1D Dynamics: Symbolic Coding

1. Full shift of two symbols

– Let

X = {s =(s0s1s2 ··· si ··· ); si =0 or 1} be the collection of all sequences of 0 and 1.

– For s =(s0s1 ··· si ··· ), t =(t0t1 ··· ti ··· ), let ∞ |s − t | d(s, t) = i i X i i=0 2

Claim: (i) (X,d) is a metric space.

s t 1 (ii) d( , ) ≤ 2n iff si = ti for all i < n. Proof: (i) d(s, t) ≥ 0 and d(s, t) = d(t, s) are obvious. For the triangle inequality, observe ∞ ∞ ∞ |si − ti| |si − vi| |vi − ti| d(s, t) = ≤ + X 2i X 2i X 2i i=0 i=0 i=0 = d(s, v)+ d(v, t).

(ii) is straight forward from definition.

– Let σ : X → X be define by

σ(s0s1 ··· si ··· )=(s1s2 ··· si+1 ··· ).

Claim: σ : X → X is continuous. Proof: 1 For ε > 0 given let n be such that 2n < ε. Let 1 s t s t δ = 2n+1 . For d( , ) <δ, d(σ( ), σ( )) < ε.

– (X,σ) as a dynamical system:

Claim (a) σ : X → X have exactly 2n periodic points of period n.

(b) Union of all periodic orbits are dense in X. (c) There exists a transitive orbit. Proof: (a) The number of periodic point in X is the same as the number of strings of 0, 1 of length n, which is 2n.

(b) ∀s =(s0s1 · · · sn · · · ), let sn =(s0s1 · · · sn; s0s1 · · · sn; · · · ). We have 1 d(sn, s) < . 2n−1 So sn → s as n →∞. (c) List all possible strings of length 1, then of length 2, then of length 3, and so on. We obtain a orbit that is clearly dense in X.

2. Dynamical Equivalence

Let f : X → X, g : Y → Y be two given dy- namical system.

Question: When can we regard (X,f) and (Y, g) as the same?

Definition: Two systems (X,f) and (Y, g) are topologically conjugate if there exists h : X → Y , such that (a) h is a homeomorphism (Both h and h−1 are well-defined and continuous); and

(b) g ◦ h = h ◦ f. Remarks: Item (a) requires that, as topological spaces, X and Y are equivalent. Item (b) requires that h maps orbit to orbit. X Y h

x y 1 h 1 f g x2 h y2 f g x3 y3 f h g

x4 y4

Item (b) is also represented by the following commuting diagram: f X X

h h

Y Y g Claim: If (X,f) and (Y, g) are conjugated by h. Then

(i) They have the same number of periodic orbits of any given period.

(ii) The stability of the corresponding periodic orbits are the same.

(iii) If one has a transitive orbit so does the other.

The proof of this Claim is left as a homework. Ex1: 1 f(x) = 2x, g(x)= 2x are not conjugate. Ex2: f(θ)= θ + π and f(θ)= θ + 1 are not conjugate.

Ex3: How about f(θ) = θ + r1 and f(θ) = θ + r2, r1 6= r2?

Not conjugate if r1 6= r2mod(2π). Need more sophisti- cated tool to show it.

Ex 4: f(x)=2x and g(x)=3x are conjugate. Proof: Construct h : R → R By conjugating fundamen- tal domains. – For f(x) = 2x, let U = [−2, −1) ∪ (1, 2]. All orbit of f(x) passing U one and only one time. – For g(x) = 3x, V = [−3, −1)∪(1, 3], is the correspond- ing fundamental domain. – To construct h : R → R, we • Construct k : (1, 2] → (1, 3]: k(x) = 2x − 1 • For x ∈ (2n, 2n+1], let h(x)= gn ◦ k ◦ f −n(x). • h(x) on x< 0 are defined similarly using [−2, −1) and [−3, −1). • Check that h : R → R so defined conjugates (R,f) and (R, g).

3. Dynamics of T (x)=7x(1 − x)

Let Λ = {x ∈ [0, 1]; T i(x) ∈ [0, 1] for all i > 0}.

Claim: (Λ, T ) and (X,σ) conjugate where X is the space of sequences of two symbols and σ is the shift operator. Proof: 1 1 Let I0 = [0, 2],I1 = [2, 1].

– Address: let a(x)=0 if x ∈ I0, and a(x)=1 if x ∈ I1. – Coding of orbit. For x ∈ Λ, let s(x)= a(x)a(T (x))a(T 2(x)) · · · .

– s :Λ → X satisfies s ◦ T = σ ◦ s: s(Tx)= a(Tx) · a(T nx) · · · = σ(s(x)).

– s :Λ → X is one to one; onto; continuous.

Key observation: for any given sequence a0a1 · · · an, ai = 0 or 1, there exists a unique interval Ia0a1···an , such that

• Ia0a1···anan+1 ⊂ Ia0a1···an ;

i • for all x ∈ Ia0a1···an , T (x) ∈ Iai ,i ≤ n;

n n ′ n • T (Ia0a1···an ) = [0, 1], |(T ) (x)| > 2 .

n • T ([0, 1] \ ∪a0a1···an Ia0a1···an ) 6∈ (0, 1).

Proof of the key observation: Inductively assume the above for Ia0a1···an .

∃ a sub-interval Ia0a1···an,0 ⊂ Ia0a1···an such that

n 1 f(f (Ia0a1···an,0)) = [0, ] 2 and Ia0a1···an,1 ⊂ Ia0a1···an such that

n 1 f(f (Ia0a1···an,1))=[ , 1]. 2

The properties of s :Λ → X claimed (1-1, onto, contin- uous) follows easily from this observation. The details are left as a homework.

– Conclusion: s :Λ → X conjugates (Λ, T ) to (X, σ).

Corollaries: (i) (Λ, T ) has 2n periodic orbits of period n; (ii) the collection of all periodic orbits of T is dense in Λ; and (iii) T has a transitive orbit.

4. Kneading theory

Let I = [a, b], and f : I → I, continuous, is such that

(i) f(a)= f(b)= a;

(ii) ∃ a unique c ∈ I, such that f(x) is monotonically increasing on [a, c), monotonically decreasing on (c, b]. We call f : I → I a unimodal map.

– Let I0 = [a, c), I1 =(c, b] and define a(x)=0 for x ∈ I0 and a(x) = 1 for x ∈ I1. Let a(c) = C.

– Let s(x) = a(x)a(f(x)) ··· a(f i(x)) ··· .

– Let K(f) = s(f(c)): the kneading sequence for f.

Question: which sequence is allowed and which is not?

A partial answer: The kneading theory.

The kneading order

(1) 0 < C < 1. (2) For s = s0s1 ··· sn ··· , t = t0t1 ··· tn ··· , let n be the first integer such that tn 6= sn. Let

τ(s, t) = # of 1’s among {s0s1 ··· sn−1}.

(3) The kneading order: s ⊳ t if

(a) τ(s, t) = even, and sn

(b) τ(s, t) = odd, and sn >tn.

Ex: (0101 · · · ) ⊳ (010C · · · ) ⊳ (0100 · · · )

Ex: (110 · · · ) ⊳ (11C · · · ) ⊳ (111 · · · ).

Note: To compare s and t is to compare s0s1 · · · sn and t0t1 · · · tn where n is the smallest integer such that tn = sn.

Theorem: Assume that K(f), the kneading sequence for f, is not periodic. If t is such that σnt ⊳ K(f) for all n ≥ 1, then there is an x ∈ [a, b] such that s(x) = t. Observation 1: x ≤ y iff s(x) ⊳ s(y).

Proof: If x and y are in I0 = (a, c), then f(x) ≤ f(y) iff x ≤ y.

If x and y are in I1 =(c, b) then f(x) ≤ f(y) iff x>y. So as long as the addresses of x and y are kept the same, then the number of switches in order before time n is the same as the number of 1’s before time n in s(x). This observation is now the same as the definition of the kneading order.

Observation 2: Let A = {x ∈ [a, b], s(x) ⊳ t}, B = {x ∈ [a, b], s(x) ⊲ t}. Then A and B are both non-empty.

Proof: This is because s(a) = 00 · · · ⊳ t so a ∈ A. Similarly, s(b) = 10 · · · ⊲ t so b ∈ B.

Observation 3: Both A and B are open in [a, b].

First note that the theorem follows from ob- servation 3 since [a, b] \ A ∪ B is a non-empty and closed subset. Proof: We argue that A is open (The reason for B open is similar).

– Let t = t0t1 · · · ti · · · , then ti 6= C for all i. i (If ti = C then σ (t) = K(f), against the assumption that σit ⊳ K(f)). – Let z ∈ [a, b] be such that s(z) = s ⊳ t. Let n be the first index such that sn 6= tn. Either tn = 0 or tn = 1. Let us consider only the case tn = 1, leaving the other case tn = 0 as a homework problem. It suffices for us to prove

Lemma: Let z ∈ [a, b] be such that s(z) ⊳ t as in the above. Then there exists a sufficiently small open neighborhood V (z) around z such that ∀x ∈ V (z), s(x)⊳ t.

Proof of this lemma: We have two possible cases : (i) sn = 0, or (ii) sn = C.

(i) If sn = 0, then there is a small open set V (z) around i z such that for any x ∈ V (z), si(x) := a(T (x)) = si for all i ≤ n. (This is because si = ti 6= C for all i ≤ n) We have

s(x)= s1s2 · · · sn · · · ⊳ t1t2 · · · tn · · · .

n (ii) If sn = C, then T z = c. Let V (z)=(z − ε, z + ε). Then T nV (z)=(T n(z − ε), T n(z + ε)) is a small interval around c. Note that we have T n(z − ε) < T n(z + ε) because

s1 · · · sn−1C ⊳ t1 · · · tn−11, implicating that there are even number of 1’s in s1 · · · sn−1. Now for all x ∈ (z − ε, z], s(x) ⊳ t is again from

s1 · · · sn−1sn(x) ⊳ t1 · · · tn−11.

For x ∈ (z, z + ε), first we have

s1(x) · · · sn(x)= t1 · · · tn (1) and t1 · · · tn now have odd number of 1. Let α be the first integer such that

tn+1 · · · tn+α ⊳ sn+1 · · · sn+α (2) (The existence of the integer α is from the fact that K(f) = sn+1sn+2 · · · because sn = C. We must have σn(t) ⊳ K(f).)

Putting (1) and (2) together we conclude

s1 · · · sn−11sn+1 · · · sn+α ⊳ t1 · · · tntn+1 · · · tn+α.

Note that the left is the first n + α symbols for s(x). Homework

1. Prove that, if (X,f) and (Y, g) conjugate, then (a) They have the same number of periodic orbits of any given period. (b) The stability of the corresponding periodic orbits are the same. (c) If one has a transitive orbit so the other.

2. Prove that h : R → R constructed in Ex4 conjugates f = 2x and g = 3x. 3. 1 Prove that f(x)= 2x is structurally table in following sense: f(x) conjugate to g(x)= f(x)+ k(x) where k(x) R R is any given function from → such that kk(x)kC1 < 1 16. 4. Finish the writing that s :Λ → X is one-to-one, onto and continuous.

5. Find all coding sequences allowed by T (x) = 2x(1 − x). How about T (x) = 4x(1 − x)?

6. Prove that the set A in observation 3 above is open in the case tn = 0.