1. Dynamical Subbases of I and I2

2. Itinerary of the tent function1 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)

3. 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

4. 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.

5. 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.

6. 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)

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

8. 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)

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

10. 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

11. 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.

12. How many other dynamical subbases of I2 ?
two cases.
two cases.
two cases.
two cases.
two cases.

13. The diameter of some component S(σ) does not decrease.
It does not form a subbase, for any
arrangement of c0,…

14. Only two cases a0  b1, a1  b0 Peano Subbase is a typical example.
Gray x Gray subbase is a typical example.

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

16. 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.

17. 1212 c=121 1210 1211 Three cases for the 4-th code. 1 2 B f(B) 3 1 2 B2 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.

18. 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

19. 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.

20. 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.

21. 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

22. 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

23. 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.

24. 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

25. 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

26. 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.

27. {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.

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

29. 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.

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