Dynamical Subbases of I and I2

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)

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

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

Dynamical Subbase of I2 (1) flip g0 X0=S(0) f B X1=S(1) g1 X f(B)

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)

Dynamical Subbase of I2 (2) flip g0 X0=S(0) f B X1=S(1) g1 f(B) X

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

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. Gray x Gray subbase is a typical example.

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

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 1 2 3 1 2 3 1 2 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.

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

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 1 2 3 4

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 1 2 3 4 1 2 3 4 5 6 1 2 3 4 5 U means no choice at that level. Only U appears after this. It is not a subbase.

1 3 0 4 2 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 0 4 2 4 3 1 5 1 5 3 5 4 2 5 1 5 3 5 0 4 2 2 1 12 102 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.

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 1535042

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 1535042

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.

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

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 1 2 3 4 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.

{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.

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

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.

Conclusion Study computation-related topological property. How many recursive structures we can consider in I2 ?