(Gn)′(H(xnm)).

So H is differentiable at every point in S with non-zero derivatives. Thus we can take x0 ∈ (0, 1).

Let

x0 ∈ · · · ⊂ Iwk⊂ Iwk−1

⊂ · · · ⊂ Iw1 ⊂ [0, 1]

be a sequence of nested intervals in the sequence of nested Markov partitions ηk∞k=0 by f . Since H is

differentiable at x0 with bound, there is an integer n0 > 0 such that

C−11 6 H ′(x) 6 C1, ∀x ∈ S ∩ Iwn0

,

where C1 > 0 is a constant.

Assume I = Iwn0. Consider the set S(I) of all intervals J ∈ ηn0+k such that J ⊂ I and F k(J) = I

(mod 1) for k = 1, 2, . . . Let Ω(I) be the union of all these intervals. Then, just by the expanding property

of f , the set Ω(I) has a full Lebesgue measure in I.

For any J ∈ S(I), F k(J) = I (mod 1) and Gk(H(J)) = H(I) (mod 1) for some k > 1. We have

|H(J)||J |

=(F k)′(ξ)

(Gk)′(η)

|H(I)||I|

.

Take x ∈ S ∩ J . Then y = F k(x) ∈ S ∩ I and

(F k)′(x)

(Gk)′(H(x))=H ′(x)

H ′(y).

Thus

C−21 6 (F k)′(x)

(Gk)′(H(x))6 C1

2 .

This implies

C1−2

(F k)′(x)

(F k)′(ξ)

(Gk)′(η)

(Gk)′(x)

|H(I)||I|

6 |H(J)||J |

6 C12

(F k)′(ξ)

(F k)′(x)

(Gk)′(x)

(Gk)′(η)

|H(I)||I|

.

From the Holder distortion property (3.4), there is a constant C2 > 1 such that

C−12 6 |H(J)|

|J |6 C2.

Since both Ω(I) and H(Ω(I)) have full measures in I and H(I), respectively, from the additive formula,

this implies that H|I is bi-Lipschitz.

Since H | I is bi-Lipschitz, H ′ exists a.e. in I and is integrable. Since (H | I)′(x) is measurable and

H | I is a homeomorphism, we can find a point y0 in I and a subset E0 containing y0 such that

(1) H | I is differentiable at every point in E0;

(2) y0 is a density point of E0;

(3) H ′(y0) = 0; and

(4) the derivative H ′ | E0 is continuous at y0.

Since [0, 1] is compact, there is a subsequence Fnk(y0) (mod 1)∞k=1 converging to a point z0 in [0, 1].

Without loss of generality, assume z0 ∈ (0, 1). Let I0 = (a, b) be an open interval about z0. There is a

sequence of interval Ik∞k=1 such that y0 ∈ Ik ⊆ I and Fnk : Ik → I0 (mod 1) is a C1+α diffeomorphism.

Then |Ik| goes to zero as k tends to infinity.

From the Holder distortion property (3.4), there is a constant C3 > 0, such that∣∣∣∣ log( |(Fnk)′(w)||(Fnk)′(z)|

)∣∣∣∣ 6 C3, ∀w, z ∈ Ik, ∀ k > 1.



Since y0 is a density point of E0, for any integer s > 0, there is an integer ks > 0 such that

|E0 ∩ Ik||Ik|

> 1− 1

s, ∀ k > ks.

Let Ek = Fnk(E0 ∩ Ik) (mod 1). Then H is differentiable at every point in Ek and, from the Holder

distortion property (3.4), there is a constant C4 > 0 such that

|Ek ∩ I0||I0|

> 1− C4

s, ∀ k > ks.

Let

E =∞∩s=1

∪k>ks

Ek.

Then E has the full measure in I0 and H is differentiable at every point in E with the non-zero derivative.

Next, we are going to prove that H ′ | E is uniformly continuous. For any x and y in E, let zk and

wk be the preimages of x and y under the diffeomorphism Fnk : Ik → I0 (mod 1). Then zk and wk are

in E0. From H F = G H (mod 1), we have

H ′(x) =(Gnk)′(H(zk))

(Fnk)′(zk)H ′(zk)

and

H ′(y) =(Gnk)′(H(wk))

(Fnk)′(wk)H ′(wk).

So ∣∣∣∣ log(H ′(x)

H ′(y)

)∣∣∣∣ 6 ∣∣∣∣ log ∣∣∣∣ (Gnk)′(H(zk))

(Gnk)′(H(wk))

∣∣∣∣∣∣∣∣+ ∣∣∣∣ log ∣∣∣∣ (Fnk)′(wk)

(Fnk)′(zk)

∣∣∣∣∣∣∣∣+ ∣∣∣∣ log(H ′(zk)

H ′(wk)

)∣∣∣∣.Suppose both f and g are C1+α for some 0 < α 6 1. From the Holder distortion property (3.4), there

is a constant C5 > 0 such that ∣∣∣∣ log ∣∣∣∣ (Fnk)′(wk)

(Fnk)′(zk)

∣∣∣∣∣∣∣∣ 6 C5|x− y|α

and ∣∣∣∣ log ∣∣∣∣ (Gnk)′(H(zk))

(Gnk)′(H(wk))

∣∣∣∣∣∣∣∣ 6 C5|H(x)−H(y)|α

for all k > 1. Therefore,∣∣∣∣ log(H ′(x)

H ′(y)

)∣∣∣∣ 6 C5(|x− y|α + |H(x)−H(y)|α) +∣∣∣∣ log(H ′(zk)

H ′(wk)

)∣∣∣∣for all k > 1. Since H ′|E0 is continuous at y0, the last term in the last inequality tends to zero as k goes

to infinity. Hence ∣∣∣∣ log(H ′(x)

H ′(y)

)∣∣∣∣ 6 C5(|x− y|α + |H(x)−H(y)|α).

This means that H ′ | E is uniformly continuous. So it can be extended to a continuous function ϕ on I0.

Because H | I0 is absolutely continuous and E has the full measure,

H(x) = H(a) +

∫ x

a

H ′(x)dx = H(a) +

∫ x

a

ϕ(x)dx

on I0. This implies that H | I0 is actually C1. (This, furthermore, implies that H | I0 is C1+α.)

Now for any x ∈ [0, 1], let J be an open interval about x. By the expanding condition on f , there are

an integer n > 0 and an open interval J0 ⊂ I0 such that Fn : J0 → J (mod 1) is a C1+α diffeomorphism.

By the equation H F = G H, we have that H | J is C1+α. Therefore, H is C1+α. We finish the proof

of the theorem.



8 Completeness

We prove the following theorem in this section.

Theorem 8.1 (Completeness). The Teichmuller space (G, dG(·, ·)) is a complete space and, moreover,

is the completion of (HG, dG(·, ·)), i.e.,

(G, dG(·, ·)) = (HG, dG(·, ·)).

Proof. Suppose gn(w)∞n=1 = [(fn, hfn)]∞n=1 is a Cauchy sequence in G, i.e., for any ϵ > 0, we have

an integer N > 0 such that

dG(gn, gm) < ϵ, ∀n,m > N.

By using the metric (7.2), this means that

d(Shfn ,Shfm) < ϵ, ∀n,m > N.

Thus we have a homeomorphism hn ∈∪h∈S Ehhfn

fixing three points 1, i, and −1 for each n such that

logKhmh−1n< 2ϵ, ∀n,m > N.

This implies that hn is a Cauchy sequence in the universal Teichmuller space. Since the universal

Teichmuller space is complete, this sequence (by choosing a subsequence if it is necessary) is convergent,

i.e., we have a quasiconformal homeomorphism h fixing three points 1, i, and −1 such that

logKHH−1n

= logKhh−1n

→ 0 as n→ ∞.

Let h = h | T and define f = h−1 qd h, which has a quasi-regular extension f = h−1 qd h.Since fn has a quasi-regular extension satisfying (3.9), we can pick hn such that fn = h−1

n qd hndiffers from this extension by conjugating an asymptotically conformal homeomorphism of D, and thus fnsatisfies (3.9). To complete the proof, we need to prove that f also satisfies (3.9). Since

F−m = H−1 Hn F−mn H−1

n H,

by the composition formula (3.3), f also satisfies (3.9). This says that [(f, h)] ∈ T US. This completes

the proof that (G, dG(·, ·)) is completed.

Now we prove that G is the completion of HG under dG(·, ·). For any g = [(f, h)] ∈ G, we construct a

sequence gn = [(fn, hn)] ⊂ HG approaching g under dG(·, ·). Suppose f is an extension of f and F

is the lift of f satisfying (3.9).

Let z = x + yi and let Gd = dz. Suppose F = H−1 Gd H. Let η(z) > 0 be the function with

η(z) → 0 as y → 0 such that

|µF−n(H−1(z))| =|µH−1(d

−nz)− µH−1(z)||1− µH−1(z)µH−1(d−nz)|

6 η(z) for a.e. z and all n. (8.1)

So we have a constant C > 0 such that

|µH−1(z)− µH−1(d−nz)| 6 Cη(z) for a.e. z and all n > 0.

Let

A0 = z = x+ yi ∈ C | 0 6 |z| < 1 and An = G−nd (A0).

For a fixed n > 0, define µn(z) = µH−1(z) for z ∈ C \ (∪∞k=n+1Ak) and µn(z) = µH−1(G

kd(z)) for

z ∈ Ak and k > n+ 1. Then µn is a Beltrami coefficient defined on the upper-half plane H and satisfies

µn(z + 1) = µn(z). Let H−1n be the quasiconformal homeomorphism with the Beltrami coefficient µn.

Then we have that H−1n (z + 1) = H−1

n (z) + 1. Define

Fn = H−1n Gd Hn.



Then we have Fn(z + 1) = Fn(z) + d and Fn = Fn | R is a circle endomorphism. Since

|µF−mn

(H−1n (z))| =

|µH−1n

(z)− µH−1n

(d−mz)||1− µH−1

n(z)µH−1

n(d−mz)|

,

we have that |µF−mn

(H−1n (z))| = 0 for a.e. z ∈

∪∞k=n+1Ak. So Fn is an analytic circle endomorphism.

Since Hn = Hn | R is a Holder continuous circle homeomorphism, from Theorem 5.9, Fn is an analytic

expanding circle endomorphism.

Since |µH−1n

(z)− µH−1(z)| = 0 for a.e. z ∈ C \ (∪∞k=n+1Ak) and

|µH−1n

(z)− µH−1(z)| 6 supk>0

|µH−1(z)− µH−1(d−kz)| 6 Cη(z)

for a.e. z ∈∪∞k=n+1Ak, we get that the sequence gn = [(fn, hn)] ⊂ HG tends to g = [(f, h)] under

dG(·, ·) as n goes to ∞. It completes the proof of the theorem.

9 Maximum metric and Teichmuller’s metric

Since G is a subspace of the space of all continuous functions on Σ, it has the maximum norm

∥g∥ = maxw∈Σ

|g(w)|, ∀ g ∈ G.

This defines the maximum metric

dmax(g, g) = ∥g − g∥, g, g ∈ G.

The space G under dmax(·, ·) is incomplete. Thus this metric is not equivalent to Teichmuller’s metric

dG(·, ·). The following theorem says that both of them induce the same topology on G.Theorem 9.1 (Continuity of the identity). The identity map

idTM : (G, dG(·, ·)) → (G, dmax(·, ·))

is uniformly continuous and the identity map

idMT : (G, dmax(·, ·)) → (G, dG(·, ·))

is only continuous.

Proof. Suppose g = [(f, hf )] and g = [f , hf )] ∈ G. Let K = exp(2dG(g, g)) > 1. For any ϵ > 0, we have

two marked circle endomorphisms (f, hf ) and (f , hf ) such that

h = hf h−1f : T → T

can be extended to a K(1 + ϵ)-quasiconformal map h defined on an annulus z ∈ C | 1r < |z| < r for

some r > 1. This implies that there are δ > 0 and M = M(K, ϵ) > 0 such that M → 1 as K → 1 and

ϵ→ 0 and such that H, which is a lift of h, is (δ,M)-quasisymmetric, i.e.,

M−1 6|H(y)−H(x+y2 )||H(x+y2 )−H(x)|

6M

for any x, y with |x − y| 6 δ. Note that h f = g h and H F = G H (mod 1), where F and G are

the lifts of f and g.

For any point w = · · ·wn ∈ Σ, we have that

Iwn,f ∈ ηn,f and Iσ(wn),f ∈ ηn−1,f



and

Iwn,f= H(Iwn,f

) ∈ ηn,f and Iσ(wn),f= H(Iσ(wn),f

) ∈ ηn−1,f .

Note that

Iwn,f ⊂ Iσ(wn),f and Iwn,f⊂ Iσ(wn),f

.

Let n0 > 0 be an integer such that

|Iσ(wn),f | 6 δ

for all n > n0. Then H | Iσ(wn),f is an M -quasisymmetric homeomorphism. Corollary 6.3 implies that

|gn(w)− gn(w)| =∣∣∣∣ |H(Iwn,f )||H(Iσ(wn),f )|

− |Iwn,f ||Iσ(wn),f |

∣∣∣∣ 6 2(M − 1).

Since g(w) = limn→∞ gn(w) and g = limn→∞ gn(w), this implies that

|g(w)− g(w)| 6 2(M − 1).

Therefore,

dmax(g, g) 6 2(M(exp(2dG(g, g)), ϵ)− 1).

This proves that idTM is uniformly continuous.

Now we prove that idMT is continuous (but not uniformly continuous). We prove it by contradiction.

Suppose idMT is not continuous, i.e., we have a real number ϵ > 0 and a point g = [(f, hf )] and a sequence

of points gm = [(fm, hfm)]∞m=1 in the Teichmuller space G such that

dmax(gm, g) = ∥gm − g∥ → 0 as m→ ∞

but

dG(gm, g) > ϵ, ∀m.Let ηn∞n=0 and ηn,m∞n=0 be the corresponding sequences of nested Markov partitions by f and fm.

Since ∥gm− g∥ → 0 as m→ ∞, we have a constant A = A(g) > 0 such that A 6 gm(w) 6 1 for sufficient

large m and all w ∈ Σ. Let us assume this is true for all m. Since Σ is a compact set, there is another

constant B = B(A) > 0 such that B 6 (gm)n(w) 6 1 for all m and all n and all w. This says that

the collection of the sequences (ηm)n∞n=0 of nested Markov partitions by fm has uniformly bounded

geometry. This implies that the quasisymmetric dilatations of all lifts Hm of hm are uniformly bounded

by a constant M > 0, i.e.,

QSm = supx∈R,t>0

|Hm(x+ t)−Hm(x)||Hm(x)−Hm(x− t)|

6M, ∀m > 0.

Assume all hm fix three points 1, i and −1. Then hmm is in a compact subset in the universal

Teichmuller space. This implies that [(fm, hm)]∞m=1 is in a compact set in T US. So it has a convergent

subsequence. Let us assume that [(fm, hm)]∞m=1 itself is convergent and converges to [(f0, h0)] as

m→ ∞. This implies that

(gm)n(w) → (g0)n(w)

as m→ ∞ uniformly on w. We also have two uniformly convergent sequences

(gm)n(w) → gm(w), (g0)n(w) → g0(w) as n→ ∞.

This further implies that

g(w) = limm→∞

gm(w) = limm→∞

limn→∞

(gm)n(w)

= limn→∞

limm→∞

(gm)n(w) = limn→∞

(g0)n(w) = g0(w).

From Theorem 6.7, this implies that [(f, h)] = [(f0, h0)] = g. This leads to a contradiction to our original

assumption. The contradiction says that idMT is continuous at each point g. The constants in the proof

depend on g. Since (G, dG(·, ·)) is complete and (G, dmax(·, ·)) is incomplete, idMT cannot be uniformly

continuous. We complete the proof.



Corollary 9.2 (The same topology). The topology on G induced by the Teichmuller metric dG(·, ·) andthe topology on G induced by the maximum metric dmax(·, ·) are the same.

10 Geometric Gibbs measures

Recall from Section 4 that the space Σ+ is a symbolic representation of T and m0 denotes the Lebesgue

probability measure on T as well as on Σ+. The symbolic dynamical system σ+ : Σ+ → Σ+ is the

symbolic representation of all circle endomorphisms of degree d > 2. Remember that M+(σ+) is the

space of all σ+-invariant probability measures.

The space Σ and the shift map σ are the dual symbolic representation of all circle endomorphisms of

degree d > 2. Remember that M(σ) is the space of all σ-invariant probability measures.

We say a measure µ ∈ M(σ) (resp. µ+ ∈ M(σ+)) is non-atomic if µ(v) = 0 for any point v ∈ Σ

(resp. µ+(w) = 0 for any point w ∈ Σ+). We say a measure µ ∈ M(σ) (resp. µ+ ∈ M(σ+)) has full

support if µ([v]n) = 0 for any n-cylinder [v]n in Σ (resp. µ+([w]n) = 0 for any n-cylinder [w]n in Σ+).

For any v = j0j1 · · · jn−1 · · · ∈ Σ+, let x = x(v) ∈ [0, 1] in (4.1) and z = e2πix ∈ T. Consider a

cylinder [v]+n in Σ+. Let vn = j0j1 · · · jn−1 and define wn = in−1 · · · i1i0 where i0 = jn−1, . . . , in−1 = j0.

Then we have the corresponding n-cylinder [w]n in Σ where w = · · ·wn ∈ Σ.

Suppose µ ∈ M(σ) is non-atomic and has full support. For each n > 0, take dn+1 intervals labeled

J00 · · · 0︸︷︷︸n

, . . . , Jvn , . . . , J(d − 1)(d − 1) · · · (d − 1)︸︷︷︸n

,

each with the length

|Jvn | = 2πµ([w]n). (10.1)

Arrange them counter-clockwise in numerical order of the angle of z ∈ Jvn on the unit circle T beginning

at 1 (see Figure 3).

Since µ is σ-invariant, we have

|Jvn | = 2πµ([w]n) = 2πd−1∑j=0

µ([wj]n+1) =d−1∪j=0

|Jvnj |.

This implies that

Jvn =

d−1∪j=0

Jvnj (10.2)

and that · · · ⊂ Jvn ⊂ Jvn−1⊂ · · · ⊂ Jv1 . Since µ is non-atomic and has full support,

∩∞k=1 Jvn contains

only one point which we denote as hµ(z). This defines a circle homeomorphism hµ(z) : T → T.

1

0 1

Ivn I0···00 I0···01 I(d−1)···(d−1)(d−1)

(d−1)···(d−1)(d−1)

−1

Jvn

J0···01

J0···00

Figure 3 Arrangement of intervals on the unit circle



Definition 10.1 (Geometric Gibbs measure). Suppose g ∈ G. A non-atomic σ-invariant probability

measure µ ∈ M(σ) is a geometric Gibbs measure associated with g if fµ = hµ qd h−1µ is a uniformly

symmetric circle endomorphism and

limn→∞

µ([w]n)

µ([σ(w)]n−1)= g(w) (10.3)

uniformly on w ∈ Σ. Furthermore, if fµ is C1+α and (10.3) holds, then we call µ a smooth geometric

Gibbs measure associated with g.

Note that under the assumption that fµ is a uniformly symmetric circle endomorphism, hµ must be a

quasisymmetric circle homeomorphism (see Corollary 5.6).

We say that g ∈ G is locally constant if there is an n0 > 0 such that g takes a constant value on every

n0-cylinder [w]n0 in Σ. We prove the following theorem.

Theorem 10.2 (Existence and uniqueness for geometric Gibbs measure). For any potential g ∈ G,there is a geometric Gibbs measure µ associated with it. There is only one local constant potential

g0(w) ≡ 1/d in G. For this potential g0, there is a unique geometric Gibbs measure associated with it.

Moreover, if g ∈ HG is a smooth potential, there is a unique smooth geometric Gibbs measure associated

with it and this unique smooth geometric Gibbs measure is indeed the Gibbs measure in the classical sense.

For a geometric Gibbs measure µ in Theorem 10.2, we have

RNDµ,µ(w) ≡ g(w),

as we expected in Section 2.

We will divide the proof of Theorem 10.2 into several subsections (see Subsections 10.1–10.4). Before

the proof, let us mention the equilibrium property for a geometric Gibbs measure.

Theorem 10.3 (Equilibrium state). Suppose g ∈ G. Then the geometric Gibbs measure µ associated

with g is an equilibrium state in the following sense:

0 = entµ(σ) +

∫Σ

log g dµ = supν∈M(σ)

entν(σ) +

∫Σ

log gdν

.

Proof. From the proof of Theorem 2.2, we already know that for any ν ∈ M(σ),

entν(σ) +

∫Σ

log g(w)dν 6 0.

Thus we only need to prove that for any geometric Gibbs measure µ associated with g, we have

entµ(σ) +

∫Σ

log g(w)dµ = 0.

The metric entropy entµ(σ) can be calculated as

entµ(σ) = limn→∞

1

n

∑wn

(−µ([w]n) logµ([w]n))

= limn→∞

∑wn

(− µ([w]n) log

µ([w]n)

µ(σ([w]n−1))

),

where the summation is over all n-cylinders. Since µ is the geometric Gibbs measure, by definition,

limn→∞

µ([w]n)

µ(σ([w]n−1))= g(w)

uniformly on Σ and there is a constant A > 0 such that A 6 g(w) 6 1 for all w ∈ Σ. This implies that

limn→∞

∑wn

(− µ([w]n) log

µ([w]n)

µ(σ([w]n−1))

)= −

∫Σ

log g(w)dµ

as we desired. This completes the proof.



10.1 Existence of geometric Gibbs measures, Part 1 of the proof of Theorem 10.2

Suppose g = [(f, hf )] ∈ G. Let F be the lift of f on R with F (0) = 0 and F (x+ 1) = F (x) + d. For each

integer n > 0, consider

Hm(x) =

dm−1∑k=0

(F−m(x+ k)− F−m(k)).

Then we have that Hm(0) = 0 and Hm(1) = 1 and

Hm(x+ 1) =

dm−1∑k=0

(F−m(x+ 1 + k)− F−m(k))

=dm−1∑k=1

(F−m(x+ k)− F−m(k)) + F−m(x+ dm) = Hm(x) + 1

since F−m(x+ dm) = F−m(x) + 1. Now define

Hn(x) =1

n

n−1∑m=0

Hm(x).

We have again Hn(0) = 0 and Hn(1) = 1 and Hn(x + 1) = Hn(x) + 1. Thus Hn is the lift of a circle

homeomorphism hn satisfying hn(1) = 1.

Since f is uniformly symmetric, following (7.5) the sequence hn∞n=1 consists of symmetric circle

homeomorphisms with a uniform varnishing bounded function ϵ(t) → 0+ as t → 0+ in Definition 3.3.

Moreover, we have a fixed number M > 0 such that hn∞n=1 is a subset of Homeo(M) in Lemma 3.2.

This implies that we have, by passing to a subsequence, a limiting homeomorphism h ∈ Homeo(M) for

hn∞n=1. Since all hn satisfy Definition 3.3 with the same vanishing function ϵ(t), h satisfies Definition 3.3

with the vanishing function ϵ(t) too. This says that h is a symmetric circle homeomorphism.

Let H be the lift of h. Then the further calculation shows that

d−1∑k=0

(H(F−1(x+ k))−H(F−1(k))) = H(x), ∀x ∈ [0, 1]. (10.4)

Let

mh = (h−1)∗m0 (10.5)

be the push-forward probability measure by h−1, i.e., mh(A) = m0(h(A)) for all Borel subsets A ⊂ T.Equation (10.4) is equivalent to say that mh is f -invariant. Note that mId = m0. We will use mh to

construct a geometric Gibbs measure µ as follows.

Let ϖn = Jwn∞n=0 be the sequence of nested Markov partitions by f . For any w = · · · in−1 · · · i1i0∈ Σ, let [w]n be the n-cylinder. Then

[w]n =d−1∪k=0

[· · · kwn]n+1.

Recall the definition of the interval Jwn in T in Section 6. Define µ([w]n) = mh(Jwn). Since

f−1(Jwn) =

d−1∪k=0

Jkwn

and mh is f -invariant, we have

µ([w]n) =d−1∑k=0

µ([· · · kwn]n+1).



By using Carateodory’s extension theorem, µ can be extended to a measure on Σ, which we still denote

as µ.

Since

σ−1([w]n) =d−1∪k=0

[wk]n+1, Jwn =d−1∪k=0

Jwnk, mh(Jwn) =d−1∑k=0

mh(Jwnk),

we have

µ(σ−1([w]n)) = µ([w]n).

This implies that µ is σ-invariant. From the construction, for hµ = h hf ,

fµ = hµ qd h−1µ = h f h−1

is a uniformly symmetric circle endomorphism and preserves the Lebesgue measure m0 (see Lemma 10.5).

Furthermore, by applying Corollary 6.3, we get

limn→∞

µ([w]n)

µ([σ(w)]n−1)= limn→∞

|h(Jwn)||h(Jw′

n−1)|

= limn→∞

|Jwn ||Jw′

n−1|= g(w).

Thus, µ is indeed a geometric Gibbs measure associated with g. This completes the proof of the existence

part of Theorem 10.2.

10.2 Geometric Gibbs measures and symmetric invariant probability measures, Part 2 of

the proof of Theorem 10.2

In the proof of the existence in the previous subsection, we first constructed a symmetric circle homeo-

morphism h and used it to induce a symmetric invariant probability measure mh and then to construct

a geometric Gibbs measure µ. Thus, a symmetric invariant probability measure has a deep relationship

with a geometric Gibbs measure. We will explore this relationship further in this subsection.

Suppose f ∈ US. Suppose m is a probability measure on T. We say that m is f -invariant if

m(f−1(A)) = m(A) (10.6)

for any Borel subset A in T. Suppose m is non-atomic and has full support. Then the distribution

function hm(z) = m([1, z]) for z ∈ T is a homeomorphism of T, where [1, z] is the arc on T from 1 to z

counter-clockwise and m = (h−1m )∗m0.

Definition 10.4. We say m is a symmetric f -invariant probability measure if (1) m(T) = 1, (2) m is

f -invariant, and (3) hm is a symmetric circle homeomorphism. Furthermore, if we change (3) to (3′) hmis a C1-diffeomorphism, then we call m a smooth f -invariant probability measure. We say f preserves

the Lebesgue measure if m0 itself is an f -invariant probability measure.

Lemma 10.5. Suppose f ∈ US. Then m is a symmetric f -invariant probability measure if and only

if fm = hm f h−1m ∈ US preserves the Lebesgue measure m0.

Proof. Suppose m is a symmetric f -invariant probability measure. The distribution function hm is a

symmetric circle homeomorphism. The composition formula (3.3) implies that fm = hm f h−1m ∈ US.

For any Borel subset A on T, since m(A) = m0(hm(A)), we have that

m0(f−1m (A)) = m0(hm(f−1(h−1

m (A)))) = m(f−1(h−1m (A))) = m(h−1

m (A)) = m0(A).

This says that fm preserves the Lebesgue measure.

Now suppose fm preserves the Lebesgue measure m0. For any Borel set A on T,

m(f−1(A)) = m0(hm(f−1(h−1m (hm(A))))) = m0(f

−1m (hm(A))) = m0(hm(A)) = m(A).

This says that m is f -invariant.



Theorem 10.6. Suppose g = [(f, hf )] ∈ G. Then the following are equivalent:

(1) There is a unique geometric Gibbs measure µ on Σ associated with g.

(2) There is a unique symmetric circle homeomorphism h ∈ S such that h f h−1 preserves the

Lebesgue measure m0.

(3) The uniformly symmetric circle endomorphism f has a unique symmetric f -invariant probability

measure mh on T.(4) There is a unique representation (f , hf ) in the symmetric conjugacy class [(f, hf )] preserving the

Lebesgue measure m0.

Proof. (2) ⇔ (3) ⇔ (4) are easily checked. We will give a proof of (1) ⇔ (2).

First, let us prove (2) ⇒ (1): Suppose µ1 and µ2 are two geometric Gibbs measures associated with g.

Then both fµ1 = hµ1 qd h−1µ1

and fµ1 = hµ2 qd h−1µ2

are uniformly symmetric circle endomorphisms

and preserve the Lebesgue measure m0. The normalized continuous potential in Theorem 6.4 associated

with fµ1 and fµ2 and f are g. Since g is a complete symmetric invariant (see Theorem 6.7), we have

symmetric circle homeomorphisms h1 and h2 such that fµ1 = h1 f h−11 and fµ1 = h2 f h−1

2 . This

implies that h1 = h2 = h. Thus hµ1 = hµ2 = h hf and µ1 = µ2.

Secondly, let us prove (1) ⇒ (2): Suppose h1 and h2 are two symmetric circle homeomorphisms such

that f1 = h1f h−11 and f1 = h1f h−1

1 both preserve the Lebesgue measurem0. Thenmh1 = (h−11 )∗m0

and mh2 = (h−12 )∗m0 are both symmetric f -invariant probability measures on T.

For an n-cylinder [w]n in Σ, let Jwn be the corresponding interval on T (see Section 6). Define

µ1([w]n) = mh1(Jwn) and µ2([wn]) = mh2(Jwn). As we have shown in Subsection 10.1, µ1 and µ2 can be

extended to σ-invariant probability measures on Σ, which we still denote as µ1 and µ2. Both of them are

non-atomic and have full support. Since h1 and h2 are both symmetric homeomorphisms, Corollary 6.3

implies that both of them are geometric Gibbs measures associated with g. Thus µ1 = µ2. This implies

that mh1(Jwn) = mh2(Jwn) for any n-cylinder [w]n and any n. This further implies that h1 = h2.

We now can show that the only local constant potential in G is the constant one g0(w) ≡ 1/d for all

w ∈ Σ.

Theorem 10.7. Suppose g ∈ G is a local constant function. Then

g(w) ≡ g0(w) ≡1

d, ∀w ∈ Σ.

Proof. Suppose g = [(f, hf )] and suppose n0 > 1 is an integer such that g|[w]n0 = awn0for every

n0-cylinder [w]n0 , where awn0is a constant only depending on the first n0 digits of w. From the existence

part of Theorem 10.2 (see Subsection 10.1), [(f, hf )] contains a representation preserving the Lebesgue

measure m0. Assume that (f, hf ) is such a representation.

Remember that ϖn = Jwn∞n=0 is the sequence of nested Markov partitions by f . Then we have

|f−1(Jwn)| =d−1∑k=0

|Jkwn |

for all wn = in−1 · · · i1i0. This implies that for any w = · · ·wn0∈ Σ,

gn0(w) =|Jwn0

||Jw′

n0−1|=

∑d−1k=0 |Jkwn0

|∑d−1k=0 |Jkw′

n0−1|= · · · =

∑in0+m

· · · in0 |Jin0+m···in0wn0|∑d−1

k=0 |Jin0+m···in0w′n0−1

|.

Each ratio of the corresponding terms in the last summation

|Jin0+m···in0wn0|

|Jin0+m···in0w′n0−1

|

tends to awn0as m goes to ∞. From (7.5), we get gn0(w) = awn0

. This further implies that gn(w) = awn0

for any n > n0. Thus for any n > n0, we have

|f(Jwn)||f(Jw′

n−1)|

=|Jwn ||Jw′

n−1|.



This implies that

|f(Jw′n−1

)||Jw′

n−1|

=|f(Jwn)||Jwn

|, equivalently,

|F (Iw′n−1

)||Iw′

n−1|

=|F (Iwn)||Iwn

|.

This says that F on every interval Iwn0must be a linear map αwn0

x+ βwn0. Since F−1 is a symmetric

homeomorphism, all αwn0must be the same d. Since F (0) = 0 and F (x+ 1) = F (x) + d, all βwn0

are 0.

This implies that F (x) = dx and f(z) = qd(z) = zd. Thus, g is the one in G associated with qd which is

g0(w) ≡ 1/d. We complete the proof.

10.3 Uniqueness of geometric Gibbs measures, Part 3 of the proof of Theorem 10.2

For g0 ≡ 1/d in G, the Lebesgue measure m0 on Σ is a geometric Gibbs measure associated with it since

m0([w]n) = 1/dn for all n-cylinders [w]n.

We prove that the Lebesgue measure m0 on Σ is the only geometric Gibbs measure associated with g0.

Suppose µ is another geometric Gibbs measure associated with g0. Then we have that fµ = hµ q h−1µ

is a uniformly symmetric circle endomorphism and that

limn→∞

µ([w]n)

µ([σ(w)]n−1)=

1

d

uniformly on w ∈ Σ. (Actually, since g0 is a complete symmetric invariant, hµ must be symmetric in this

case.)

Now for any fixed n-cylinder [w]n, let w = · · ·wn and σ(w) = · · ·w′n−1. Then

[w]n =∪wm

[· · · wmwn]n+m and [σ(w)]n−1 =∪wm

[· · · wmw′n−1]n+m−1,

where wm = in+m−1 · · · in runs over all strings of 0, . . . , d− 1 of length m > 0. Thus we have

µ([w]n)

µ([σ(w)]n−1)=

∑wm

µ([· · · wmwn]n+m)∑wm

µ([· · · wmw′n−1]n+m−1)

.

Each ratio of the corresponding terms in the last summations

µ([· · · wmwn]n+m)

µ([· · · wmw′n−1]n+m−1)

tends to the constant 1/d as m goes to ∞, and from (7.5) we get

µ([w]n)

µ([σ(w)]n−1)=

1

d.

This implies that µ([w]n) = 1/dn for all n-cylinders [w]n and thus µ = m0.

From Theorem 10.6, the proof above gives another proof of the uniqueness of the symmetric qd-invariant

measure as we previously proved in [18].

Corollary 10.8 (Symmetric invariant measure). The Lebesgue measure m0 on T is the unique sym-

metric qd-invariant probability measure.

Corollary 10.9 (Symmetric rigidity). Suppose f ∈ US preserves the Lebesgue measure and suppose

h ∈ S such that h f = qd h. Then h = id.

We conjecture that there is only one geometric Gibbs measure associated with every g ∈ G. More

precisely, we have the following conjecture.

Conjecture 10.10 (Uniqueness for geometric Gibbs measure). The geometric Gibbs measure con-

structed in Theorem 10.2 is unique.

Equivalently, in view of Theorem 10.6, we have the following two equivalent conjectures. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Conjecture 10.11 (Uniqueness for symmetric invariant probability measure). Suppose f ∈ US. Thesymmetric f -invariant probability measure constructed in Theorem 10.2 is unique.

Conjecture 10.12 (Symmetric rigidity). Suppose both f1, f2 ∈ US preserve the Lebesgue measure

and suppose h ∈ S such that h f1 = f2 h. Then h = id.

10.4 Smooth geometric Gibbs measures, Part 4 of the proof of Theorem 10.2

Suppose f ∈ US is C1+Dini. From Theorem 5.9, it is expanding. Let ω(t) be the Dini modulus of

continuity for the derivative F ′. Let ω(t) be the one defined in (3.7). A measure m on T is C1+ω if

m(A) =∫Aρ(z)|dz| for any Borel subsets of T and ρ has the modulus of continuity ω(t).

Theorem 10.13. A C1+ω map f ∈ US has a unique C1+ω f -invariant probability measure mf .

Proof. Remember that F is the lift of f . Then F ′(x) is a periodic function of period 1. Let Cω be the

space of all continuous periodic functions on R of period 1 with the modulus of continuity ω. Let z = e2πix.

For any A ⊂ T, let A ⊂ [0, 1] such that exp(A) = A. Then one can check that m(A) = exp(∫Aρ(e2πix)dx)

is an f -invariant measure if and only if

d−1∑k=0

ρ(e2πiF−1(x+k))

F ′(F−1(x+ k))= ρ(e2πix). (10.7)

Theorem 5.9 implies that f is expanding. Without of loss of generality, we assume λ = minx∈R F′(x)

> 1. Let M = maxx∈R F′(x) < +∞. Then ψ(x) = M

F ′(x) > 1 has the modulus of continuity ω(t). Let

K > 0 be a constant such thatψ(x)

ψ(y)6 eKω(|x−y|), |x− y| 6 1.

Define the Ruelle operator L with weight ψ as

Lϕ(x) =d−1∑k=0

ϕ(F−1(x+ k))ψ(F−1(x+ k)) : Cω → Cω.

We first claim that L has a positive eigenfunction ϕ in Cω with eigenvalue M . Let us prove this claim:

Define

HωK =

ϕ ∈ Cω

∣∣∣∣ϕ(x) > 1,ϕ(x)

ϕ(y)6 eKω(|x−y|), |x− y| 6 1

.

For any ϕ ∈ HωK , it is clear that Lϕ(x) > 1. In addition,

Lϕ(x) =d−1∑k=0

ψ(F−1(x+ k))ϕ(F−1(x+ k))

6d−1∑k=0

ψ(F−1(y + k))ϕ(F−1(y + k))eK(ω(|F−1(x+k)−F−1(y+k)|)+ω(|F−1(x+k)−F−1(y+k)|))

6 Lϕ(y) · eKω(λ−1|x−y|)+Kω(λ−1|x−y|) = Lϕ(y) · eKω(|x−y|).

This implies that LHωK ⊆ Hω

K .

Let S be the set of positive numbers ξ for which there is an element ϕ inHωK satisfying Lϕ > ξϕ. Then S

is non-empty because ξ = 1/∥ϕ∥ ∈ S for any ϕ ∈ HωK , and bounded because maxx∈R

∑d−1k=0 ψ(F

−1(x+k))

is an upper bound. Let a = supS > 0. There is a sequence ξn∞n=1 in S converging to a. Let ϕn be a

function inHωK such that Lϕn > ξnϕn. We normalize it by minx∈Xϕn(x) = 1. ConsiderA = ϕn∞n=1 as

a family in the space C0 of all continuous functions of period 1 equipped with the supremum norm. Since

A ⊂ HωK , the normalization condition implies that A is a bounded family. Moreover, A is equicontinuous

and thus the Arzela-Ascoli theorem implies A is in a compact subset in C0. Thus we have a convergent

subsequence ϕnk(x) → ϕ∞(x) uniformly as k → ∞. This implies that ϕ∞ ∈ Hω

K and Lϕ∞ > aϕ∞. Since http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


a = supS, we further get Lϕ∞ = aϕ∞. Integrating on both sides we get a = M . This completes the

proof of the claim.

Using the strong mixing property on f , we have that for any eigenfunction ϕ ∈ Cw with Lϕ = Mϕ,

there is a real number b such that ϕ = bϕ∞ (refer to [13]). This further implies that we have a unique

ρ(e2πix) = cϕ∞(x) such that∫[0,1]

ρ(e2πix)dx = 1 satisfies Equation (10.7). This completes the proof of

the theorem.

From Theorems 10.13 and 6.7, we have the following corollary.

Corollary 10.14 (Symmetric rigidity in the smooth case). Suppose f1, f2 ∈ US are both C1+Dini and

preserve the Lebesgue measure m0. Suppose h ∈ S such that h f1 = f2 h. Then h = id.

Proof. From Theorem 5.9, both f1 and f2 are expanding. From h f1 = f2 h and the proof of

Theorem 7.1, h is C1. Let ρ(z) = h′(z). Since f1 and f2 both preserve the Lebesgue measure, the

conjugacy equation h f1 = f2 h implies that 1 and ρ both satisfy (10.7). Now Theorem 10.13 implies

that ρ(z) ≡ 1. This says h = id.

Let h(e2πix) = exp(∫ x0ρ(e2πix)dx) in Theorem 10.13. One conclusion from the proof of Theorem 10.13

is that if f ∈ US is C1+Dini, then H ′m(x) → ρ(e2πix) uniformly as m→ ∞ since

H ′m(x) =

dm−1∑k=0

1

(Fm)′(F−m(x+ k))=

1

MmLm1.

This further implies that H ′n(x) → ρ(e2πix) uniformly as n→ ∞ and hn(z) → h(z) uniformly as n→ ∞.

Thus h(z) is a C1+w circle diffeomorphism and mf = mh is a smooth f -invariant probability measure.

Consider g = [(f, hf )] ∈ HG where f is C1+α. Then ω(t) = Ctα and ω(t) = Ctα. Thus h(z) is a C1+α

circle diffeomorphism such that

f = h f h−1 = h hf qd h−1f h−1

is also C1+α and preserves the Lebesgue measure. Let µ = ((h hf )−1)∗m0. Since h is a C1+α-

diffeomorphism, we have two constants C > 0 and 0 < τ < 1 such that∣∣∣∣ |h(Iwn)||h(Iw′

n−1)|

− |Iwn ||Iw′

n−1|

∣∣∣∣ = ∣∣∣∣ µ([w]n)

µ([σ(w)]n−1)− gn(w)

∣∣∣∣ 6 Cτn.

This implies that µ is a smooth geometric Gibbs measure associated with g. Corollary 10.14 now makes

sure that such a smooth Gibbs measure is unique.

From the Holder distortion (3.4), we have constants, which we still denote as C > 0 and 0 < τ < 1,

such that

|gn(w)− g(w)| 6 Cτn, ∀w ∈ Σ, n > 0.

Thus we have ∣∣∣∣ µ([w]n)

µ([σ(w)]n−1)− g(w)

∣∣∣∣ 6 2Cτn.

This implies that we have another constant, which we still denote as C > 0, such that

µ([σ(w)]n−1)(1− Cτn) 6 µ([w]n)

g(w)6 µ([σ(w)]n−1)(1 + Cτn)

for all w ∈ Σ and all n > 0. Since∏∞n=0(1 + Cτn) < ∞, we have a constant, which we still denote as

C > 0, such that

C−1 6 µ([w]n)

exp(∑n−1i=0 log g(σi(w)))

6 C

for any n-cylinder [w]n. This implies that the smooth geometric Gibbs measure µ is indeed the Gibbs

measure in the classical sense.

Now arguments in Subsections 10.1–10.4 complete the proof of Theorem 10.2. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


11 Complex Banach manifold structure

Remember that we use QS to denote the space of all quasisymmetric circle homeomorphisms fixing 1

and S to denote the space of all symmetric circle homeomorphisms fixing 1. The quotient space

AQS = QS/S

is called asymptotical universal Teichmuller space as we have seen in Section 6. Gardiner and Sullivan [8]

proved that AQS has a complex manifold structure following Bers’ embedding (refer to [1]).

In Section 6, we have also shown that the space G can be embedded into AQS as a subspace. In this

section, we show that it is actually a complex submanifold of AQS following Bers’ embedding again.

Let H be the space of all holomorphic periodic functions ϕ of period 1 defined on the lower half plane

L = z = x+ yi ∈ C | y < 0 such that the norm

∥ϕ∥ = supz∈L

|y2ϕ(z)| <∞. (11.1)

Let H0 be the subspace of B such that every element ϕ in it satisfies

|y2ϕ(z)| → 0, as y → 0 uniformly on x. (11.2)

Bers’ embedding embeds the universal Teichmuller space into a bounded open ball in H as follows:

For any h ∈ QS, let h be a quasiconformal extension of h on the unit disk D and let H be its lift on the

upper-half plane H. Let µH = Hz/Hz be its Beltrami coefficient defined on H. It is a periodic function

of period 1. Define

µ =

µH(z), z ∈ H,0, z ∈ L.

Let wµ be the normalized quasiconformal homeomorphism with the Beltrami coefficient µ. Then it is

analytic on L and wµ(z + 1) = wµ(z) + 1. The Schwarzian derivative

ϕ(z) = S(wµ)(z) =(wµ)′′′

(wµ)′− 3

2

((wµ)′′

(wµ)′

)2

on the lower-half plane is the Bers’ embedding.

For two analytic functions wµ and wν on L, we have a cocycle condition for Schwarzian derivatives,

S(wν wµ) = S(wν) wµ · [(wµ)′]2 + S(wµ).

Thus if wν is a Mobius transformation, then we have S(wν wµ) = S(wµ). So ϕ is independent of choice

of any element in the equivalent class [h] in the universal Teichmuller space. The Bers’ embedding

ι : [h] → ϕ

embeds the universal Teichmuller space into an open set in H containing the ball of radius 1/2 centered

at 0 and contained in the ball of radius 3/2 centered at 0.

The Bers’ embedding ι on the ball of radius 1/2 centered at 0 inH has a holomorphic section s following

from Ahlfors-Weill’s technique (refer to [1, 8]) as follows.

For any analytic function ϕ ∈ H with ∥ϕ∥ < 1/2, consider two independent solutions η1 and η2 of the

equation η′′ = −12ϕη. Define

β =

η1(z) + (z − z)η′1(z)

η2(z) + (z − z)η′2(z), z ∈ H,

η1(z)

η2(z), z ∈ L.



The Beltrami coefficient of β is

µ =

1

2(z − z)2ϕ(z), z ∈ H,

0, z ∈ L.

Thus ∥µ∥∞ < 1 and is a periodic function of period 1. Define

µ(z) =

µ(z), z ∈ H,µ(z), z ∈ L.

It is still a periodic function of period 1. Let wµ be the normalized quasiconformal homeomorphism with

the Beltrami coefficient µ. Then H = wµ | R is a homeomorphism of R with H(x+ 1) = H(x) + 1. So it

is the lift of a circle homeomorphism h. The local section

s : ϕ→ [h]

is the local right inverse of ι defined on a small disk U centered at 0 in H, i.e., ι s = id on U . The

Bers’ embedding ι, the local section s, and changing basepoint isomorphisms give an altar of holomorphic

charts on the universal Teichmuller space which makes the universal Teichmuller space a complex Banach

manifold.

According to [8], if h is a symmetric circle homeomorphism, then Bers’ embedding ι([h]) = ϕ ∈ H0 and

the holomorphic section s maps the ball of radius 1/2 centered at 0 in H0 into S. Thus Bers’ embedding

induces an embedding, which we still denote as ι, from AQS into the quotient space H/H0. The local

section s induces a local section, which we still denote as s defined on the quotient space U/H0 such that

ι s = id. The embedding ι, the local section s, and changing basepoint isomorphisms give an altar of

holomorphic charts on AQS. Thus AQS is also a complex Banach manifold.

What we are going to prove is that Bers’ embedding also induces an embedding, which we still denote

as ι from G into a bounded open subset in the quotient space H1/H0, where

H1 = ϕ ∈ H | y2(ϕ(d−nz)− ϕ(z)) → 0 as y → 0 uniformly on x and n, z = x+ yi.

For any point g = [(f, hf )] ∈ G, f = hf qd h−1f is a uniformly symmetric circle endomorphism. From

Lemma 3.15, we have a quasi-regular extension f whose lift F = Hf Gd H−1f satisfies (3.9). From the

formula in (8.1), we have

µH−1f

(z)− µH−1f

(d−nz) → 0 as y → 0 uniformly on x and n. (11.3)

Define

µ =

µH−1f

(z), z ∈ H,

0, z ∈ L.

Let wµ be the normalized quasiconformal map of the Riemann sphere C whose Beltrami coefficient is µ.

Then wµ is analytic on the lower-half plane. Let ϕ = S(wµ) ∈ H be Bers’ embedding of (f, hf ). For any

other representations of [(f, hf )], the difference between Bers’ embeddings of them and ϕ are in H0. So

we actually embed the equivalence class [(f, hf )] into an equivalence class [ϕ] ∈ H/H0. We still use ι to

denote this embedding.

We further define

µn =

µF−n(z), z ∈ H,0, z ∈ L.

Let wµn be the normalized quasiconformal map of C whose Beltrami coefficient is µn. Since wµn =

wµ G−nd (wµ)−1,

S(wµn)((wµ)−1(z)) = S(wµ)(d−nz)− S(wµ)(z) = ϕ(d−nz)− ϕ(z). http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Since µF−n(z) → 0 as y → 0 uniformly on x and n, we get ϕ ∈ H1 and, furthermore, we have

ι([(f, hf )]) = [ϕ] ∈ H1/H0.

For any [ϕ] ∈ H1/H0 with ∥ϕ∥ < 1/2, the corresponding h = s(ϕ) satisfies (11.3). This implies that

f = h qd h−1 is a uniformly symmetric circle endomorphism. Let g = gf ∈ G. It is independent of thechoice of representation in [ϕ]. Thus the local section s induces a local section, which we still denote as

s defined on the open ball U of radius 1/2 centered at 0 in H1/H0 such that ι s = id. The embedding

ι and the local section s and changing basepoint isomorphisms give an altar of holomorphic charts on G.Thus G is a complex manifold and we have proved the following theorem.

Theorem 11.1 (Complex Banach manifold). The space (G, dG(·, ·)) is a complete complex Banach

manifold.

Remember that g0 ≡ 1/d is the basepoint of G. The tangent space Tg0G of G at the basepoint now

can be thought of as a subspace of the tangent space T[id]AQS of the universal asymptotical Teichmuller

space AQS at the basepoint.

Let ht be a smooth path in QS such that h0 = id. Let Ht be the corresponding path of lifts. According

to [23], the tangent vector

E(x) =dHt(x)

dt

∣∣∣∣t=0

is a periodic Zygmund continuous function of period 1 vanishing at all integers n, i.e.,

E(x+ y) + E(x− y)− 2E(x) = O(y), E(x+ 1) = E(x), E(n) = 0, (11.4)

and if ht is a smooth path in S, then E is a little Zygmund continuous function of period 1 vanishing at

all integers n, i.e.,

E(x+ y) + E(x− y)− 2E(x) = o(y), E(x+ 1) = E(x), E(n) = 0. (11.5)

Let Z be the space of functions satisfying (11.4) and Z0 be the space of functions satisfying (11.5). Then

T[id]AQS = Z/Z0.

Now consider a path gt = [(ft, ht)] ∈ G such that g0 is the basepoint. Let Ft and Ht be the lift of ftand ht. Then we have E ∈ Z such that

Ht(x) = x+ tE(x) + o(t) and H−1t (x) = x− tE(x) + o(t).

Now consider

F−nt (x) = Ht G−n

d H−1t (x) = Ht(d

−nx− td−nE(x) + o(t))

= d−nx− td−nE(x) + o(t) + tE(d−nx− td−nE(x) + o(t))

= d−nx+ (E(d−nx)− d−nE(x))t+ o(t) = d−nx+ Vn(x)t+ o(t),

where

Vn(x) = E(d−nx)− d−nE(x).

Since all F−nt are uniformly symmetric, according to [23], the functions Vn(x) are uniformly little Zygmund

functions, i.e., there is a bounded continuous function ε(y) such that ε(y) → 0 as y → 0+ and such that

|Vn(x+ y) + Vn(x− y)− 2Vn(x)| < ε(y)y (11.6)

for all x ∈ R, y > 0, and all n. Moreover,

Vn(x+ dn) = Vn(x). (11.7)

So we have proved the following theorem. http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


Theorem 11.2. Let Z1 be the space of all functions E ∈ Z satisfying (11.6) and (11.7). Then we have

Tg0G = Z1/Z0.

The Hilbert transformation J on Z is defined by

JE(x) =1

π

∫ ∞

−∞E(y)R(x, y)dy,

where

R(x, y) =1

y − x− x

y − 1+x− 1

y.

Let ζ = ξ + ηi be a complex number. Then ζ = ξ − ηi. By Stokes’ formula,

JE(x) =1

π

∫ ∞

−∞E(y)R(x, y)dt =

2i

π

∫ ∫H∂E(ζ)R(x, ζ)dξdη + iE(x),

where H is the upper-half plane and E is a continuous extension of E to the complex plane and ∂E =

∂E/∂ζ.

Let µ(ζ) = ∂E(ζ). Then

JE(x) =2i

π

∫ ∫Hµ(ζ)R(x, ζ)dξdη + iE(x)

=−2i

π

∫ ∫Lµ(ζ)R(x, ζ)dξdη − iV (x),

where L is the lower-half plane.

Define

µ(ζ) =

−iµ(ζ), ζ ∈ H,iµ(ζ), ζ ∈ L.

Then

JE(x) =−1

π

∫ ∫Cµ(ζ)R(x, ζ)dξdη.

In addition, we also have

∂(JE) = −iµ and ∂(J 2E) = i2µ = −µ.

This implies

J 2E = −E.

Hence the Hilbert transformation J gives an almost complex structure on Z.

Now we will prove that the Hilbert transformation J also gives an almost complex structure on the

tangent space Tg0G.

Theorem 11.3. Let [JE] be the Hilbert transformation of [E] ∈ Z1/Z0. Then [JE] satisfies the

equations (11.6) and (11.7). Therefore, J [E] := [JE] ∈ Z1/Z0. Since J : Z1/Z0 → Z1/Z0 and

J 2 = −I, it is an almost complex structure on Tg0G.Proof. Note that

JE(x) =1

π

∫ ∞

−∞E(y)

(1

y − x− x

y − 1+x− 1

y

)dy.

From E(x+ 1) = E(x), we have

1

π

∫ ∞

−∞

xE(y)

y − 1dy =

1

π

∫ ∞

−∞

xE(y + 1)

ydy =

1

π

∫ ∞

−∞

xE(y)

ydy.

Hence

JE(x) =1

π

∫ +∞

−∞

E(y)

y − xdy − 1

π

∫ ∞

−∞

E(y)

ydy.



From this form of the Hilbert transform, it is easy to check JE(0) = 0 and JE(x + 1) = JE(x) since

E(y + 1) = E(y). This is the equations in (11.4).

Now

Vn(x) = JE(d−nx)− d−nJE(x)

=1

π

∫ +∞

−∞

E(y)

y − d−nxdy − 1

π

∫ +∞

−∞

d−nE(y)

y − xdy − (1− d−n)

1

π

∫ ∞

−∞

E(y)

y

=1

π

∫ +∞

−∞

E(d−ny)− d−nE(y)

y − xdy − (1− d−n)

1

π

∫ ∞

−∞

E(y)

y

=1

π

∫ +∞

−∞

Vn(y)

y − xdy − (1− d−n)

1

π

∫ ∞

−∞

E(y)

y.

Thus we have that Vn satisfies (11.6) and (11.7). This proves the theorem.

Let GM be the space of all geometric Gibbs measures and HGM be the space of all smooth geometric

Gibbs measures. Then we have a projection

P : GM → G, P(µ) = g,

where µ is the geometric Gibbs measure associated with g. Theorem 10.2 says that P is surjective and

P(HGM) = HG. If Conjecture 10.10 holds, then P is also injective and induces a natural complex

Banach manifold structure on GM from G. Otherwise, we can introduce an equivalence relation between

all geometric Gibbs measures as follows: Two geometric Gibbs measures µ and ν are said to be equivalent,

denoted as µ ∼s ν, if there is a symmetric homeomorphism h such that hµ = h hν . From Theorem 6.7,

we know that µ ∼s ν if and only if they are geometric Gibbs measures associated with the same g ∈ G,i.e.,

GM(g) = P−1(g)

is an equivalence class. It is a convex subset. Then GM/ ∼s= G has a complex Banach manifold

structure.

12 Kobayashi’s metric

In the previous section (see Section 11), we gave a complete complex Banach manifold structure on G as

well as on GM/ ∼s. Thus we can define Kobayashi’s metric on it.

Consider the unit disk D as a hyperbolic Riemann surface. Its hyperbolic metric

dD(z, w) = tanh−1(z, w) =1

2log

1 + |z−w||1−zw|

1− |z−w||1−zw|

, z, w ∈ D (12.1)

is Kobayashi’s metric on it. Since G is a connected complete Banach complex manifold, we can consider

the space H(D,G) of all holomorphic maps from D into G.Kobayashi’s pseudo-metric dK(·, ·) on G is defined to be the largest pseudo metric on G such that

dK(f(z), f(w)) 6 ρD(z, w), ∀ z, w ∈ D and ∀ f ∈ H(D,G). (12.2)

Another way to describe Kobayashi’s metric on G is as follows. Given p, q ∈ G, consider the sub-

space Hp,q consisting of all f ∈ H(D,G) such that f(0) = p and f(s) = q for some s ∈ D. Let

r = inff∈Hp,q |s| and

d1(p, q) =1

2log

1 + r

1− r. (12.3)

Note that if Hp,q = ∅, then d1(p, q) = ∞. Now consider the space

Cn = p0 = p, p1, . . . , pn−1, pn = q http://engine.scichina.com/doi/10.1007/s11425-019-1638-6


of n-chains connecting p, q ∈ G and define

dn(p, q) = infCn

n∑i=1

d1(pi−1, pi). (12.4)

It is clear that dn+1(p, q) 6 dn(p, q) for all n > 0 and p, q ∈ G. Kobayashi’s pseudo-metric is

dK(p, q) = limn→∞

dn(p, q), p, q ∈ G. (12.5)

By using this description of Kobayashi’s metric, one can show that for the complex plane C, its Kobayashi’s

pseudo metric is 0 and that for the hyperbolic disk D, its Kobayashi’s pseudo metric is the hyperbolic

metric. Furthermore, using this description, one can see that a holomorphic map contracts Kobayashi’s

metrics. Precisely, let F : G → G be a holomorphic map. Then

dK(F (p), F (q)) 6 dK(p, q), ∀ p, q ∈ G.

This contraction property will be useful in the study of geometric Gibbs theory. We have the following

conjecture (refer to [5, 6, 11,19,24]).

Conjecture 12.1. Kobayashi’s metric dK(·, ·) and Teichmuller’s metric dG(·, ·) on G coincide, i.e.,

dK(p, q) = dG(p, q), ∀ p, q ∈ G.

13 Metric entropy function on geometric Gibbs measures

Now let us consider the metric entropy function

entµ(σ) =

∫Σ

− log g(w)dµ(w) : GM → R+ = x > 0.

It takes the maximum value log d at the basepoint m0 which is the unique geometric Gibbs measure for

g0(w) ≡ 1/d. We also knew that when restricted on HGM, entµ has no other local maximum and local

minimum points. What is the infimum of the entropy function? It used to be an important question even

in the area of the smooth dynamical systems. We have settled this question for Anosov diffeomorphisms

of a compact Riemannian manifold with respect to their SRB-measure in [9,10]. The next theorem says

that this is true in our case.

Theorem 13.1. The infimum of entµ(σ) : GM → R+ = x > 0 is zero.

Proof. Without loss of generality, we assume d = 2 in the proof. We first construct a family of C1

functions ϕt(x) | 0 < t < 1/6 on [1/2− t, 1/2 + t] satisfying the following conditions:

(i) ϕt(1/2± t) = 2;

(ii) ϕ′t((1/2− t)+) = ϕ′t((1/2 + t)−) = 0;

(iii) 2 6 ϕt(x) 6 2/t for |x− 1/2| 6 t;

(iv) ϕt(x) > 1/2t for |x− 1/2| 6 t− t2;

(v)∫ 1/2

1/2−t ϕt(ξ)dξ =∫ 1/2+t

1/2ϕt(ξ)dξ =

12 ;

(vi) ϕt depends on t smoothly.

We then define a C2 diffeomorphism F1 = Ft,1 from [1/2− t, 1/2 + t] to [1/2, 3/2] by

F1(x) =1

2+

∫ x

12−t

ϕt(ξ)dξ.

We have

F1

(1

2

)= 1, F1

(1

2− t

)=

1

2, F1

(1

2+ t

)=

3

2, F ′

1

((1

2− t

)+

)= 2 = F ′

1

((1

2+ t

)−

).



Moreover,

2 6 F1′(x) 6 2

tfor

∣∣∣∣x− 1

2

∣∣∣∣ 6 t and F1′(x) > 1

2tfor

∣∣∣∣x− 1

2

∣∣∣∣ 6 t− t2. (13.1)

Let G1(x) be the inverse function of F1. Then G1(x) is a C2 diffeomorphism from [1/2, 3/2] to

[1/2− t, 1/2 + t]. Since G′1(x) =

1ϕ(G1(x))

, the properties of F1 give that

G1(1) =1

2, G1

(1

2

)=

1

2− t, G1

(3

2

)=

1

2+ t

and

G′1

((1

2

)+

)= G′

1

((3

2

)−

)=

1

2and G′′

1

((1

2

)+

)= G′′

1

((3

2

)−

)= 0.

We note that the fact F ′1(x) 6 2/t on [1/2 − t, 1/2 + t] implies that the lengths of the intervals

F1([1/2− t, 1/2− t+ t2]) and F1([1/2 + t− t2, 1/2 + t]) are less than 2t. Hence the inequalities in (13.1)

implyt

26 G1

′(x) 6 1

2, |x− 1| 6 1

2and G1

′(x) 6 2t, |x− 1| 6 1

2− 2t. (13.2)

We now define

G0(x) = −1

2+ t+

∫ x

− 12

(1−G′1(ξ + 1))dξ = x+

1

2−G1(x+ 1), x ∈

[− 1

2,1

2

].

It is clear that

G0(0) = 0 and G0

(± 1

2

)= ±

(1

2− t

).

The definition also gives G′0(x) = 1−G′

1(x+ 1), i.e.,

G′0(x) +G′

1(x+ 1) = 1, x ∈[− 1

2,1

2

]. (13.3)

Hence we have

G′0

((− 1

2

)+

)= G′

0

((1

2

)−

)=

1

2and G′′

0

((− 1

2

)+

)= G′′

0

((1

2

)−

)= 0.

So G0 : [−1/2, 1/2] → [−1/2 + t, 1/2− t] is a C2-diffeomorphism. Moreover, by (13.2), we have

1

26 G′

0(x) 6 1− t

2, |x| 6 1

2and 1− 2t 6 G′

0(x) < 1, |x| 6 1

2− 2t. (13.4)

Let F0 = Ft,0(x) : [−1/2 + t, 1/2− t] → [−1/2, 1/2] be the inverse of G0(x). Then we have

F0(0) = 0, F0

(±(1

2− t

))= ±1

2, F ′

0

((− 1

2+ t

)+

)= F ′

0

((1

2− t

)−

)= 2.

Moreover,2

2− t6 F ′

0(x) 6 2 for |x| 6 1

2− t. (13.5)

Note that the lengths of G0([−1/2,−1/2 + 2t]) and G0([1/2 − 2t, 1/2]) are less than 2t. We have

G0([−1/2 + 2t, 1/2− 2t]) ⊃ [−1/2 + 3t, 1/2− 3t]. By the second part of (13.4) we get

1 < F ′0(x) 6

1

1− 2tfor |x| 6 1

2− 3t. (13.6)

Now we define

Ft(x) =

F0(x), x ∈

[− 1

2+ t,

1

2− t

],

F1(x), x ∈[1

2− t,

1

2+ t

].



Note that [−1/2+ t, 1/2+ t] is a fundamental interval of T in R. Extend Ft to R by Ft(x+k) = F (x)+2k

for x ∈ [−1/2 + t, 1/2 + t] and all integers k. One can check that Ft is a C2 diffeomorphism of R with

Ft(0) = 0 and Ft(x+ 1) = Ft(x) + 2. Thus it is the lift of a C2 circle endomorphism ft on T.For any z ∈ T, let π(x) = z for x ∈ (−1/2, 1/2] (a fundamental interval of T in R). Let

x0 = G0(x) ∈(− 1

2+ t,

1

2− t

]and x1 = G1(x+ 1) ∈

(1

2− t,

1

2+ t

].

Let z0 = π(x0) and z1 = π(x1) in T. Then we have z0, z1 = f−1t (z). From (13.3),

1

F ′t (x0)

+1

F ′t (x1)

= G′0(x) +G′

1(x+ 1) = 1.

This implies that ft preserves the Lebesgue measure m0 on T.Since F0(0) = 0, by (13.6) we can obtain that

|Ft(x)− x| 6∫ x

0

|F ′t (ξ)− 1|dξ 6

∣∣∣∣ 1

1− 2t− 1

∣∣∣∣ = 2t

1− 2tfor |x| 6 1

2− 3t.

This implies limt→0+ Ft(x) = x as t→ 0+ for all x ∈ R \ 12 + n.

We are left to show that limt→0+ hm0(ft)(x) = 0. This comes directly from the construction. Indeed,

if we denote I0 = (− 12 +3t, 12 − 3t] and I1 = ( 12 − 3t, 12 +3t], then I0, I1 is a partition of T lift to R. By

(13.1) and (13.5), F ′t(x) 6 2/t for x ∈ I1. Hence, by (13.6) we can get

hm0(ft) =

∫Tlog f ′t(z)dm0 =

∫I0

logF ′t (x)dx+

∫I1

logF ′t (x)dx

6 (1− 6t) · log 1

1− 2t+ 6t · log 2

t,

which converges to 0 as t→ 0+. This is what we need.

Remark 13.2. In the above construction, ft(z) converges pointwisely to z as t goes to 0 for all except

for one point z ∈ T. The convergence cannot be uniformly. More precisely, it holds that limt→0+ Ft(x) = x

for x = 1/2 + k and limt→0+ Ft(1/2 + k) = 1 + 2k for all integers k. Theorem 13.1 was adapted in the

paper [10] to show a different phenomenon on the boundary of area-preserving Anosov dynamical systems.

Acknowledgements This work was supported by National Science Foundation of USA (Grant No. DMS-

1747905), the Simons Foundation (Grant No. 523341), Professional Staff Congress of the City University of New

York Enhanced Award (Grant No. 62777-00 50) and National Natural Science Foundation of China (Grant No.

11571122). The author thanks his student John Adamski and colleague Sudeb Mitra for proofreading the abstract

and the introduction of this paper.

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