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**Dynamical Subbases of I and I2**

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**Itinerary of the tent function**

1 X0 X1 t5(x) x t(x) t3(x) t2(x) t4(x) 1 For x=3/16, φ(x)=001⊥1000… Sn,0 ={x | φ(x)(n)=0} = {x | tn(x)∈X0} = t-n(X0) Sn,1 = t-n(X1)

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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). 1 X0 B X1 1

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**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.

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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.

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**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. 1 X0 X1 B g1g0(B) g1(B) g0g0(B) B g1g0(B) g1(B) g0g1(B)

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**Dynamical Subbase of I2 (1)**

flip g0 X0=S(0) f B X1=S(1) g1 X f(B)

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**Dynamical Subbase of I2 (1)**

000 010 00 01 S(0) 001 011 101 111 S(1) 10 11 100 110 S0,0, S0,1 S3,0, S3,1 S1,0, S1,1 S2,0, S2,1 flip g0 X1=S(0) f Gray x Gray Subbase B X0=S(1) g1 : ⊥⊥11000… X Degree 2 f(B)

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**Dynamical Subbase of I2 (2)**

flip g0 X0=S(0) f B X1=S(1) g1 f(B) X

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**Dynamical Subbase of I2 (2)**

X X1=S(1) X0=S(0) flip f(B) B g1 g0 f Peano Subbase : ⊥1⊥⊥1000… Degree 3

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**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.

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**How many other dynamical subbases of I2 ?**

two cases. two cases. two cases. two cases. two cases.

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

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

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**Only two cases a0 b1, a1 b0 Peano Subbase is a typical example.**

Gray x Gray subbase is a typical example.

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**Characterization of Peano-like subbases.**

1 2 B f(B)

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0 1, 1 2, 2 3 by f-1 1 2 B f(B) 1 2 B f(B) 3 1 2 B f(B) 3 1 2 B f(B) 3 c=12 c=121 120 122 1 2 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.

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**1212 c=121 1210 1211 Three cases for the 4-th code. 1 2 B f(B) 3 1 2 B**

2 B f(B) 3 1 2 B f(B) 3 1212 4 1 2 B f(B) 3 1 2 B f(B) 3 c=121 1210 1211 Three cases for the 4-th code.

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**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 2 B f(B) 3 c=1210

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**c=1211UU c=1211U c=1211 U means no choice at that level.**

2 B f(B) 3 c=1211UU 4 5 1 2 B f(B) 3 1 2 B f(B) 3 c=1211U 4 5 4 4 4 c=1211 U means no choice at that level. Only U appears after this. It is not a subbase.

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1 2 B f(B) 3 12 12121 4 5 1 2 B f(B) 1 2 B f(B) 3 121 1 2 B f(B) 3 1212 4 5 5 2 1 12 102 1302 13042 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.

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**The order of the number on the boundary determine the subbase (i. e**

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 1 2 B f(B) 3 4 5 4 5 4 5 a=12121

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**In this way, it is a purely symbolic problem.**

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 1 2 B f(B) 3 4 5 4 5 4 5 121

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**There are three cases that we do not have a subbase**

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.

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

(therefore, c(2n)=1), then c(i)=U (i>2n). Not a subbase. 1 2n B f(B) 2 n

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**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 2 B f(B) 3 1 2 B f(B) 3 c=1211UU 4 5 c=1210

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**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.

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

1’ 1 B f(B) 0’ 2 2’ 2 1 B f(B) 1 B 3’ 3 3 1’ 1’ 2’ 3’ 3 0’ 0’ 2 {2}x{2} {2,1}x{2,2} {2,1,U}x{2,2,U} 1’0’201 1’30’201 {0,1,2,U} sequence for both 0 and 0’. Symbolic manipulation.

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**Characterization of GrayxGray-like subbases.**

1 B f(B) 1’ 0’

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**The two components interact, and a bit complicated. **

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.

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**Conclusion Study computation-related topological property.**

How many recursive structures we can consider in I2 ?

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Transformations and Tessellations Edited By: K. Stone.

Transformations and Tessellations Edited By: K. Stone.

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