Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta with: (why the case n = 2 is )

Dynamics of the family of complex maps

Paul BlanchardToni GarijoMatt HolzerU. HoomiforgotDan LookSebastian Marotta

with:€

Fλ (z) = zn +λ

zn, n ≥ 2

(why the case n = 2 is )

Mark MorabitoMonica Moreno RochaKevin PilgrimElizabeth RussellYakov ShapiroDavid Uminsky

CR AZ Y

€

Fλ (z) = z n +λ

z n

The case n > 2 is great because:

€

Fλ (z) = z n +λ

z n


There exists a McMullen domain around = 0 ....

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.



Parameter plane for n=3

€

λ

€

Fλ (z) = z n +λ

z n


... surrounded by infinitely many “Mandelpinski” necklaces...






€

λ

€

Fλ (z) = z n +λ

z n


... surrounded by infinitely many “Mandelpinski” necklaces...

... and the Julia sets behave nicely as

€

λ → 0





€

λ =.01



€

λ =.0001

€

λ =.000001


€

λ

€

Fλ (z) = z n +λ

z n

There is no McMullen domain....





The case n = 2 is crazy because:

€

Fλ (z) = z n +λ

z n


... and no “Mandelpinski” necklaces...






€

Fλ (z) = z n +λ

z n


... and the Julia sets “go crazy” as

€

λ → 0

... and no “Mandelpinski” necklaces...







€

λ =−.000001

€

λ =−.0001

€

λ =−.01


Some definitions:

Julia set of

J = boundary of {orbits that escape to }

€

∞

= closure {repelling periodic orbits}

= {chaotic set}

Fatou set

= complement of J

= predictable set

€

Fλ (z) = zn +λ

zn

Computation of J:



Color points that escape toinfinity shades of red orange yellow green blue violet Black points do not escape.J = boundary of the black region.

€

λ =.08i

€

Fλ (z ) = z 3 +λ

z 3

Easy computations:

€

λ =.08i

is superattracting, so have immediate basin Bmapped n-to-1 to itself.

€

∞€

Fλ (z ) = z 3 +λ

z 3



B

Easy computations:


€

∞€

Fλ (z ) = z 3 +λ

z 3



B

T

€

λ =.08i

0 is a pole, so havetrap door T mapped

n-to-1 to B.

Easy computations:


€

∞€

Fλ (z ) = z 3 +λ

z 3

0 is a pole, so havetrap door T mapped

n-to-1 to B.

The Julia set has 2n-fold symmetry.



B

T

€

λ =.08i

Easy computations:

€

Fλ (z ) = z 3 +λ

z 3

2n free critical points

€

cλ = λ1/2n



€

λ =.08i

Easy computations:


€

cλ = λ1/2n

€

Fλ (z ) = z 3 +λ

z 3



€

λ =.08i

Easy computations:


€

cλ = λ1/2n

Only 2 critical values

€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3



€

λ =.08i

Easy computations:


€

cλ = λ1/2n


€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i



Easy computations:


€

cλ = λ1/2n


€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i



Easy computations:


€

cλ = λ1/2n


€

vλ = ±2 λ

But really only 1 freecritical orbit

€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i



Easy computations:


€

pλ = (−λ )1/2n


€

vλ = ±2 λ

But really only 1 freecritical orbit

€

Fλ (z ) = z 3 +λ

z 3



€

λ =.08iAnd 2n prepoles

€

cλ = λ1/2n

The Escape Trichotomy

There are three possible ways that thecritical orbits can escape to infinity,

and each yields a different typeof Julia set.

(with D. Look & D Uminsky)


€

vλ∈

€

J ( Fλ

)

€

⇒B is a Cantor set



€

vλ∈

€

J ( Fλ

)

€

⇒B

€

⇒

is a Cantor set

T

€

vλ∈ is a Cantor set of

simple closed curves

€

J ( Fλ

)

(McMullen)


(n > 2)


€

vλ∈

€

J ( Fλ

)

€

⇒B

€

⇒€

⇒

is a Cantor set

T

€



€

J ( Fλ

)

€

Fλ

k(v

λ) ∈ T

€

J ( Fλ

) is a Sierpinski curve

(McMullen)


(n > 2)

In all other cases is a connected set, and if

€

J ( Fλ

)

€

vλ∈

€

J ( Fλ

)

€

⇒BCase 1:



parameter planewhen n = 3

is a Cantor set

€

vλ∈

€

J ( Fλ

)

€

⇒BCase 1:




€

λ

is a Cantor set

€

vλ∈

€

J ( Fλ

)

€

⇒BCase 1:






J is a Cantor set

€

λ

is a Cantor set

€

vλ∈

€

J ( Fλ

)

€

⇒BCase 1:






J is a Cantor set

€

λ

€

cλ

€

vλ

is a Cantor set




Case 2: T

€

vλ∈

€

⇒

€

J ( Fλ

) is a Cantor set ofsimple closed curves




€

λ

Case 2: T

€

vλ∈

€

⇒

€

J ( Fλ

) is a Cantor set ofsimple closed curves



The central disk isthe McMullen domain

€

λ

Case 2: T

€



€

⇒

€

J ( Fλ

)




€

λ

Case 2: T

€



J is a Cantor set of simple closed curves



€

⇒

€

J ( Fλ

)

T

B




€

λ

Case 2: T

€



J is a Cantor set of simple closed curves



€

⇒

€

J ( Fλ

)

€

cλ

€

vλ




Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ





Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)



A Sierpinski curve is a planarset homeomorphic to theSierpinski carpet fractal

is a Sierpinski curve

Sierpinski curves are important for two reasons:

1. There is a “topological characterization” of the carpet

2. A Sierpinski curve is a “universal plane continuum”



The Sierpinski Carpet

Topological Characterization

Any planar set that is:1. compact2. connected3. locally connected4. nowhere dense5. any two complementary domains are bounded by simple closed curves that are pairwise disjoint is a Sierpinski curve.

Universal Plane Continuum

Any planar, one-dimensional, compact, connected set can be homeomorphically embedded in a Sierpinski curve.

For example....



This set



This set



can be embedded inside




Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ




€

λ

Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ


A Sierpinski “hole”



€

λ

Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)

A Sierpinski curve




A Sierpinski “hole”



€

λ

Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)



€

cλ

€

vλ

€

Fλ (vλ )

A Sierpinski curve


A Sierpinski “hole”Escape time 3



€

λ

Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)



A Sierpinski curve


Another Sierpinski “hole”



€

λ

Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)



A Sierpinski curve


Another Sierpinski “hole”

€

cλ

€

vλ

€

Fλ2(vλ )

Escape time 4



Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)

€

λ



A Sierpinski curve


Another Sierpinski “hole”Escape time 7



Case 3:

€

⇒

€

Fλk(v λ ) ∈ T

€

J ( Fλ

)

€

λ



A Sierpinski curve


Another Sierpinski “hole”Escape time 5



So to show that is homeomorphic to





Need to show:

compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s



Need to show:


Fatou set is the union of the preimages of B; all disjoint, open disks.



Need to show:


Fatou set is the union of the preimages of B; all disjoint, open disks.



Need to show:


If J contains an open set, then J = C.



Need to show:


If J contains an open set, then J = C.



Need to show:


No recurrent critical orbits and no parabolic points.



Need to show:


No recurrent critical orbits and no parabolic points.



Need to show:


J locally connected, so theboundaries are locally connected. Need to show they are s.c.c.’s. Can only meet at (preimages of) critical points, hence disjoint.



Need to show:


So J is a Sierpinski curve.









Remark: All Julia sets drawn from Sierpinski holes are homeomorphic, but only those in symmetrically locatedSierpinski holes have the same dynamics.

The maps on all these Julia sets are dynamically different.









Remark: All Julia sets drawn from Sierpinski holes are homeomorphic, but only those in symmetrically locatedSierpinski holes have the same dynamics.

In fact, there are exactly (n-1)(2n)k-3 Sierpinski holes withescape time k, and (2n)k-3 different conjugacy classes (n odd).

The maps on all these Julia sets are dynamically different.

(with K.Pilgrim)

1. The McMullen domain

2. Mandelpinski necklaces

3. Julia sets near 0

Topics

Part 1: The McMullen Domain

Why is there no McMullen domain when n = 2?

What is the preimage of T?

First suppose

€

vλ∈ T:


First suppose


€

vλ∈ T:

Can the preimage be 2n disjoint disks, each of which contains a critical point?




First suppose


€

vλ∈ T:

Can the preimage be 2n disjoint disks, each of which contains a critical point?



No --- there would then be 4n preimages of any point in T, but the map has degree 2n.


So some of the preimages of Tmust overlap, and by 2n-foldsymmetry, all must intersect.


So some of the preimages of Tmust overlap, and by 2n-foldsymmetry, all must intersect.

By Riemann-Hurwitz, the preimage of T must then

be an annulus.



€

cλ

€

vλ


So here is the picture:

T

B

A

A is the annulus separating B and T


So here is the picture:B

TA


XF maps X 2n-to-1

onto T


So here is the picture:B

TA


X

A

A

0

1

F maps both A0

1and A as an n-to-1

covering onto A

F maps X 2n-to-1onto T


B

TA

X

A

A

0

1

So mod(A0) = 1/n mod(A)

And mod(A1) = 1/n mod(A)


B

TA

X

A

A

0

1



When n = 2,mod(A0) + mod(A1) =

mod(A)


B

TA

X

A

A

0

1



When n = 2,mod(A0) + mod(A1) =

mod(A)

So there is no room for X, i.e., does

not lie in T

€

vλ

€

vλ = ±2 λ

€

Fλ (vλ ) = 2nλn /2 +λ

2nλn /2 = 2nλn /2 +1

2nλ (n /2−1)


Here is another reason:

€

Fλ (vλ ) = 2nλn /2 +λ

2nλn /2 = 2nλn /2 +1

2nλ (n /2−1)



€

n > 2⇒ Fλ (vλ ) → ∞ as λ → 0

€

vλso lies in T when n > 2

€

vλ = ±2 λ



€

n > 2⇒ Fλ (vλ ) → ∞ as λ → 0

€

n = 2⇒ Fλ (vλ ) = 4λ +1

4€


€

vλ = ±2 λ

€

Fλ (vλ ) = 2nλn /2 +λ

2nλn /2 = 2nλn /2 +1

2nλ (n /2−1)

€

Fλ (vλ ) = 2nλn /2 +λ

2nλn /2 = 2nλn /2 +1

2nλ (n /2−1)



€

n > 2⇒ Fλ (vλ ) → ∞ as λ → 0

€

so Fλ (vλ ) →1/ 4 as λ → 0(???)€


€

vλ = ±2 λ

€

n = 2⇒ Fλ (vλ ) = 4λ +1

4

Part 2: Mandelpinski Necklaces

Parameter plane for n = 3



A “Mandelpinski necklace” is a simple closed curve in the parameter plane that passes alternately through k centersof baby Mandelbrot sets and k centers of Sierpinski holes.





C1 passes through thecenters of 2 M-sets

and 2 S-holes








and 4 S-holesQuickTime™ and a

TIFF (LZW) decompressorare needed to see this picture.

*

* only exception:2 centers of period 2 bulbs, not M-sets





and 10 S-holes






and 28 S-holes




and 82 S-holesQuickTime™ and a



Theorem: There exist closed curves Cj, surrounding the McMullen domain. Each Cj passes

alternately through (n-2)nj-1 +1 centers of baby Mandelbrot sets and centers of Sierpinski holes.€

j = 1,...,∞

C14 passes through thecenters of 4,782,969 M-sets and S-holesQuickTime™ and a





j = 1,...,∞




C1: 3 holes and M-sets



j = 1,...,∞


C2: 9 holes and M-setsC3: 33 holes and M-setsQuickTime™ and a




j = 1,...,∞

*



Easy computations:

€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i

All of the criticalpoints and prepoleslie on the “criticalcircle” : |z| = | |

€

λ 1/2n

€

γ0

€

γ0



€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i

All of the criticalpoints and prepoleslie on the “criticalcircle” : |z| = | |

€

λ 1/2n

€

γ0

€

γ0

which is mapped 2n-to-1onto the “critical value line”

connecting

€

±vλ €

vλ

€

−vλ

Easy computations:



€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i

€

γ0

€

vλ

Any other circle around 0is mapped n-to-1 to an ellipse

whose foci are

€

±2 λ

Easy computations:

€

−vλ



€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i

€

γ0

€

vλ

€

±2 λ

Easy computations:

€

−vλ


whose foci are



€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i

€

γ0

€

vλ

€

±2 λ

So the exterior of is mapped as an n-to-1 covering of the

exterior of the critical value line. €

γ0

Easy computations:

€

−vλ


whose foci are



€

Fλ (z ) = z 3 +λ

z 3

€

λ =.08i

€

γ0

€

vλ

€

±2 λ

So the exterior of is mapped as an n-to-1 covering of the

exterior of the critical value line. Same with the interior of . €

γ0

€

γ0

Easy computations:

€

−vλ


whose foci are

Simplest case: C1

Assume that both sit on the critical circle.

€

±vλ

€

vλ

€

−vλ

€

γ0

€

pλ

(i.e., )

€

| vλ | = | cλ | = | pλ |

€

⇒ 2 | λ |1/2 = | λ |1/2n


€

±vλ

€

vλ

€

−vλ

€

γ0

€

pλ

Simplest case: C1

€

⇒ 2 | λ |1/2 = | λ |1/2n

€

⇒ 22n | λ |n = | λ |

€

⇒ | λ | = 2−

2n

n−1 ⎛ ⎝

⎞ ⎠


€

±vλ

This is the “dividing circle” in the parameter plane

Parameter plane n = 3



r = 2-3

Simplest case: C1

€

⇒ 2 | λ |1/2 = | λ |1/2n

€

⇒ 22n | λ |n = | λ |

€

⇒ | λ | = 2−

2n

n−1 ⎛ ⎝

⎞ ⎠




€

±vλ

€

⇒ 2 | λ |1/2 = | λ |1/2n

€

⇒ 22n | λ |n = | λ |

This is the “dividing circle” in the parameter plane

Parameter plane n = 4

€

⇒ | λ | = 2−

2n

n−1 ⎛ ⎝

⎞ ⎠

r = 2-8/3

Simplest case: C1

Assume that lies on the dividing circle,so both sit on the critical circle.

€

±vλ

€

vλ

€

−vλ

€

γ0

€

λ

€

pλ

In this picture, is real and n = 3

€

λ

Simplest case: C1


€

±vλ

€

vλ

€

−vλ

€

γ0

As rotates around the dividing circle, rotates a half-turn, while and rotate1/2n of a turn. So each meets exactlyn - 1 prepoles and critical points.€

λ

€

±vλ

€

pλ

€

vλ

€

λ

€

pλ€

cλ


€

λ

Simplest case: C1


€

±vλ

€

vλ

€

−vλ

€

γ0

As rotates around the dividing circle, rotates a half-turn, while and rotate1/2n of a turn. So each meets exactlyn - 1 prepoles and critical points.€

λ

€

±vλ

€

pλ

€

vλ

€

λ

€

pλ€

cλ


€

λ

Simplest case: C1

So the dividing circle is C1



The dividing circle when n = 5

n-1 = 4 centers of Sierpinski holes;n-1 = 4 centers of baby Mandelbrot sets

Now assume that lies inside the critical circle:

€

±vλ

€

γ0€

vλ

€

−vλ

€

γ0


€

±vλ

The exterior of is mapped n-to-1onto the exterior of the critical value ray, so there is a preimage mapped n-to-1 to , €

γ0

€

γ0

€

γ1

€

vλ

€

−vλ


€

±vλ

€

γ0€

vλ

€

−vλ

€

γ1


γ0

€

γ0

€

γ1

then is mapped n-to-1 to ,

€

γ1


€

±vλ

€

γ2

€

γ0€

vλ

€

−vλ

€

γ1

€

γ2


γ0

€

γ0

€

γ1

and on and onout to

€

∂B


€

±vλ

then is mapped n-to-1 to ,

€

γ1

€

γ0€

vλ

€

−vλ

€

γ1

€

γ2

B


γ0

€

γ0

€

γ1

€

γ2

€

γ0 contains 2n critical points and 2n prepoles, so

€

γ1 contains 2n2 pre-critical points and pre-prepoles

€

γ0€

vλ

€

−vλ

€

γ1

€

γ0 contains 2n critical points and 2n prepoles, so

€

γ1 contains 2n2 pre-critical points and pre-prepoles

€

γ0€

vλ

€

−vλ

€

γ1

€

γ2

B

€

γk contains 2nk+1 points thatmap to the critical pointsand pre-prepoles under

€

Fλk

€

γ0€

vλ

€

−vλ

€

γk

As rotates by one turn, these 2nk+1 points on each rotate by 1/2nk+1 of a turn.

€

γk

€

λ

€

γ0€

vλ

€

−vλ

€

γk

€

Fλ (vλ ) ≈λ

2nλn /2 =1

2nλ1−n /2

Since

the second iterate of the criticalpoints rotate by 1 - n/2 ofa turn


€

γk

€

λ

€

γ0€

vλ

€

−vλ

€

γk

€

Fλ (vλ ) ≈λ

2nλn /2 =1

2nλ1−n /2

Since

the second iterate of the criticalpoints rotate by 1 - n/2 ofa turn, so this point hitsexactly

€

(n / 2 −1)(2nk+1) +1 = (n − 2)nk+1 +1

preimages of the critical pointsand prepoles on

€

γk


€

γk

€

λ

There is a natural parametrization of each

€

γk

€

γkλ (θ )

The real proof involves the Schwarz Lemma:

€

γ0€

vλ

€

−vλ

€

γk



€

γkλ (θ )



There is a natural parametrization of each

€

γk

€

γkλ (θ )

The real proof involves the Schwarz Lemma:

€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

Best to restrict to a “symmetry region” inside the dividingcircle, so that is well-defined.

€

γkλ (θ )



€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

Then we have a second map from the parameter plane to thedynamical plane, namely which is invertible on the symmetry sector

€

G(λ ) = Fλ (vλ )

€

G−1

Best to restrict to a “symmetry region” inside the dividingcircle, so that is well-defined.

€

γkλ (θ )



€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

Then we have a second map from the parameter plane to thedynamical plane, namely which is invertible on the symmetry sector

€

G(λ ) = Fλ (vλ )

€

G−1

€

G−1(γ kλ (θ ) )

a map from a “disk” to itself.

So consider the composition



€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

€

G−1

€

G−1(γ kλ (θ ) )

a map from a “disk” to itself.

So consider the composition

Schwarz implies that has a unique fixed point,i.e., a parameter for which the second iterate of the criticalpoint lands on the point , so this proves theexistence of the centers of the S-holes and M-sets.€

G−1(γ kλ (θ ) )

€

γkλ (θ )

Remarks: This proves the existence of centers of Sierpinski holes and Mandelbrot sets. Producingthe entire M-set involves polynomial-like maps;while the entire S-hole involves qc-surgery.

Part 3: Behavior of the Julia sets

n = 2: When , the Julia set of is the unit circle. But, as , the Julia set of converges to the closed unit disk

€

Fλ

€

λ → 0

€

λ =0

€

Fλ


n > 2: J is always a Cantor set of “circles” when is small.


€

Fλ

€

λ → 0

€

λ =0

€

Fλ

€

λ


n > 2: J is always a Cantor set of “circles” when is small.


€

Fλ

€

λ → 0

€

λ =0

€

Fλ

Moreover, there is a such that there is always a“round” annulus of “thickness” between two of these circles in the Fatou set. So J does not converge to the unit disk when n > 2.

€

λ

€

δ > 0

€

δ


n = 2

Theorem: When n = 2, the Julia sets converge to the unit disk as



€

λ → 0

Suppose the Julia sets do not converge to the unit disk D as

€

λ → 0

Sketch of the proof:


€

λ → 0


Then there exists and a sequence such that, for each i,there is a point such that lies in the Fatou set.

€

Bδ (zi )

€

δ > 0

€

λi → 0

€

zi ∈D


€

λ → 0



€

Bδ (zi )

€

λi → 0

€

zi ∈D

€

Bδ (z1)

€

Bδ (z2 )

€

Bδ (z3)

€

Bδ (z4 )

€

δ > 0


€

λ → 0



€

Bδ (zi )

€

λi → 0

€

zi ∈D

The must accumulate on some nonzero point, say ,so we may assume that lies in the Fatou set for all i.

€

zi

€

z*

€

Bδ (z*)

€

Bδ (z1)

€

Bδ (z2 )

€

Bδ (z3)

€

Bδ (z4 )

€

δ > 0


€

λ → 0



€

Bδ (zi )

€

λi → 0

€

zi ∈D

€

Bδ (z*)


€

zi

€

z*

€

Bδ (z*)

€

δ > 0


€

λ → 0



€

Bδ (zi )

€

λi → 0

€

zi ∈D

€

Bδ (z*)


€

zi

€

z*

€

Bδ (z*)

But for large i, so stretchesinto an “annulus” that surrounds the origin, so thisdisconnects the Julia set.€

Fλ i ≈ z2

€

Fλ ik

€

Bδ (z*)€

Fλk

€

δ > 0

So the Fatou components must become arbitrarily small:





€

λ =−0.0001

€

λ =−0.0000001





€

λ =.01



€

λ =.0001

€

λ =.000001

n > 2: Note the “round” annuli in the Fatou set; there is alwayssuch an annulus of some fixed width for small.





€

λ =.00000001

€

λ =.0000000001

€

| λ |



€

λ =.000000000001

T

B

€

| λ | small

€

⇒ T is tiny

A0

mod A0 = m is huge

Say n = 3:

T

B

€

| λ | small

€

⇒ T is tiny

A0

mod A0 = m is hugeand the boundary of A0

is very close to S1

Say n = 3:

T

B

€

| λ | small

€

⇒ T is tinymod A0 = m is hugeand the boundary of A0

is very close to S1

Say n = 3:

X1

A1

A1 and A1 mapped to A0;X1 is mapped to T

X1

A1

A1

~~

A0

mod A1 = mod A1 = mod X1 = m/3; A1 S1

€

⇒

€

≈

€

∂~

T

B

€

| λ | small

€


is very close to S1

Say n = 3:

X1

A1


X2

A1

A2

A2 is mapped to A1;X2 is mapped to X1

mod A2 = mod X2 = m/32; A2 S1

€

⇒mod A1 = mod A1 = mod X1 = m/3; A1 S1

€

⇒

€

≈

€

∂

€

≈

€

∂

~

~

B

€

| λ | small

€


is very close to S1

Say n = 3:


Xk

Ak

Ak-1

€

M

Ak is mapped to Ak-1; Xk to Xk-1

mod A2 = mod X2 = m/32; A2 S1

€

⇒mod A1 = mod A1 = mod X1 = m/3; A1 S1

~

€

⇒

€

≈

€

∂

€

≈

€

∂mod Ak = mod Xk = m/3k; Ak S1

€

⇒

€

≈

€

∂

A2 is mapped to A1;X2 is mapped to X1

~

€

N

B

€

| λ | small

€


is very close to S1

Say n = 3:

Xk

Ak

Ak-1

1 mod Ak < 3

€

≤Eventually find k sothatand Ak S1

€

≈

€

∂

€

=α

B

€

| λ | small

€


is very close to S1

Say n = 3:

Xk

Ak

Ak-1

1 mod Ak < 3

€


€

≈

€

∂

€

⇒ Ak must contain a round annulus of modulus

(Ble, Douady, and Henriksen)

€

=α

€

α −1/2 >1/2

B

€

| λ | small

€


is very close to S1

Say n = 3:

Xk

Ak

Ak-1

€

⇒ Ak must contain a round annulus of modulus

(Ble, Douady, and Henriksen)

But does this annulus have definite “thickness?”

1 mod Ak < 3

€


€

≈

€

∂

€

=α

€

α −1/2 >1/2



€

∂Ak in here



€

∂Ak in here

€

=αmod Ak

says that the innerboundary of Ak

cannot be insideor outside ,so the round annulusin Ak is “thick”€

γ0

€

γ1

€

γ1

€

γ0

Ak



€

∂Ak in here

€

μ1

€

μ0

Ak

Same argumentsays that Ak Xk

is twice as thick

€

∪

€

∪ Xk



Xk

So Xk musthave definite thickness as well

Part 4: A major application

Here’s the parameter plane when n = 2:



Qu

ickTim

e™

an

d a

TIF

F (

LZ

W)

decom

pre

sso

rare

nee

de

d t

o s

ee t

his

pic

ture

.

Rotate it by 90 degrees:

and this object appears everywhere.....



Documents

Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta with: (why the case n = 2 is )