Dynamical Subbases of I and I 2. Itinerary of the tent function 01 0 1 X0X0 X1X1 xt(x)t(x)...

30
Dynamical Subbases of I and I 2

Transcript of Dynamical Subbases of I and I 2. Itinerary of the tent function 01 0 1 X0X0 X1X1 xt(x)t(x)...

Dynamical Subbases of I and I2

Itinerary of the tent function

0 1

0

1

X0 X1

x t(x) t2(x)t3(x) t4(x)

For x=3/16,

t5(x)

φ(x)=001⊥1000…

Sn,0 ={x | φ(x)(n)=0} = {x | tn(x)∈X0} = t-n(X0)Sn,1 = t-n(X1)

Dynamical Subbase. A dyadic subbase S of a topological space X is dynamical if, for X0 = S0,0, X1 =

S0,1, and B = X -(X0 ∪X1), there is a 2-1 continuous map f:X →X such that

f|X0 ∪B :X0∪B → X , homeo.

f|X1 ∪B : X1∪B → X , homeo.

Sn,0 = t-n(X0) and Sn,1 = t-n(X1) (n=0,1,2…).

Prop. φ(f(x)) is the one-bit shift of φ(x). That is, the tail operation realizes the map f.

Prop. B coincides with {x | |f--1(f(x))|=1}. g0 = f|X0 ∪B

-1: X → X0∪B

g1 = f|X1 ∪B-1: X → X1∪B

For σ = d0d1…dn ∈ 2n, S(σ) = gd0gd1gd2…gdn(X).

0 1

0

1

X0 X1B

A condition for a dynamical subbase.

A 2-1 continuous map f:X →X such that f|X0∪B :X0∪B → X , homeo.

f|X1∪B : X1∪B → X , homeo.

is a dynamical subbase if the maximal of the diameter of S(σ) for σ ∈ 2n decreases to 0 when n goes to infinity.

Conjugacy

• Two maps f:X →X and g:X →X are conjugate if there is a homeomorphic map h:X →X such that fh=hg.

f:X → X h↓ ↓h g:X → X • If f and g are conjugate and S is a dynamical

subbase induced by f, then g also induces a dynamical subbase.

• We identify dynamical subbases which are derived from conjugate dynamical system.

Dynamical Subbases of I

• All the dynmical subbases of I are conjugate to the Gray subbase.

(proof) A 2-1 continuous map homeo. to X on [0,B] and on [B,1] is increasing on [0,B] and decreasing on [B,0] or vice versa.

B, g0(B), g1(B), g0g0(B), g0g1(B), g1g0(B), g1g1(B),…. are ordered in the same way as the Gray-subbase, and they are dense in I.

0 1

1

X0 X1B g1g0(B) g1(B) g0g0(B) B g1g0(B) g1(B) g0g1(B)

Dynamical Subbase of I2 (1)

X

X1=S(1)

X0=S(0)flip

f(B)

B

g1

g0

f

Dynamical Subbase of I2 (1)

X

X0=S(1)

X1=S(0)flip

f(B)

B

g1

g0

f

S(0)

S(1)

00 01

10 11

S0,0, S0,1 S1,0, S1,1 S2,0, S2,1S3,0, S3,1

111

110100

101

001 011

000 010

Gray x Gray Subbase: 11000…⊥⊥

Degree 2

Dynamical Subbase of I2 (2)

X

X1=S(1)

X0=S(0)

flip

f(B)

B

g1

g0

f

Dynamical Subbase of I2 (2)

X

X1=S(1)X0=S(0)

flip

f(B)

Bg1

g0

fPeano Subbase

: 1 1000…⊥ ⊥⊥Degree 3

How many other dynamical subbases of I2 ?

• B is a line segment, whose endpoints are on the boundary of I2.

• f(B) is a line segment contained in the boundary of I2.

How many other dynamical subbases of I2 ?

two cases. two cases.

two cases. two cases. two cases.

The diameter of some component S(σ) does not decrease.

It does not form a subbase, for any arrangement of c0,…

Only two cases

• a0 b1, a1 b0 Peano Subbase is a typical

example.

• a0 b1, a1 b0 Gray x Gray subbase is a

typical example.

Characterization of Peano-like subbases.

1

0

2B

f(B)

0 1, 1 2, 2 3 by f-1

1

0

2B

f(B)

1

0

2B

f(B)

3

3

1

0

2B

f(B)

3

3

1

0

2B

f(B)

3

3

Three cases, depending on the order of 3 and 0. 1:right (1-side), 2:left(2-side), 0: overlap on 0.c: code sequence of the subbase.

c=121 120 122c=12 1 2 1 2 3 1 2 3 1 2 3

1

0

2B

f(B)

3

3

Three cases for the 4-th code.

c=121 1 2 3

1

0

2 B

f(B)

3

3

1

0

2 B

f(B)

3

3

1210 1211

1

0

2 B

f(B)3

3

1212 1 2 3 4

4

4

4

1 2 3 4 1 2 3 4

• After 0, there is no choice. (code sequence terminates with 0) Some of them define subbase.Some of them do not.(c=1210 is not a subbase.)

First consider the case 0 does not appear in the code.

1

0

2 B

f(B)

3

3

c=1210 1 2 3 4

U means no choice at that level.Only U appears after this. It is not a subbase.

1

0

2 B

f(B)3

3

c=1211

4

4

4

1

0

2 B

f(B)3

3

c=1211UU

4

4

4

5

5

5

1

0

2 B

f(B)3

3

c=1211U

4

4

4

5

5

5

1 2 3 4 1 2 3 4 5 1 2 3 4 5 6

5

5

1

0

2 Bf(B)3

3

1212121

4

4

45

5

1

0

2 Bf(B)

1

0

2 Bf(B)

3

3

121

1

0

2 Bf(B)3

3

1212

4

4

4

102 1302 13042 1535042

The next sequence is obtained by,• Starting with 0, go right, and then go back to 1 discarding 0.• Increment each number.• Insert 0 at the original position. If new number is inserted there, we

have two choices. (Corresponding to 1,2).• If new number is not inserted there, it is U.

1 3 0 4 2

0 4 2 4 3 1

1 5 3 5 4 21 5 3 5 0 4 2 2 1

12

• The order of the number on the boundary determine the subbase (i.e. dynamical system) modulo conjugacy.

• We only need to code bottom half of it, from 1 to 2. (Top half is mirror image without 0).

5

5

1

0

2 B

f(B)3

3

a=12121

4

4

45

5

1535042

• Proposition. If bigger and bigger number appear on both sides of 0, then it forms a subbase.

In this way, it is a purely

symbolic problem.

5

5

1

0

2 B

f(B)3

3

121

4

4

45

5

1535042

There are three cases that we do not have a subbase. After some sequence of 1,2,U,1. UUU… occur.2. We select 1 everytime we have a choice.3. We select 2 everytime we have a choice.

• If c(n)=2 and c(i) != 2 (n<i<2n), then c(2n) is not U.

• If c(n)=2 and c(i) != 2 (n<i≦2n) (therefore, c(2n)=1), then

c(i)=U (i>2n). Not a subbase.

1

0

2n

B

f(B)n

2

The case 0 appears in the code.

(a) if c(2n)=0 and c(i)!=2 for (n < i < 2n).

not a subbase.(b) Otherwise, it forms a

subbase. (degree infinite).

(a): Replace 0 with 1UUU… (b): Replace 0 with 1222…• If we identify node k with the

adjoining node k+n, we have the same figure.

1

0

2 B

f(B)

3

3

c=1210 1 2 3 4

1

0

2 B

f(B)3

3

c=1211UU

4

4

4

5

5

5

1 2 3 4 5 6

Characterization of Peano-like dynamical subbases of I2

• We do the replacement in the previous slide for codes with 0. Code is an infinite sequence of {1,2,U}. The sequence of {1,2} obtained by removing U determines the code.

• In the Cantor Space {1,2}ω, those forming a subbase is nowhere dense, closed, continuum cardinarity.

1’1B

f(B)

0’

0

• {0,1,2,U} sequence for both 0 and 0’.• Symbolic manipulation.

1’1B

f(B)

0’

02

2

2’ 1’1B

0’

02

2

2’

3

3

3

3’

3’

1’30’2011’0’201

{2}x{2} {2,1}x{2,2} {2,1,U}x{2,2,U}

Characterization of GrayxGray-like subbases.

1’1B

f(B)

0’

0

• Similar result around 0.– If c(n)=2 and c(i) != 2 (n<i<2n), then c(2n) is not U.– If c(n)=2 and c(i) != 2 (n<i≦2n) (therefore, c(2n)=1), then c(i)=U (i>2n). Not a subbase.

• The two components interact, and a bit complicated.

• The Peano-like cases as extreme case. {2,2,2,…} as the first component.

Conclusion

• Study computation-related topological property.

• How many recursive structures we can consider in I2 ?