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
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
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.
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.
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}
• 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.