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Dynamical Subbases of I and I 2

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Itinerary of the tent function 01 0 1 X0X0 X1X1 xt(x)t(x) t2(x)t2(x) t3(x)t3(x) t4(x)t4(x) For x =3/16, t5(x)t5(x) φ ( x )=001 ⊥ 1000… S n,0 ={x | φ ( x )( n ) =0} = {x | t n ( x ) ∈ X 0 } = t -n ( X 0 ) S n,1 = t -n ( X 1 )

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Dynamical Subbase. A dyadic subbase S of a topological space X is dynamical if, for X 0 = S 0,0, X 1 = S 0,1, and B = X -(X 0 ∪ X 1 ), there is a 2-1 continuous map f:X → X such that f| X 0 ∪ B : X 0 ∪ B → X, homeo. f| X 1 ∪ B : X 1 ∪ B → X, homeo. S n,0 = t -n ( X 0 ) and S n,1 = t -n ( X 1 ) (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}. g 0 = f| X 0 ∪ B -1 : X → X 0 ∪ B g 1 = f| X 1 ∪ B -1 : X → X 1 ∪ B For σ = d 0 d 1 …d n ∈ 2 n, S( σ ) = g d0 g d1 g d2 …g dn (X). 01 0 1 X0X0 X1X1 B

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A condition for a dynamical subbase. A 2-1 continuous map f:X → X such that f| X 0 ∪ B : X 0 ∪ B → X, homeo. f| X 1 ∪ B : X 1 ∪ B → X, homeo. is a dynamical subbase if the maximal of the diameter of S( σ ) for σ ∈ 2 n 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, g 0 (B), g 1 (B), g 0 g 0 (B), g 0 g 1 (B), g 1 g 0 (B), g 1 g 1 (B),…. are ordered in the same way as the Gray- subbase, and they are dense in I. 01 1 X0X0 X1X1 B g 1 g 0 (B) g 1 (B) g 0 g 0 (B) B g 1 g 0 (B) g 1 (B) g 0 g 1 (B)

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Dynamical Subbase of I 2 (1) X X 1 =S(1) X 0 =S(0) flip f(B) B g1g1 g0g0 f

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Dynamical Subbase of I 2 (1) X X 0 =S(1) X 1 =S(0) flip f(B) B g1g1 g0g0 f S(0) S(1) 00 01 1011 S 0,0, S 0,1 S 1,0, S 1,1 S 2,0, S 2,1 S 3,0, S 3,1 111 110100 101 001 011 000 010 Gray x Gray Subbase : ⊥⊥ 11000… Degree 2

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Dynamical Subbase of I 2 (2) X X 1 =S(1) X 0 =S(0) flip f(B) B g1g1 g0g0 f

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Dynamical Subbase of I 2 (2) X X 1 =S(1) X 0 =S(0) flip f(B) B g1g1 g0g0 f Peano Subbase : ⊥ 1 ⊥⊥ 1000… Degree 3

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How many other dynamical subbases of I 2 ? B is a line segment, whose endpoints are on the boundary of I 2. f(B) is a line segment contained in the boundary of I 2.

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How many other dynamical subbases of I 2 ? 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. a0 b1, a1 b0 Gray x Gray subbase is a typical example.

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Characterization of Peano-like subbases. 1 0 2 B f(B)

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0 1, 1 2, 2 3 by f -1 1 0 2 B f(B) 1 0 2 B 3 3 1 0 2 B 3 3 1 0 2 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=121120122c=12 1 2 1 2 3

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1 0 2 B 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 3 3 12101211 1 0 2 B f(B) 3 3 1212 1 2 3 4 4 4 4 1 2 3 4

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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 0 2 B f(B) 3 3 c=1210 1 2 3 4

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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

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5 5 1 0 2 B f(B) 3 3 1212121 4 4 4 5 5 1 0 2 B f(B) 1 0 2 B 3 3 121 1 0 2 B f(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 2 1 5 3 5 0 4 2 2 1 12

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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 4 5 5 1535042

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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 4 5 5 1535042

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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*2n). Not a subbase. 1 0 2n B f(B) n 2
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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 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

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Characterization of Peano-like dynamical subbases of I 2 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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1’ 1 B f(B) 0’ 0 {0,1,2,U} sequence for both 0 and 0’. Symbolic manipulation. 1’ 1 B f(B) 0’ 0202 2 2’ 1’ 1 B 0’ 0202 2 2’ 3 3 3 3’ 1’30’201 1’0’201 {2}x{2}{2,1}x{2,2} {2,1,U}x{2,2,U}

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Characterization of GrayxGray-like subbases. 1’ 1 B f(B) 0’ 0

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Similar result around 0. – If c(n)=2 and c(i) != 2 (n*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 I 2 ?

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