Dynamical systems Expanding maps on the circle. Classification Jana - - PowerPoint PPT Presentation

coding classification

Dynamical systems

Expanding maps on the circle. Classification Jana Rodriguez Hertz

ICTP

2018

coding classification semiconjugacy

Index

1

coding semiconjugacy points of non-injectivity

2

classification theorem classification proof

coding classification semiconjugacy

expanding maps are factors of σ

theorem

coding classification semiconjugacy

expanding maps are factors of σ

theorem

f : S1 → S1 expanding map

coding classification semiconjugacy

expanding maps are factors of σ

theorem

f : S1 → S1 expanding map deg(f) = 2

coding classification semiconjugacy

expanding maps are factors of σ

theorem

f : S1 → S1 expanding map deg(f) = 2 ⇒ f is a factor of σ on Σ+

2

coding classification semiconjugacy

expanding maps are factors of σ

theorem

f : S1 → S1 expanding map deg(f) = 2 ⇒ f is a factor of σ on Σ+

2

∃ h : Σ+

2 → S1 such that f n(h(x)) ∈ ∆xn for all n ≥ 0

coding classification points of non-injectivity

Index

1

coding semiconjugacy points of non-injectivity

2

classification theorem classification proof

coding classification points of non-injectivity

points of non-injectivity

points of non-injectivity

coding classification points of non-injectivity

points of non-injectivity

points of non-injectivity

if h(x) = h(y) = x

coding classification points of non-injectivity

points of non-injectivity

points of non-injectivity

if h(x) = h(y) = x then there exists n ≥ 0

coding classification points of non-injectivity

points of non-injectivity

points of non-injectivity

if h(x) = h(y) = x then there exists n ≥ 0 such that f n(x) = p

coding classification points of non-injectivity

points of non-injectivity

points of non-injectivity

if h(x) = h(y) = x then there exists n ≥ 0 such that f n(x) = p where f(p) = p

coding classification points of non-injectivity

proof

comments on the proof - points of non-injectivity

coding classification points of non-injectivity

proof

comments on the proof - points of non-injectivity

f is injective on ∆o

i

coding classification points of non-injectivity

proof

comments on the proof - points of non-injectivity

f is injective on ∆o

i

f(∂∆i) = p

coding classification points of non-injectivity

proof

comments on the proof - points of non-injectivity

f is injective on ∆o

i

f(∂∆i) = p if x ∈ ∆o

0 ∪ ∆o 1

coding classification points of non-injectivity

proof

comments on the proof - points of non-injectivity

f is injective on ∆o

i

f(∂∆i) = p if x ∈ ∆o

0 ∪ ∆o 1

then first symbol of x such that h(x) = x

coding classification points of non-injectivity

proof

comments on the proof - points of non-injectivity

f is injective on ∆o

i

f(∂∆i) = p if x ∈ ∆o

0 ∪ ∆o 1

then first symbol of x such that h(x) = x is 0 or 1 (no ambiguity)

coding classification points of non-injectivity

proof

comments on the proof - points of injectivity

∆0 ∆1 p q p p

coding classification points of non-injectivity

proof

comments on the proof - f 2

∆00 ∆01 ∆10 ∆11 p p p q p p

coding classification points of non-injectivity

proof

proof - points of injectivity

coding classification points of non-injectivity

proof

proof - points of injectivity

let h(x) = h(y) = x with x = y

coding classification points of non-injectivity

proof

proof - points of injectivity

let h(x) = h(y) = x with x = y let N be the first integer such that xN = yN

coding classification points of non-injectivity

proof

proof - points of injectivity

let h(x) = h(y) = x with x = y let N be the first integer such that xN = yN then h(x) = h(y) = x ∈

N−1

• n=0

f −n(∆xn)

coding classification points of non-injectivity

proof

proof - points of injectivity

let h(x) = h(y) = x with x = y let N be the first integer such that xN = yN then h(x) = h(y) = x ∈

N−1

• n=0

f −n(∆xn) which is an interval ∆x0...xN−1

coding classification points of non-injectivity

proof

proof - points of injectivity

coding classification points of non-injectivity

proof

proof - points of injectivity

now N is the first integer such that xN = yN

coding classification points of non-injectivity

proof

proof - points of injectivity

now N is the first integer such that xN = yN then h(x) = h(y) = x ∈ f −N(∆0) ∩ f −N(∆1)

coding classification points of non-injectivity

proof

proof - points of injectivity

now N is the first integer such that xN = yN then h(x) = h(y) = x ∈ f −N(∆0) ∩ f −N(∆1) ⇒ x ∈ f −N(∆0 ∩ ∆1)

coding classification points of non-injectivity

proof

proof - points of injectivity

now N is the first integer such that xN = yN then h(x) = h(y) = x ∈ f −N(∆0) ∩ f −N(∆1) ⇒ x ∈ f −N(∆0 ∩ ∆1) but ∆0 ∩ ∆1 = {p, q}

coding classification points of non-injectivity

proof

proof - points of injectivity

now N is the first integer such that xN = yN then h(x) = h(y) = x ∈ f −N(∆0) ∩ f −N(∆1) ⇒ x ∈ f −N(∆0 ∩ ∆1) but ∆0 ∩ ∆1 = {p, q} ⇒ f N+1(x) = p

coding classification points of non-injectivity

proof

proof - points of injectivity

now N is the first integer such that xN = yN then h(x) = h(y) = x ∈ f −N(∆0) ∩ f −N(∆1) ⇒ x ∈ f −N(∆0 ∩ ∆1) but ∆0 ∩ ∆1 = {p, q} ⇒ f N+1(x) = p

coding classification theorem

Index

1

coding semiconjugacy points of non-injectivity

2

classification theorem classification proof

coding classification theorem

classification

theorem

coding classification theorem

classification

theorem

let f, g : S1 → S1 be expanding maps

coding classification theorem

classification

theorem

let f, g : S1 → S1 be expanding maps such that deg(f) = deg(g) = 2

coding classification theorem

classification

theorem

let f, g : S1 → S1 be expanding maps such that deg(f) = deg(g) = 2 ⇒ f and g are topologically conjugate

coding classification classification

Index

1

coding semiconjugacy points of non-injectivity

2

classification theorem classification proof

coding classification classification

corollary

classification

coding classification classification

corollary

classification

f : S1 → S1 expanding map

coding classification classification

corollary

classification

f : S1 → S1 expanding map deg(f) = 2

coding classification classification

corollary

classification

f : S1 → S1 expanding map deg(f) = 2 ⇒ f topologically conjugate to E2

coding classification proof

Index

1

coding semiconjugacy points of non-injectivity

2

classification theorem classification proof

coding classification proof

proof

proof - theorem

coding classification proof

proof

proof - theorem

f, g : S1 → S1 expanding maps

coding classification proof

proof

proof - theorem

f, g : S1 → S1 expanding maps with deg(f) = deg(g) = 2

coding classification proof

proof

proof - theorem

f, g : S1 → S1 expanding maps with deg(f) = deg(g) = 2 ⇒ ∃ hf, hg : Σ+

2 → S1 semiconjugacies

coding classification proof

proof

proof - theorem

f, g : S1 → S1 expanding maps with deg(f) = deg(g) = 2 ⇒ ∃ hf, hg : Σ+

2 → S1 semiconjugacies

such that

coding classification proof

proof

proof - theorem

f, g : S1 → S1 expanding maps with deg(f) = deg(g) = 2 ⇒ ∃ hf, hg : Σ+

2 → S1 semiconjugacies

such that

1

f ◦ hf = hf ◦ σ

coding classification proof

proof

proof - theorem

f, g : S1 → S1 expanding maps with deg(f) = deg(g) = 2 ⇒ ∃ hf, hg : Σ+

2 → S1 semiconjugacies

such that

1

f ◦ hf = hf ◦ σ

2

g ◦ hg = hg ◦ σ

coding classification proof

proof

definition of h

coding classification proof

proof

definition of h

let us define h : S1 → S1 such that

coding classification proof

proof

definition of h

let us define h : S1 → S1 such that h(x) = hg(h−1

f

(x))

coding classification proof

proof

h is well-defined, case 1

coding classification proof

proof

h is well-defined, case 1

Case 1: h−1

f

(x) consists of a single point

coding classification proof

proof

h is well-defined, case 1

Case 1: h−1

f

(x) consists of a single point ⇒ h is well-defined

coding classification proof

proof

h is well-defined, case 1

Case 1: h−1

f

(x) consists of a single point ⇒ h is well-defined √

coding classification proof

proof

h is well-defined, case 2

coding classification proof

proof

h is well-defined, case 2

∃x = y such that hf(x) = hf(y) = x

coding classification proof

proof

h is well-defined, case 2

∃x = y such that hf(x) = hf(y) = x ⇒ ∃N such that f N(x) = pf, with f(pf) = pf

coding classification proof

h is well-defined, case 2

coding classification proof

h is well-defined, case 2

f N(x) = pf

coding classification proof

h is well-defined, case 2

f N(x) = pf hf(x) = x

coding classification proof

h is well-defined, case 2

f N(x) = pf hf(x) = x ⇐ ⇒ σN(x) = 0000000 . . . or σN(x) = 1111111 . . .

coding classification proof

h is well-defined, case 2

f N(x) = pf hf(x) = x ⇐ ⇒ σN(x) = 0000000 . . . or σN(x) = 1111111 . . .

• therwise

f n(x) = p ∈ ∆01 ∪ ∆10 for some n ≥ N

coding classification proof

h is well-defined, case 2

f N(x) = pf hf(x) = x ⇐ ⇒ σN(x) = 0000000 . . . or σN(x) = 1111111 . . .

• therwise

f n(x) = p ∈ ∆01 ∪ ∆10 for some n ≥ N→ contradiction

coding classification proof

proof

h is well defined, case 2

coding classification proof

proof

h is well defined, case 2

hf(x) = hf(y) = x with x = y

coding classification proof

proof

h is well defined, case 2

hf(x) = hf(y) = x with x = y N ≥ 0 the first such that f N(x) = p

coding classification proof

proof

h is well defined, case 2

hf(x) = hf(y) = x with x = y N ≥ 0 the first such that f N(x) = p ⇒ xn = 0 and yn = 1 for all n ≥ N

coding classification proof

proof

h is well defined, case 2

hf(x) = hf(y) = x with x = y N ≥ 0 the first such that f N(x) = p ⇒ xn = 0 and yn = 1 for all n ≥ N xN−1 = 1 and yN−1 = 0

coding classification proof

proof

h is well defined, case 2

hf(x) = hf(y) = x with x = y N ≥ 0 the first such that f N(x) = p ⇒ xn = 0 and yn = 1 for all n ≥ N xN−1 = 1 and yN−1 = 0 xn = yn for all n ≤ N − 2

coding classification proof

proof

h is well defined, case 2

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y)

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y) hg(x) ∈ ∆g

x0...xN−210000...

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y) hg(x) ∈ ∆g

x0...xN−210000...

hg(y) ∈ ∆g

x0...xN−2011111...

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y) hg(x) ∈ ∆g

x0...xN−210000...

hg(y) ∈ ∆g

x0...xN−2011111...

⇒ hg(x), hg(y) ∈ ∆x1...xN−2 = [aN−2, bN−2]

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y) hg(x) ∈ ∆g

x0...xN−210000...

hg(y) ∈ ∆g

x0...xN−2011111...

⇒ hg(x), hg(y) ∈ ∆x1...xN−2 = [aN−2, bN−2] gN−1 is injective in (aN−2, bN−2)

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y) hg(x) ∈ ∆g

x0...xN−210000...

hg(y) ∈ ∆g

x0...xN−2011111...

⇒ hg(x), hg(y) ∈ ∆x1...xN−2 = [aN−2, bN−2] gN−1 is injective in (aN−2, bN−2) ∃! r ∈ (aN−2, bN−2) such that gN−1(r) = qg

coding classification proof

proof

h is well defined, case 2

let us show hg(x) = hg(y) hg(x) ∈ ∆g

x0...xN−210000...

hg(y) ∈ ∆g

x0...xN−2011111...

⇒ hg(x), hg(y) ∈ ∆x1...xN−2 = [aN−2, bN−2] gN−1 is injective in (aN−2, bN−2) ∃! r ∈ (aN−2, bN−2) such that gN−1(r) = qg hg(x) ∈ [r, bN−2] and hg(y) ∈ [aN−2, r]

coding classification proof

proof

h is well defined, case 2

coding classification proof

proof

h is well defined, case 2

hg(x) ∈ [r, bN−2] ∩

n≥N g−N(∆0)

coding classification proof

proof

h is well defined, case 2

hg(x) ∈ [r, bN−2] ∩

n≥N g−N(∆0) = r

coding classification proof

proof

h is well defined, case 2

hg(x) ∈ [r, bN−2] ∩

n≥N g−N(∆0) = r

hg(y) ∈ [aN/2, r] ∩

n≥N g−N(∆1)

coding classification proof

proof

h is well defined, case 2

hg(x) ∈ [r, bN−2] ∩

n≥N g−N(∆0) = r

hg(y) ∈ [aN/2, r] ∩

n≥N g−N(∆1) = r

coding classification proof

proof

h is well defined, case 2

hg(x) ∈ [r, bN−2] ∩

n≥N g−N(∆0) = r

hg(y) ∈ [aN/2, r] ∩

n≥N g−N(∆1) = r

⇒ h is well-defined.

coding classification proof

proof

h is continuous

coding classification proof

proof

h is continuous

let x be such that f n(x) = pf for all n ≥ 0

coding classification proof

proof

h is continuous

let x be such that f n(x) = pf for all n ≥ 0 take N > 0 such that d(x, y) <

1 3N ⇒ d(hg(x), hg(y)) < ε

coding classification proof

proof

h is continuous

let x be such that f n(x) = pf for all n ≥ 0 take N > 0 such that d(x, y) <

1 3N ⇒ d(hg(x), hg(y)) < ε

x = ∞

n=0 f −n(∆xn) is in the interior of N n=0 f −n(∆xn)

coding classification proof

proof

h is continuous

let x be such that f n(x) = pf for all n ≥ 0 take N > 0 such that d(x, y) <

1 3N ⇒ d(hg(x), hg(y)) < ε

x = ∞

n=0 f −n(∆xn) is in the interior of N n=0 f −n(∆xn)

⇒ there is δ > 0 such that d(x, y) < δ ⇒ d(h(x), h(y)) < ε

coding classification proof

proof

h is continuous

coding classification proof

proof

h is continuous

let x be such that f K(x) = pf for some K > 0

coding classification proof

proof

h is continuous

let x be such that f K(x) = pf for some K > 0 ⇒ h−1

f

(x) = {x, y} such that

coding classification proof

proof

h is continuous

let x be such that f K(x) = pf for some K > 0 ⇒ h−1

f

(x) = {x, y} such that x = x0 . . . xK−2011111 . . . and y = x0 . . . xK−2100000 . . .

coding classification proof

proof

h is continuous

let x be such that f K(x) = pf for some K > 0 ⇒ h−1

f

(x) = {x, y} such that x = x0 . . . xK−2011111 . . . and y = x0 . . . xK−2100000 . . . take ε > 0 and N > 0 and take y > x

coding classification proof

proof

h is continuous

let x be such that f K(x) = pf for some K > 0 ⇒ h−1

f

(x) = {x, y} such that x = x0 . . . xK−2011111 . . . and y = x0 . . . xK−2100000 . . . take ε > 0 and N > 0 and take y > x if y ∈ ∆x0...xK−21000 (with N subsymbols)

coding classification proof

proof

h is continuous

let x be such that f K(x) = pf for some K > 0 ⇒ h−1

f

(x) = {x, y} such that x = x0 . . . xK−2011111 . . . and y = x0 . . . xK−2100000 . . . take ε > 0 and N > 0 and take y > x if y ∈ ∆x0...xK−21000 (with N subsymbols) then d(h(x), h(y)) < ε

coding classification proof

proof

h is continuous

coding classification proof

proof

h is continuous

analogously, we take y < x

coding classification proof

proof

h is continuous

analogously, we take y < x if y ∈ ∆x0...xK−2011111 (with N subsymbols)

coding classification proof

proof

h is continuous

analogously, we take y < x if y ∈ ∆x0...xK−2011111 (with N subsymbols) then d(h−1

f

(y), y) <

1 3N and then

coding classification proof

proof

h is continuous

analogously, we take y < x if y ∈ ∆x0...xK−2011111 (with N subsymbols) then d(h−1

f

(y), y) <

1 3N and then

d(h(y), h(x)) = d(hg(h−1

f

(y)), hg(y)) < ε

coding classification proof

proof

conclusion

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous.

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous. and h ◦ f =

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous. and h ◦ f = hg ◦ h−1

f

• f =

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous. and h ◦ f = hg ◦ h−1

f

• f = hg ◦ σ ◦ h−1

f

=

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous. and h ◦ f = hg ◦ h−1

f

• f = hg ◦ σ ◦ h−1

f

= g ◦ hg ◦ h−1

f

=

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous. and h ◦ f = hg ◦ h−1

f

• f = hg ◦ σ ◦ h−1

f

= g ◦ hg ◦ h−1

f

= g ◦ h

coding classification proof

proof

conclusion

it is easy to see that h−1 is well defined and continuous. and h ◦ f = hg ◦ h−1

f

• f = hg ◦ σ ◦ h−1

f

= g ◦ hg ◦ h−1

f

= g ◦ h