investigate the dynamics of fully chaotic systems. · proposition 7.12: the circle map is ergodic...

7

ERGODICITY AND MIXING

• Problem: Investigate the dynamics of fully chaotic systems.

• Ergodic systems.

• Mixing systems.

• A paradigmatic scheme: Symbolis Dynamics.

• The Smale horseshoe.

• The recurrence theorem of Poincare.

7.1 General definitions

In the framework of ergodic theory a dynamical system may be characterized as follows:

• A differentiable manifold G , and:(i) there is a measure µ on G defined by a positive and smooth density;(ii) there is a flow on G , i.e., a one parameter group φt of diffeomorphisms;(iii) the flow φt preserves the measure µ, i.e., for any measurable subset A ⊂ G one has

µ(φ−tA) = µ(A) .

(iv) We assume that the measure is normalized, i.e., µ(G ) = 1.

• Variazioni:

The continuos flow may be replaced by a discrete one (a map): just set t ∈ Z.

Don’t need the map be a diffeomorphism: e.g., it may be non differentiable, even not continuousand not invertible (in this case set t ∈ N0), but . . .

the existence of an invariant measure is essential.

(see, however, the model of the Smale horseshoe).

Ergodicity and mixing – 186

7.2 Ergodicity

• A concept first introduced in Statistical Mechanics. In very rough terms:

A mechanical system with a large number of particles (e.g., a gas, 1023 molecules).

Want to relate a macroscopic physical quantities (e.g., temperature) with microscopic mechanicalquantity (e.g., kinetic energy).

The measure of a macroscopic quantity considered as a time average of microscopic quantities,but . . .

requires to be able to deal with a large system of differential equations.

Ergodic hypothesis (Boltzmann): assuming that there are no first integrals beside energy, supposethat an orbit visits all the energy surface;

then replace the time average over an orbit with the average over all possible states (over thewhole available phase space).

• Problem: justify the ergodic hypothesis, with all its consequences, in Statistical Mechanics.

An overwhelming, essentially unsolved problem.

• The viewpoint of dynamical system theory (still in very rough terms):

Provide a mathematically well established theory of chaotic systems.

• Most of the discussion here concerned with maps.


7.2.1 Time average and phase average

• Let x ∈ G and Ω(x) ⊂ G the orbit through x, i.e., Ω(x) =⋃

t φtx where

t ∈ R for a continuos flow; t ∈ Z for an invertible map; t ∈ N0 for non invertible map;

Definition 7.1: For x ∈ G the time average of a function f : G → R is defined as

Flow: Map:

f(x) = limT→+∞

1

T

∫ T

0

f(φt(x)dt . f(x) = limN→+∞

1

N

N∑

t=0

f(φt(x)) ,

provided the limit exists.

Definition 7.2: For a measurable function f : G → R the phase average is defined as

〈f〉 =

∫

G

f dµ .

Proposition 7.3: (Birkhoff.) Let (G , µ, φ) be a dynamical system, and let f be a measurable function.Then the time average

f(x) = limT→+∞

1

T

∫ T

0

f(φt(x)dt or f(x) = limN→∞

1

N

N∑

t=0

f(φt(x))

exists almost everywhere.

Corollary 7.4: With the same hypotheses of proposition 7.3 one has

f(φtx) = f(x) ,

whenever the average exists, for any t ∈ Z or t ∈ Z+, as appropriate.


7.2.2 Sojourn time and invariant functions

• Characteristic function: for a measurable subset A ∈ G

χA(x) = 1 for x ∈ A , otherwise χA(x) = 0 .

Definition 7.5: For an orbit x , φ(x) , φ2(x) , . . . the average sojourn time of a measurable set A ⊂ G isdefined as

τA(x) = limN→∞

1

N#t : 0 ≤ t ≤ N ∧ φt(x) ∈ A .

Equivalently:

τA(x) = limN→∞

1

N

N∑

t=0

χA(φt(x)) = χA(x) ,

where χA is the characteristic function of A .

• Measures the time spent by the orbit in the subset A.

Definition 7.6: An invariant function for φ is a function f : G → R such that

f(φ(x)) = f(x) for all x ∈ G .

• Example: take any measurable function g(x). Define f(x) = g(x). This is an invariant function (seecorollary 7.4).

Definition 7.7: A subset A ⊂ G is invariant in case one has φ−1(A) = A .


7.2.3 Equivalent definitions of ergodicity

Definition 7.8: A system (G , µ, φ) is said to be ergodic if the time average of any measurable functioncoincides with the phase average almost everywhere (i.e., except for a subset of points of zero measure). Thatis:

f = 〈f〉 a.e. in G .

Definition 7.9: A system (G , µ, φ) is said to be metrically indecomposable if every invariant subset A ⊂ G

has either full or zero measure. That is,

φ−1(A) = A =⇒ µ(A) = 0 ∨ µ(A) = 1 .

Proposition 7.10: Let the measure µ be normalized, i.e., µ(G ) = 1. The following properties are equivalentdefinitions of ergodicity:

(i) For every measurable subset A ⊂ G one has τA = µ(A) a.e. in G .

(ii) The system is metrically indecomposable: if φ−1(A) = A then we have either µ(A) = 0 or µ(A) = 1 .

(iii) If f is a measurable invariant function then f = const a.e. in G .

(iv) If f is an invariant L2 function with respect to the measure µ then f = const a.e. in G .

Proof. Denote by (0) the property of definition 7.8. Proceed according to the scheme

(0) ⇒ (i) ⇒ (ii) ⇒ (iii) ⇒ (0)

⇓

(iv) ⇒ (ii) .


• (0)⇒ (i). Take the characteristic function χA of A, with phase average 〈χA〉 = µ(A). by definition of sojourn time τA = χA. By ergodicity, τA = χA = 〈χA〉 = µ(A).

• (i)⇒ (ii). By contradiction. Let φ−1(A) = A and 0 < µ(A) < 1. Then τA(x) = 1 for all x ∈ A, and τA(x) = 0 for x /∈ A. Contradicts (i), because it should be τA(x) = χA = µ(A) 6= 0, 1.

• (ii)⇒ (iii). By contradiction. Let be invariant and not constant a.e.; then there exists A ⊂ G measurable such that f(x) > 〈f〉 and 0 < µ(A) < 1. By corollary 7.4 A is invariant. Contradicts (ii).

• (iii)⇒ (0). By contradiction. Let g(x) be such that g(x) 6= 〈g〉 for x ∈ A, with 0 < µ(A) < 1. Define f(x) = g(x), invariant function by corollary 7.4. Trivially, f(x) = g(x) and f(x) 6= 〈f〉. Contradicts (iii), because f is invariant and not constant a.e. in G .

• (iii)⇒ (iv). Obvious.

• (iv)⇒ (ii). Let A ⊂ G be measurable and invariant. The characteristic function χA is invariant and L2. χA a.e. constant implies either χA = 0 or χA = 1 on all of G . Hence µ(A) =

∫

Gχadµ is either 0 or 1.

Q.E.D.


Proposition 7.11: Let two different measures µ and ν be given on a phase space G , and let both of thembe invariant for the map φ. If φ is ergodic with respect to both measures then there exist disjoint subsets Aµ

and Aν such thatµ(Aµ) = ν(Aν) = 1 , µ(Aν) = ν(Aµ) = 0 .

Proof.

• µ 6= ν implies that there is a measurable function f such that 〈f〉µ 6= 〈f〉ν .

• By ergodicity, there are subsets Aµ and Aν with measure µ(Aµ) = ν(Aν) = 1 such that f(x) = 〈f〉µfor x ∈ Aµ and f(x) = 〈f〉ν for x ∈ Aν .

• Show that Aµ ∩ Aν = ∅. By contradiction:

Let x ∈ Aµ ∩ Aν . Then f(x) should take two different values, contradicting the hypothesis ofergodicity.

• Since Aµ ∩ Aν = ∅, conclude that µ(Aν) = ν(Aµ) = 0.Q.E.D.


7.2.4 The circle map

Let:• G = T a circle with the angular coordinate ϑ;• µ the Lebesgue measure on T, based on the length of the shorter arc joining two points ϑ0, ϑ1 ;• φ is the rotation by an angle ω ∈ R, namely

(7.1) ϑ→ ϑ+ ω mod(2π) .

The map is invertible and measure preserving.

Proposition 7.12: The circle map is ergodic if and only if ω2π is an irrational number.

Lemma 7.13: An orbit of the circle map is periodic if and only if ω2π is a rational number. In that case all

orbits are periodic. If ω2π is an irrational number then every orbit is dense on the torus.

Proof.

• Periodic means φrϑ0 = ϑ0 for some ϑ0 ∈ T and r ∈ N0. translational invariance: φrϑ0 = ϑ0 implies φrϑ = ϑ for all ϑ ∈ T; implies also rϑ = 2sπ for some integer m. Hence ω

2π = sr.

• Suppose ω2π irrational

The orbit Ω(ϑ0) is a infinite sequence of distinct point on the compact phase space T. For every δ > 0 there are r, s such that dist(φrϑ0, φ

sϑ0) < δ. Let α = (s− r)ϑ mod(2π) By translational invariance, φr+jϑ0 − φs+jϑ0 = α < δ for all j ∈ Z; the sequence ϑ0, ϑ0 + α, ϑ0 + 2α . . . mod(2π) is a subsequence of Ω(ϑ0). Every interval of lenght δ contains at least one point of the subsequence. Since δ is arbitrary, every interval contains at least a point of the orbit.

Q.E.D.


Proof of proposition 7.12.

• Let ω2π = s

r, a rational number.

f(ϑ) = sin(2πrϑ) is an invariant function which is not a.e. constant.

Property (iii) of proposition 7.10 is not satisfied; the map is not ergodic.

• Let ω2π be irrational. Use property (iv) of proposition 7.10.

Expand f ∈ L2 in Fourier series as

f(ϑ) =∑

k∈Z

fkei2πkϑ .

Apply the map:

f(

φtϑ)

=∑

k∈Z

(

fkei2πktω

)

ei2πkϑ

fϑ = f(

φtϑ)

implies that for all t ∈ Z

fk(

ei2πktω − 1)

= 0 .

Hence fk = 0 for k 6= 0, and f(ϑ) = f0 is constant.Q.E.D.


7.2.5 The Kronecker flow on a torus

• Consider first a torus T2. Phase space: T2 with angular coordinates ϕ1, ϕ2 . µ: the Lebesgue measure. Continuos flow: φt(ϕ1, ϕ2) = (ϕ1 + ω1t, ϕ2 + ω2t),

with (ω1, ω2) ∈ R2.

The flow is measure preserving.

• If ω1/ω2 = s/r is a rational number then all orbits areperiodic.

• If ω1/ω2 is an irrational numebr all orbits are dense onT2.

• The system is ergodic if and only if ω1/ω2 is an irrationalnumber.

• Poincare section: the successive sections of an orbit withthe segment ϕ1 = 0. Let (0, ϑ) be the inital point. Let α = 2π ω2

ω1.

The Poincare section is the circle map

ϑ→ ϑ+ α mod(2π) .

2π

2π0

α


• The n–dimensional torus Phase space: Tn with angular coordinates ϕ = (ϕ1, . . . , ϕn) . µ: the Lebesgue measure. Continuos flow: φtϕ = ϕ+ ωt, with ω ∈ Rn. The flow is measure preserving.

• Resonance module:Mω =

k ∈ Zn : 〈k, ω〉 = 0

.

If dimMω = n− 1 then every orbit is periodic (full resonance). If 0 < dimMω =< n − 1 then every orbit is dense on a torus of dimension n − dimMω (partial

resonance). If dimMω = 0 then every orbit is dense on Tn (non resonant case).

• If 〈k, ω〉 = 0 with 0 6= k ∈ Zn then F (ϕ) = sin〈k, ω〉 is an invariant function. There are dimMω independent invariant functions for the flow.

• The system is ergodic if and only if dimMω = 0.

• May reduce the flow to a map on Tn−1 by considering a Poincare section. E.g., use ϕn = 0.


7.2.6 The translation map on the torus

The map generated by a Poincare section of the Kronecker flow on a torus.

Phase space: the torus Tn with coordinates ϕ = (ϕ1, . . . , ϕn). µ: the Lebesgeue measure on Tn. Map: φϕ = ϕ+ ω mod(2π) with ω ∈ Rn. The map is measure preserving.

• Invariant functions: try with a trigonometric monomial F (ϕ) = ei〈k,ϕ〉. Calculate:

F (ϕ+ ω) = ei〈k,ϕ+ω〉 = ei〈k,ω〉 · ei〈k,ϕ〉 .

F (ϕ) = ei〈k,ϕ〉 is an invariant function if and only if ω is resonant, i.e.,

ei〈k,ω〉 − 1 = 0 with 0 6= k ∈ Zn .

Equivalent resonance condition:

〈k, ω〉+ 2rπ = 0 with 0 6= k ∈ Zn , r ∈ Z .

• The system is ergodic if and only if ω is not resonant.


7.2.7 The doubling map on the circle

• Let x ∈ [0, 1) be a coordinate on the torus T. Phase space: G = [0, 1). µ: the Lebesgue measure. Map:

φx = 2x (mod 1) .

• Equivalent formulation: Phase space: the unit circle in C, represented as

ei2πx with x ∈ [0, 1). Map:

φ ei2πx = ei4πx .

• The map is continuous, not invertible, preserves themeasure, and has x = 0 as the unique fixed point.• Say that an orbit is definitely periodic in case it be-comes a periodic orbit after a transient.

I

φ−1(I)

0

1

01

Exercise 7.14: Calculate the iterates of thedoubling map of the circle on a digital computer(an elementary program).• Question: what happens?• If you find the result puzzling, try the nextexercise.

Exercise 7.15: Calculate the iterates of themap of the circle

x → 3x mod 1 .

• Question: any difference with the doublingmap of the circle?• Explain the results.


Exercise 7.16: Prove the following properties for the doubling map of the circle.

(i) The subset Q ∈ [0, 1) of rational numbers is invariant.

(ii) The subset R ∈ [0, 1) of irrational numbers is invariant.

(iii) If the initial point x is a rational number then the orbit is either periodic or definitely periodic.

(iv) If the initial point may be written as a dyadic fraction (i.e., a fraction of the form j/2k with a powerof 2 as denominator) then the orbit falls down on the fixed point x = 0 in at most k iterates.

(v) If the initial point x is an irrational number then the orbit is not periodic.

(vi) The map possesses a dense orbit (non periodic, of course). Produce an example.

(vii) There exists a non periodic orbit which is not dense. Produce an example.

(viii) The map is ergodic.


7.3 Mixing

Mixing is a stronger property than ergodicity.

7.3.1 Equivalent definitions of mixing

Definition 7.17: A discrete and invertible dynamical system (G , µ, φ) is said to be mixing in case for anytwo measurable subsets A, B ⊂ G one has

(7.2) limt→∞

µ(φ−tA ∩B) = µ(A) · µ(B) .

Definition 7.18: For two real measurable functions f , g the time correlation is defined as

corr(f, g) = limt→∞

∫

G

(

f φt)

gdµ .

• In shorter notation:corr(f, g) = lim

t→∞

⟨

(f φt)g⟩

.

Proposition 7.19: A discrete and invertible dynamical system (G , µ, φ) is mixing if and only if for anytwo measurable real functions f, g on G one has

(7.3) corr(f, g) =

∫

G

fdµ ·

∫

G

gdµ .


Proof. Proceed in three steps:(i) Prove that (7.2) and (7.3) are equivalent for characteristic functions.

This also proves that (7.3) implies (7.2).(ii) Prove that (7.2) implies (7.3) for sums of characteristic functions.(iii) Prove that (7.2) implies (7.3) for measurable functions.

• (i) let A, B ⊂ G be beasurable and let f = χA and g = χB . Remarks

(

χA φt)

(x) = χA

(

φt(x))

= χφ−tA(x) and χφ−tAχB = χφ−tA∩B .

Hence (7.3) writeslimt→∞

⟨

(χA φt)χB

⟩

= limt→∞〈χφ−tAχB〉 = 〈χA〉〈χB〉 .

By the second remark, for characteristic functions get

limt→∞

µ(φ−tA ∩B) = µ(A)µ(B) ,

namely, the definition of mixing.

• (ii) Take two partitions Aj and Bk of G into disjoint subsets, and let

f =∑

j

fjχAj, g =

∑

k

gkχBk, fj , gj ∈ R .

Using (i) calculate⟨

(f φt)g⟩

=∑

j,k fjgk〈χφ−tAjχBk〉 =

∑

j,k fjgkµ(φ−tAj ∩Bk).

Letting t→∞, in view of (7.2) get

limt→∞

⟨

(f φt)g⟩

=∑

j,k

fjgkµ(Aj)µ(Bk) =∑

j

fj〈χAj〉∑

k

gk〈χBk〉 = 〈f〉〈g〉 ,

as claimed.


• (iii) Let f, g be measurable functions.

Write f = f + f ′ and g = g + g′ with f , g sums of characteristic functions and ‖f ′‖ < ε, ‖g′‖ < ε(use the L2 norm).

Calculate ∣

∣

∣〈(f φt)g〉 − 〈(f φt)g〉∣

∣

∣ < Cε ,∣

∣

∣〈f〉〈g〉 − 〈f〉〈g〉∣

∣

∣ < Cε

with some constant C and uniformly in t.

By (ii) the claim is true for f , g.

Since ε is arbitrary the claim is true for f, g.Q.E.D.

Proposition 7.20: A mixing system is also ergodic.

Proof. Prove that an invariant measurable subset A has either full or zero measure.• Let φ−tA = A; hence µ(φ−tA ∩ A) = µ(A).• By definition of mixing

µ(A) = limt→∞

µ(φ−tA ∩B) =[

µ(A)]2

.

• Implies either µ(A) = 0 or µ(A) = 1.Q.E.D.

Example 7.21: Ergodicity and mixing. The circle map, the translation on the torus and the Kroneckerflow in non rsonant case are ergodic but not mixing. The doubling map of the circle is mixing, and so alsoergodic.


7.3.2 The baker transformation

• Stretching and shrinking the square. Phase space: the square [0, 1)× [0, 1). µ: the Lebesgue measure. Map: φ(x, y) = (x1, y1) with

(7.4) x1 = 2x− ⌊2x⌋ , y1 =1

2(y + ⌊2x⌋) .

The map is not continuous, but it is invertible and pre-serves measure. The map is mixing.

• Qualitative considerations

Main action: horizontal stretching and vertical shrink-ing.

A small area transformed into a very long and thin strip.

The process of cutting and superimposing rectanglesproduces the mixing effect.

φ φ

φ φ

φ

φ

φ φ


7.4 Isomorphism between dynamical systems

Definition 7.22: The dynamical systems (G , µ ,Φ) and (N , ν ,Ψ) are said to be isomorphic in case thereis a one-to-one measure preserving function h : G → N such that the following diagram is commutative,i.e., if Ψ h = h Φ ,

Gh−→←−h−1

N

Φ

y

yΨ

Gh−→←−h−1

N

• Identifies two dynamical systems.• Enough to ask the diagram be commutative almost everywhere (i.e., except for a subset of zero measure).


7.5 Symbolic dynamics

An abstract, very general scheme. Need:• An alphabet A of N > 1 symbols, N finite.• Phase space: two choices.

The set S of doubly infinite sequences

s = sjj∈Z = . . . , s−2 , s−1 , s0 , s1 , s2 , . . . , sj ∈ A .

The set S+ of one-sided sequences

s = sjj∈N0= s0 , s1 , s2 , . . . , sj ∈ A .

• Want to introduce a topology, a metric and a measure.

7.5.1 Topology

Define a basis of neighbourhoods:• For s∗ ∈ S the basis

Uk

k≥0is

Uk(s∗) = s ∈ S : sj = s∗j for |j| ≤ k .

• For s∗ ∈ S+ the basis

Uk

k≥0is

Uk(s∗) = s ∈ S+ : sj = s∗j for j = 0, . . . , k .


7.5.2 Metric

Let s, t ∈ S. Two different metrics.• Let D > 1; define

(7.5) dist(s, t) =∑

j∈Z

δ(sj , tj)

D|j|, δ(x, y) =

1 if x 6= y

0 if x = y.

May choose D = N = #A .• Identify the alphabet with A = 0, . . . , N − 1. Define

(7.6) dist(s, t) =∑

j∈Z

|sj − tj |

D|j|.

• For the space S+ of one–sided sequences just make the sum to run over N0.

Lemma 7.23: Let D > 3, and let dist(s, s∗) = ε for some ε > 0. If 1Dk+1 ≤ ε < 1

Dk then s ∈ Uk(s∗) and

s /∈ Uk+1(s∗)

• Hint for the proof. For s ∈ Uk(s

∗) the distance (7.5) is

dist(s, s∗) =∑

|j|>k

δ(sj , s∗j )

D|j|≤∑

|j|>k

1

D|j|=

2

(D − 1)Dk,

If s /∈ Uk+1(s∗) then at least one of s−j 6= s∗−j or sj 6= s∗j holds true; get

1

Dk+1≤ dist(s, s∗) ≤

2

(D − 1)Dk.

The completion of the proof is left to the reader.


7.5.3 Measure

• Introducing a σ–algebra on S (or S+). Cylinders (sometimes called rectangles): the subsets

Cβj = s ∈ S : sj = β , j ∈ Z , β ∈ A .

(similar for S+, with j ∈ N0). Extension to elements of the σ–algebra: let j1, . . . , jk be arbitrary distinct integers; let β1, . . . , βk

arbitrary symbols from A . Define

Cβ1,...,βk

j1,...,jk= Cβ1

j1∩ . . . ∩ Cβk

jk.

• Introducing the product measure. To every symbol in A = (α0, . . . , αN−1) associate a positive number, or weight, pαj

with

pα0+ . . .+ pαN−1

= 1 .

• Define the product measure asµ(

Cβ1,...,βk

j1,...,jk

)

= pβ1· . . . · pβk

.

Lemma 7.24: The following properties hold true.(i) For β 6= γ and for any j one has Cβ

j ∩ Cγj = ∅, and so also

(7.7) µ(Cβj ∪ Cγ

j ) = µ(Cβj ) + µ(Cγ

j ) = pβ + pγ .

(ii) The complement of the cylinder Cβj is a union of disjoint cyliders; precisely

(7.8) S \ Cβj =

⋃

β′∈A \β

Cβ′

j , with Cβ′

j ∩ Cβ′′

j = ∅ for β′ 6= β′′ .


Proof.

(i) By definition of cylinder: let s be a sequence with sj = γ 6= β. Then s /∈ Cβj . The cylinders being disjoint

the measures add together.

(ii) Clearly S =⋃

β′∈ACβ′

j , the cylinders Cβ′

j being disjoint by (i). Just subtract Cβj .

Q.E.D.

Corollary 7.25: Lemma 7.24 generalizes as follows:(iii) Let j1, . . . , jn ∩ k1, . . . , kn 6= ∅, and let β1, . . . , βn and γ1, . . . , γn be the corresponding symbols. If

there is a pair jl = km such that βl 6= γm then Cβ1,...,βn

j1,...,jn∩Cγ1,...,γn

k1,...,kn= ∅, and so also

(7.9)µ(Cβ1,...,βn

j1,...,jn∪ Cγ1,...,γn

k1,...,kn) = µ(Cβ1,...,βn

j1,...,jn) + µ(Cγ1,...,γn

k1,...,kn)

= pβ1· . . . · pβn

+ pγ1· . . . · pγn

.

(iv) The complement of the cylinder Cβ1,...,βn

j1...,jnis the union of disjoint cylinders

(7.10) S \ Cβ1,...,βn

j1...,jn=

⋃

β′

1,...,β′

n

Cβ′

1,...,β′

n

j1...,jn,

the union being made over all β′1 ∈ A \ β1, . . . , β′

n ∈ A \ βn.

• The proof is just an adaptation of the proof of lemma 7.24


Lemma 7.26: Let A1, . . . , An and B1, . . . , Bm be two sets of cylinders, all of them being pairwisedisjoint, i.e., Aj ∩ Ak = Aj ∩Bk = Bj ∩Bk = ∅ for every allowed pairs j, k. Then

(7.11) µ(

(A1 ∪ . . . ∪ An) ∩ (B1 ∪ . . . ∪Bm))

= µ(A1 ∪ . . . ∪ An)µ(B1 ∪ . . . ∪Bm) .

Proof.

• Recall(A1 ∪ . . . ∪An) ∩ (B1 ∪ . . . ∪Bm) =

⋃

j,k

Aj ∩Bk

where Aj ∩Bkj=1,...,n, k=1,...,m is a set of pairwise disjoint cylinders.

• Evaluate

µ(

(A1 ∪ . . . ∪ An) ∩ (B1 ∪ . . . ∪Bm))

=∑

j,k

µ(Aj ∩Bk)

=∑

j

µ(Aj)×∑

k

µ(Bk)

= µ(A1 ∪ . . . ∪An)µ(B1 ∪ . . . ∪Bm) .

Q.E.D.

Example 7.27: Sets of zero measure. Let A0 ∈ S+ be the set of sequences which do not contain the symbolα0. Then we have µ(A0) = 0.


• The base subsets of the topology are the cylinders

C(n,β) = Cβ1,...,β2n+1

−n,...,0,...,n , β = (β1, . . . , β2n+1) ∈ A2n+1 .

Lemma 7.28: Let A = Cγ1

j1∪ . . .∪Cγn

jnbe a finite union of cylinders. Then A may be decomposed into an

union of disjoint sets of cyliders.

Proof. May assume that j1 ≤ . . . ≤ jn and max(

|j1|, |jn|)

≤ k. Consider first the union of two cylindersCγ1

j1∪ Cγ2

j2. A (non unique) decomposition is (using the notation for the base subsets of the topology)

A =⋃

β∈B

C(k,β) , B = β ∈ A2k+1 : βj1 = γ1 , βj2 = γ2 .

For n > 2 set, e.g.,

A =⋃

β∈B

C(k,β) , B = β ∈ A2k+1 : βj1 = γ1, . . . , βjn = γn .

Q.E.D.

Lemma 7.29: For any measurable subset A ∈ S and for any positive ε there exists A ∈ S which is a finiteunion of disjoint cylinders and satisfies

(7.12) µ(A A) < ε .

• The symbol denotes the symmetric difference of sets, i.e., AB = (A∪B)\ (A∩B), or, equivalently,AB = (A \B) ∪ (B \A).• A known statement in Lebesgue measure theory. The proof will be included . . . sooner or later.


7.5.4 The Bernoulli shift

• Introducing a dynamics in S or S+. The (left) Bernoulli shift σ on S is defined as

σ : S → S

s 7→ σ(s) : (σ(s))j = sj+1 for j ∈ Z .

The (left) Bernoulli shift σ on S+ is defined as

σ : S+ → S+

s 7→ σ(s) : (σ(s))j = sj+1 for j ∈ N0 .

s−4 s−3 s−2 s−1 s0 s1 s2 s3

s−4 s−3 s−2 s−1 s0 s1 s2 s3

. . .

. . .

s−4 s−3 s−2 s−1 s0 s1 s2 s3

s−4 s−3 s−2 s−1 s0 s1 s2 s3

. . .

. . .

• The shift transformation σ on both S and S+ is continuous with respect to the topology of subsec-tion 7.5.1. For S: the inverse image of an open subset is just a right translation of the subset itself. For S+: the inverse image of an open subset is the union of open subsets.

• Recalling some definitions. For s ∈ S the orbit through s is denoted Ω(s). Ω(s) is periodic of period τ > 0 in case φτx = x for every x ∈ Ω(s). Say also that s is a periodic

point. Ω(s) is definitely periodic in case φT

(

Ω(s))

is periodic for some T ≥ 0 (applies to S+).• Elementary properties.

A periodic orbit possesses a minimal period τ , and is also periodic with period 2τ, 3τ, . . . A periodic orbit contains a finite number of points. A non periodic orbit contains a countable set of distinct points.


• There are infinitely many periodic orbits. The set of periodic orbits is dense in S (S+). Denote a periodic point s as (semicolons just mark identical repeated sequences)

s = s1, . . . , sn := . . . ; s1, . . . , sn; s1, . . . , sn; s1, . . . , sn; . . .

For any given open set Uk(s∗) of the basis of the topology construct the periodic point s ∈ Uk(s

∗)

s = s∗−k, . . . , s∗0, . . . , s

∗k .

The points s obviously belongs to both Uk(s∗) and Ω(s).

For S+ just adapt the construction:

s = s∗0, . . . , s∗k .

Construct more examples.

• There is a dense set of definitely periodic points in S+. For any given open set Uk(s

∗) of the basis of the topology construct the definitely periodic orbit

s = s∗−k, . . . s∗0, . . . , s

∗k, x1, . . . , xm ,

with an arbitray finite sequence x1, . . . , xm of m > 0 symbols. The initial point s satisfies s ∈ Uk(s

∗). After m iterations, φm

(

Ω(s))

is a periodic orbit that possibly will never visit again Uk(s∗).

Another answer: the initial point

s = x1, . . . , xm, s∗−k, . . . , s∗0, . . . s

∗k .

Construct further examples.


• There are non periodic points s such that Ω(s) is dense in S (S+). Consider all the finite sequences of length 1, 2, 3, . . . obtained by taking all possible combinations

of symbols in A :

α0 , . . . , αN−1 of length 1 ,

α0, α0 , . . . , α0, αN , . . . , αN−1, α0 , . . . , αN−1, αN−1 of length 2 ,

α0, α0, α0 , . . . , αN−1, αN−1, αN−1 , of length 3 ,

. . . . . . . . . . . . . . .

the n–th line containing Nn sequences. Concatenate the sequences as (semicolons separate the partial sequences)

s = . . . αN−1, αN−1, αN−1 ; . . . ; α0, α0, α0 ; αN−1, αN−1 ; . . . ; α0, α0 ; αN−1 ; . . . ; α0 ;

α0 ; . . . ; αN−1 ; α0, α0 ; . . . ; αN−1, αN−1 ; α0, α0, α0, ; . . . ; αN−1, αN−1, αN−1 ; . . .

The orbit clearly visits every neighbourhood of every point of S. Construct further examples.

• There are non periodic points s such that Ω(s) is non dense in S (S+). Concatenate the finite sequences

α0, α1 , . . . , αN−1 of length 1 ,

α0, α0 , α0, α1 , . . . , α0, αN−1 of length 2 ,

α0, α0, α0 , . . . , α0, α0, αN−1 of length 3 ,

. . . . . . . . . . . . . . .

every line containing N subsequences. Another example: take any random, non periodic sequence and remove all occurrences of a given

symbol. Construct further examples.


7.5.5 Mixing

Proposition 7.30: The Bernoulli shift is mixing for the measure µ defined by any arbitrary choice ofpositive weights pα0

, . . . , pαN−1, with pα0

+ . . .+ pαN−1= 1.

Proof. Apply the definition of mixing. Proceed in three steps:(i) it is true for cylinders;(ii) it is true for disjoint unions of cylinders;(iii) it is true for any measurable set A ∈ S.

(i) Let Cβ1,...,βm

j1,...,jmand Cγ1,...,γn

k1,...,knbe two cylinders.

May assume j1 < . . . < jm and k1 < . . . < kn.

Straightforward remark: σ−t(

Cβ1,...,βn

j1,...,jn

)

= Cβ1,...,βn

j1+t,...,jn+t.

Let j1 + t > kn . Then

µ(

σ−t(

Cβ1,...,βn

j1,...,jn

)

∩ Cγ1,...,γn

k1,...,kn

)

= µ(

Cβ1,...,βn

j1+t,...,jn+t ∩ Cγ1,...,γn

k1,...,kn

)

= µ(

Cβ1,...,βn

j1+t,...,jn+t

)

µ(

Cγ1,...,γn

k1,...,kn

)

= µ(

Cβ1,...,βm

j1,...,jm

)

µ(

Cγ1,...,γn

k1,...,kn

)

.

The equality holds true for any choice of positive weights p0, . . . , pN−1 , the proof being indepen-dent of them.

(ii) Follows from (i), with enough patience.


(iii) Use lemma 7.29.

Let µ(A A) < ε and µ(B B) < ε with A and B disjoint unions of cylinders. By (ii) for t big enough get

µ(

σ−tA ∩ B)

= µ(

σ−tA)

µ(B) = µ(A)µ(B) .

Using A ⊂(

A ∪ (A A))

and the similar relation for B, B get also∣

∣µ(

σ−tA ∩B)

− µ(

σ−tA ∩ B)∣

∣ < aε ,∣

∣µ(A)µ(B) − µ(A)µ(B)∣

∣ < aε

Hence for t big enough get∣

∣µ(

(σ−tA) ∩B)

− µ(A)µ(B)∣

∣ < 2aε .

Since ε is arbitrary the definition of mixing applies.

Again, the proof is independent of the choice of the weights p1, . . . , pN−1 .Q.E.D.


Example 7.31: The Bernoulli scheme and the circle map. Let the alphabet be A = 0, 1. Let the phasespace be the set S+ of one–sided sequences

s = s0, s1, s2, . . . = sjj∈Z+.

• Describes the action of repeatedly tossing a coin. E.g., associate 0 → head and 1 → tail.

• Assign the weights p0 = p1 = 1/2 (unbiased coin): this is named Bernoulli scheme B(12 ,12 ).

• Define the isomorphism h:

To every x ∈ [0, 1) associate sequence of digits of the binary representation of x.

If x is rational then it has two equivalent binary representations, namely (〈xxx〉 denotes any finitesequence of digits)

0. 〈xxx〉 1 0 0 0 0 0 0 0 0 . . .

0. 〈xxx〉 0 1 1 1 1 1 1 1 1 . . .

Exclude from S+ all sequences that end with infinite ones; a set of measure zero, since it iscountable.

• Show that the map [0, 1)→ S+ is one-to-one and preserves the measure (exercise).

• The Bernoulli shift reproduces the circle map.


Example 7.32: The Bernoulli scheme and the baker map. Let the alphabet be A = 0, 1. Let the phasespace be the set S of two–sided sequences

s = s−2, s−1, s0, s1, s2, . . . = sjj∈Z .

• Assign the weights p0 = p1 = 1/2.

• Define the isomorphism h:

Represent (x, y) ∈ [0, 1)× [0, 1) as

x = 0.a1a2a3 . . . , y = 0.b1b2b3 . . .

with binary digits aj , bj.

Associate (the semicolon separating the reversed y sequence form the x sequence)

(x, y) −→ s = . . . b3 b2 b1 ; a1 a2 a3 . . . ,

Again: rational coordinates have two equivalent binary representaions.

Again, exclude from S all sequences that end with infinite ones either on the left or on the right;a set of measure zero, since it is countable.

• Show that the map is one-to-one and preserves the measure.

• The Bernoulli shift reproduces the baker map.


Example 7.33: More on the baker map. A general method of constructionof the isomorphism.• Partition the square Q = Q0 ∪ Q1 with Q0 = [0, 1/2)× [0, 1) and Q1 =[1/2, 1)× [0, 1) . Associate to a point (x, y) the sequence in S

sj =

0 if φj(x) ∈ Q0 ,

1 if φj(x) ∈ Q1 , j ∈ Z .

0

1

01Q0 Q1

x φ(x)

φ3(x)

φ4(x)

φ5(x)

φ2(x)

• The inverse of the isomorphism, for given s.• Forward steps, j ≥ 0: vertical strips (figure)

φj(x) ∈ Qsj

φj−1(x) ∈ Qsj−1∩ φ−1(Qsj )

φj−2(x) ∈ Qsj−2∩ φ−1(Qsj−1

∩ φ−1(Qsj ))

φj−3(x) ∈ Qsj−3∩ φ−1

(

Qsj−2∩ φ−1(Qsj−1

∩ φ−1(Qsj )))

. . . ,

hence

x ∈ Qs0 ∩ φ−1(Qs1 ∩ φ−1(Qs2 ∩ . . . ∩ φ−1(Qsj ) ∩ . . .)) .

• Similar procedure for backward steps: horizontal strips.• The point x is found as the countable intersection of strips.

Qsj Qsj−1∩ φ−1Qsj

Qsj−2 ∩ φ−1(Qsj−1∩ φ−1Qsj) Qsj−3

∩ φ−1(Qsj−2∩ φ−1(Qsj−1

∩ φ−1Qsj))


7.6 The Smale horseshoe

Models the dynamics in a neighbourhood of stableand unstable intersecting manifolds. Concentrate ontopological properties; forget the measure.• Take a rectangle Q.(i) Shrink the rectangle by a factor less than 1/2 in

the vertical direction.(ii) Stretch the resulting rectangle by a factor

greater than 2 in the horizontal direction.(iii) Bend the resulting strip into the shape of a

horseshoe.(iv) Superimpose the resulting horseshoe to the orig-

inal rectangle so that both ends and the curvedpart exceed the sides of the rectangle.

U1

U0

• Keep only the intersection of the strip with therectangle: two horizontal strips U0, U1 .• Map the strips U0, U1 back to the square Q. The inverse image of U0, U1 is well deter-mined: the vertical strips V0, V2 .

V0 V1

U1

U0


• Intersect the vertical strips with the horizontalstrips.• The forward map is well defined in the resultingsquares U0 ∩ V0 , U1 ∩ V0 , U0 ∩ V1 and U1 ∩ V1 .• With a little abuse of notation (i.e., ignore every-thing that fall outside the square) write

U0 ∪ U1 = φ(Q) ∩Q , V0 ∪ V1 = φ−1(Q) ∩Q .

V1V0

U0

U1

• Next step. Forward map: the four squares are mappedinto the strips

U0,0 = φ(U0 ∩ V0) , U0,1 = φ(U0 ∩ V1) ,

U1,0 = φ(U1 ∩ V1) , U1,1 = φ(U1 ∩ V1) .

Again with a little abuse of notation write

U0,0 ∪ U0,1 ∪ U1,0 ∪ U1,1 = φ(U0 ∪ U1) ∩Q

= φ2(Q) ∩Q .

U0,1

U0,0

U1,0

U1,1


With a backward iteration construct the preimageof V0, V1 : the four vertical strips. Write

V0,0 ∪ V0,1 ∪ V1,0 ∪ V1,1 = φ−1(V0 ∪ V1) ∩Q

= φ−2(Q) ∩Q

• The map φ2 and its inverse φ−2 are well defined in the16 curvilinear squares

U0,0 ∩ V0,0 , U0,0 ∩ V0,1 , U0,0 ∩ V1,0 , U0,0 ∩ V1,1 ,

U0,1 ∩ V0,0 , U0,1 ∩ V0,1 , U0,1 ∩ V1,0 , U0,1 ∩ V1,1 ,

U1,0 ∩ V0,0 , U1,0 ∩ V0,1 , U1,0 ∩ V1,0 , U1,0 ∩ V1,1 ,

U1,1 ∩ V0,0 , U1,1 ∩ V0,1 , U1,1 ∩ V1,0 , U1,1 ∩ V1,1 .

• By iteration. After n steps find: 2n disjoint horizontal strips syntetically written as

⋃

j1,...,jn=0,1

Uj1,...,jn = φn(Q) ∩Q .

2n disjoint vertical strips syntetically written as⋃

j1,...,jn=0,1

Vj1,...,jn = φ−n(Q) ∩Q .

• The map φn and its inverse φ−n are well defined on4n curvilinear closed squares generated by the intersec-tions of the horizontal strips with the vertical ones.

U1,1

U0,0

U0,1

U1,0

V0,0 V0,1 V1,0 V1,1


• The horseshoe map is well defined on the subset of Q

Λ =⋂

j∈Z

φj(Q) ,

which is clearly invariant.• The horseshoe map restricted to Λ is associated to the Bernoulli shift on S with alphabet A = 0, 1.

Associate to x in Λ the sequence s ∈ S defined as

sj =

0 if φj(x) ∈ U0 ,

1 if φj(x) ∈ U1 .

Prove that the inverse is weel defined, i.e., every s ∈ S determines a unique point x ∈ Λ. Hint: follow the proof for the baker’s map.

• Further properties (exercise): Λ is a closed set similar to the Cantor set. the mapping h : Λ→ S is continuous if one considers in Λ the topology induced by the metrics

dist(s, s′) =∑

j∈Z

|sj − s′j |

2|j|.


• The horseshoe model describes the dynamics in aneighbourhood of a homoclinic orbit. Consider a (curvilinear) rectangle Q around

a homoclinic point P on the stable manifoldW (s) .

By continuity, the image of Q will cover the un-stable manifold W (u), being stretched in the di-rection of W (u) and shrinked in the direction ofW (s) .

For some k > 0 the image φk(Q) must intersectQ forming the horseshoe.

W (s)

W (u)

P

O

φk(Q)

Q


7.7 Poincare’s recurrence theorem

Consider an abstract dynamical system (G , φ, µ) with φ invertible and µ(G ) = 1.• Let A ⊂ G be measurable. Define:

Recurrent set:RA = x ∈ A : ∀K0 > 0 , ∃k > K0 such that Ψk(x) ∈ A .

Wandering set: the complement of the recurrent set, i.e.,

VA = A\RA ,

Remark: a wandering point may return to A a finite number of times; a recurrent point returnsto A an infinite number of times.

• Define alsoVA,K0

= x ∈ A : Ψk(x) 6∈ A for all k ≥ K0 .

Prove that VA,K0, and so also VA , are measurable sets.

GetVA,K0

= x ∈ A : Ψk(x) ∈ G \A for all k ≥ K0 ,

hence

VA,K0= A ∩

⋂

k≥K0

Ψ−k(G \A)

;

Get alsoVA =

⋃

K0>0

VA,K0.


Theorem 7.34: (Poincare, 1889) We have

µ(VA) = 0 .

Proof.

• By the definition of VA,K0

ΨnK0(VA,K0) ∩ A = ∅ , ∀n > 0 .

• Implies that for every n1 > n2 > 0 we also have

Ψn1K0(VA,K0) ∩Ψn2K0(VA,K0

) = ∅ .

By contradiction: assume the latter set contains a point x1 ; then

Ψ−n2K0(x1) ∈ VA,K0∩Ψ(n1−n2)K0(VA,K0

)

⊂ Ψ(n1−n2)K0(VA,K0) ∩A = ∅ .

• Hence all sets ΨnK0(VA,K0) are disjoint.

• Since the map is measure preserving,

µ

(

⋃

n>0

ΨnK0(VA,K0)

)

=∑

n>0

µ(

ΨnK0(VA,K0))

=∑

n>0

µ(VA,K0) ,

From µ(G ) <∞ conclude µ(VA,K0) = 0 .

Q.E.D.


investigate the dynamics of fully chaotic systems. · proposition 7.12: the circle map is ergodic...

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