dynamic classification of escape time sierpinski curve julia sets dynamics of the family of complex...

Dynamic Classification of Escape TimeSierpinski Curve Julia Sets

Dynamics of the family of complex maps

Paul BlanchardToni GarijoMatt HolzerU. HoomiforgotDan LookSebastian Marotta

with:

€

Fλ (z) = zn +λ

zn

Mark MorabitoMonica Moreno RochaKevin PilgrimElizabeth RussellYakov ShapiroDavid Uminsky

, n > 1

A Sierpinski curve is any planar set that is homeomorphic to the Sierpinski carpet fractal.

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

The Sierpinski Carpet

Sierpinski Curve



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.

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

More importantly....

A Sierpinski curve is a universal plane continuum:

For example....



This set



This set



can be embedded inside



This set



can be embedded inside

Moreover, Sierpinski curves occur all the time as Julia sets.

Dynamics of

complex and

€

Fλ (z) = z n +λ

z n

€

λ,z

€

n ≥ 2

A rational map of degree 2n.

Also a “singular perturbation” of zn.



€

λ = 0

€

Fλ

( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0



€

λ = 0

€

Fλ

( z ) = z2

+λ

z2

€

λ ≠ 0

€

λ = −1 / 16



But when , theJulia set explodes


€

λ = 0



€

λ = 0

€

Fλ

( z ) = z2

+λ

z2

€

λ ≠ 0

€

λ = −1 / 16



But when , theJulia set explodes

A Sierpinski curve


€

λ = 0



€

λ = 0

€

Fλ

( z ) = z2

+λ

z2

€

λ ≠ 0But when , theJulia set explodes

€

λ = −0 . 01

Another Sierpinski curve




€

λ = 0



€

λ = 0

€

Fλ

( z ) = z2

+λ

z2

€

λ ≠ 0But when , theJulia set explodes

€

λ = −0 . 2

Also a Sierpinski curve




€

λ = 0

Easy computations:



€

Fλ (z ) = z 3 +λ

z 3

€

λ=.036+.026i

2n free critical points

€

cλ = λ1/2n

Easy computations:



€

λ=.036+.026i


€

cλ = λ1/2n€

Fλ (z ) = z 3 +λ

z 3

Easy computations:



€

λ=.036+.026i


€

cλ = λ1/2n

Only 2 critical values

€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3

Easy computations:

€

λ=.036+.026i


€

cλ = λ1/2n

Only 2 critical values

€

vλ = ±2 λ

But really only 1 freecritical orbit since

the map is symmetricunder

€

Fλ (z ) = z 3 +λ

z 3



€

z → −z

Easy computations:



€

λ=.036+.026i

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

€

∞ B€

Fλ (z ) = z 3 +λ

z 3

Easy computations:



€

λ=.036+.026i


B

T

€

Fλ (z ) = z 3 +λ

z 3

€

∞

0 is a pole, so havetrap door T mapped

n-to-1 to B.

Easy computations:



€

λ=.036+.026i


B

T

€

Fλ (z ) = z 3 +λ

z 3

€

∞

So any orbit that eventuallyenters B must do so by

passing through T.

0 is a pole, so havetrap door T mapped

n-to-1 to B.

The Escape Trichotomy

There are three distinct ways the critical orbit can enter B:


€

vλ∈

€

J ( Fλ

)

€

⇒B is a Cantor set



€

vλ∈

€

J ( Fλ

)

€


T

€

vλ∈ is a Cantor set of

simple closed curves

€

J ( Fλ

)


(this case does not occur if n = 2)

€

⇒

(McMullen)


€

vλ∈

€

J ( Fλ

)

€


T

€

vλ∈ is a Cantor set of

simple closed curves

€

J ( Fλ

)

€

Fλ

k(v

λ) ∈ T

€

J ( Fλ

) is a Sierpinski curve


(this case does not occur if n = 2)

€

⇒

€

⇒

(McMullen)



€

vλ∈

€

J ( Fλ

)

€


parameter planewhen n = 3

Case 1:





€

vλ∈

€

J ( Fλ

)

€



J is a Cantor set

€

λ



€

vλ∈

€

J ( Fλ

)

€



J is a Cantor set



€

λ





€

vλ∈

€

J ( Fλ

)

€



J is a Cantor set

€

λ




Case 2: the critical values lie in T, not B



€

vλ∈

€

⇒T


€

λ lies in the McMullen domain

€

λ



€

vλ∈

€

⇒T


J is a Cantor set of simple closed curves

€


€

λQuickTime™ and a

TIFF (LZW) decompressorare needed to see this picture.

Remark: There is no McMullen domain in the case n = 2.



€

vλ∈

€

⇒T


J is a Cantor set of simple closed curves

€


€

λQuickTime™ and a

TIFF (LZW) decompressorare needed to see this picture.



€

⇒T


€

λ lies in a Sierpinski hole

€

Fλ

k(v

λ) ∈

Case 3: the critical orbit eventually lands in the trap door.



€

⇒T


J is an escape time Sierpinski curve

€


€

λ

€

Fλ

k(v

λ) ∈





€

⇒T


€


€

λ

€

Fλ

k(v

λ) ∈



J is an escape time Sierpinski curve



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:


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



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.

Have an exact count of the number of Sierpinski holes:

Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)


Reason: The equation reduces to a polynomial of degree (n-1)(2n)(k-3) ;we have a Bottcher coordinate on each Sierpinski hole;and so all the roots of this polynomial are distinct. So we have exactly that many “centers” of Sierpinski holes, i.e., parameters for which the critical points all land on 0 and then on .

€

Fλk−1(λ1/2n ) = 0

€

∞





n = 3escape time 32 Sierpinski holes

parameter planen = 3



n = 4escape time 12 402,653,184 Sierpinski holes



Sorry. I forgot to indicate their locations. parameter plane

n = 4


Given two Sierpinski curve Julia sets, when do we know that the dynamics on them are the same, i.e., the maps are conjugate on the Julia sets?

Main Question:





These sets are homeomorphic, but are the dynamics on them the same?

#1: If and are drawn from the same Sierpinski hole, then the correspondingmaps have the same dynamics, i.e., they are topologically conjugate on their Julia sets.€

λ

€

μ




#1: If and are drawn from the same Sierpinski hole, then the correspondingmaps have the same dynamics, i.e., they are topologically conjugate on their Julia sets.


are needed to see this picture.So all these parametershave the same dynamicson their Julia sets.


€

λ

€

μ


λ

€

μ




This uses quasiconformalsurgery techniques


λ

€

μ

#2: If these parameters come from Sierpinski holes with different “escape times,” then the maps cannot be conjugate.





€

Fλ (z ) = z 3 +λ

z 3

Two Sierpinski curve Julia sets, so they are homeomorphic.

€

cλ

€

cμ

€

Fμ (z ) = z 3 +μ

z 3





escape time 3 escape time 4

So these maps cannot be topologically conjugate.

€

cλ

€

cμ

€

Fλ (z ) = z 3 +λ

z 3

€

Fμ (z ) = z 3 +μ

z 3





is the only invariant boundary of an escapecomponent, so must be preserved by any conjugacy.

€

∂B

€

Fλ (z ) = z 3 +λ

z 3

€

Fμ (z ) = z 3 +μ

z 3





is the only preimage of , so this curve must also be preserved by a conjugacy.

€

∂B

€

∂T

€

Fλ (z ) = z 3 +λ

z 3

€

Fμ (z ) = z 3 +μ

z 3





If a boundary component is mapped toafter k iterations, its image under the conjugacy must also have this property,and so forth.....

€

∂T

€

Fλ (z ) = z 3 +λ

z 3

€

Fμ (z ) = z 3 +μ

z 3



2-11-1

3-1

The curves around c are special; they are the only other ones in Jmapped 2-1 onto their images.

c

€

Fλ (z ) = z 3 +λ

z 3





2-11-1

3-1

2-1

3-11-11-1

This bounding region takes 3 iterates to land on the

boundary of B.

But this bounding region takes 4 iterates to land, so

these maps are not conjugate.

€

Fλ (z ) = z 3 +λ

z 3

€

Fμ (z ) = z 3 +μ

z 3

For this it suffices to consider the centers of the Sierpinski holes; i.e., parameter values

for which for some k 3.

€

Fλk(cλ ) = ∞

#3: What if two maps lie in different Sierpinski holes that have the same escape time?




Two such centers of Sierpinski holes are “critically finite” maps, so by Thurston’s Theorem, if they are topologically

conjugate in the plane, they can be conjugated by a Mobius transformation (in the orientation preserving case).

€

Fλk(cλ ) = ∞




Two such centers of Sierpinski holes are “critically finite” maps, so by Thurston’s Theorem, if they are topologically

conjugate in the plane, they can be conjugated by a Mobius transformation (in the orientation preserving case).

Since and under the conjugacy,the Mobius conjugacy must be of the form .

€

∞ a ∞

€

0 a 0

€

z a αz

€

Fλk(cλ ) = ∞

€

h(Fλ (z)) = Fμ (h(z))then:

€

α zn +λ

zn

⎛

⎝ ⎜

⎞

⎠ ⎟=αzn +

αλ

zn=α nzn +

μ

α nzn

If we have a conjugacy

€

h(z) =αz


€

h(Fλ (z)) = Fμ (h(z))

€

h(z) =αz

then:

€

α zn +λ

zn

⎛

⎝ ⎜

⎞

⎠ ⎟=αzn +

αλ

zn=α nzn +

μ

α nzn

Comparing coefficients:

€

αn−1 =1

€


€

α zn +λ

zn

⎛

⎝ ⎜

⎞

⎠ ⎟=αzn +

αλ

zn=α nzn +

μ

α nzn


€

αn−1 =1

€

μ =α2λ


€

h(z) =αz

€


€

α zn +λ

zn

⎛

⎝ ⎜

⎞

⎠ ⎟=αzn +

αλ

zn=α nzn +

μ

α nzn


€

αn−1 =1

€

μ =α2λ

Easy check --- for the orientation reversing case:

is conjugate to via

€

Fλ

€

Fλ

€

h(z) = z


€

h(z) =αz

Theorem. If and are centers of Sierpinski holes, then iff or whereis a primitive root of unity; then any twoparameters drawn from these holes have the samedynamics.

€

Fλ ≈ Fμ

€

α

€

(n −1)st

€

μ

€

λ

n = 3: Only and are conjugatecenters since

€

λ

€

λ

€

αλ

€

α2λ

€

λ

€

λ

€

αλ

€

α2λ, , , , ,n = 4: Only

are conjugate centers where .

€

α 3 =1

€

μ =α2 jλ

€

μ =α2 jλ

€

α =−1



n = 3, escape time 4, 12 Sierpinski holes,but only six conjugacy classes

€

λ

€

λconjugate centers: ,



€

αλ

€

α2λ

€

λ

€

λ

€

αλ

€

α2λ, , , , , where

€

α 3 =1

n = 4, escape time 4, 24 Sierpinski holes,but only five conjugacy classes

conjugate centers:

Theorem: For any n there are exactly (n-1) (2n) Sierpinski holes with escape time k. The number ofdistinct conjugacy classes is given by:

k-3

a. (2n) when n is odd;k-3

b. (2n) /2 + 2 when n is even.k-3 k-4





For n odd, there are no Sierpinski holes along the real axis,so there are exactly n - 1 conjugate Sierpinski holes.

n = 3 n = 5

For n even, there is a “Cantor necklace” along the negativeaxis, so there are some “real” Sierpinski holes,

n = 4




n = 4 magnification




are needed to see this picture. M


n = 4 magnification




are needed to see this picture. M34 4

5 5

For n even, there is a “Cantor necklace” along the negativeaxis, so we can count the number of “real” Sierpinski holes,

and there are exactly n - 1 conjugate holes in this case:

n = 4 magnification




are needed to see this picture. M34 4

5 5

For n even, there are also 2(n - 1) “complex” Sierpinski holes that have conjugate dynamics:

n = 4 magnification




are needed to see this picture. M



n = 4: 402,653,184 Sierpinski holes with escape time 12; 67,108,832 distinct conjugacy classes.

Sorry. I again forgot to indicate their locations.



n = 4: 402,653,184 Sierpinski holes with escape time 12; 67,108,832 distinct conjugacy classes.

Problem: Describe the dynamics on these different conjugacy classes.

Other ways that Sierpinski curve Julia sets arise:

1. Buried points in Cantor necklaces;

2. Main cardioids in buried Mandelbrot sets;

3. Structure around the McMullen domain;

4. Other families of rational maps;

5. The difference between n = 2 and n >2 ;

6. Major applications

1. Cantor necklaces in the parameter plane

parameter plane n = 4


are needed to see this picture.MThe necklace is the Cantor middle thirds set with disks replacing removed intervals.




are needed to see this picture.MThere is a Cantor necklace along the negative real axis




The open disks are Sierpinski holes

34 4

5 5





The open disks are Sierpinski holes;the buried points in the Cantor setalso correspond to Sierpinski curves;

34 4

5 5





The open disks are Sierpinski holes;the buried points in the Cantor setalso correspond to Sierpinski curves;and all are dynamically different.

34 4

5 5


The “endpoints” in the Cantor set (parameters on the boundaries ofthe Sierpinski holes) do not correspond to Sierpinski curves.


A “hybrid” Sierpinski curve;some boundary curves meet.




There are lots of other Cantornecklaces in the parameter planes.




2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.



n = 4

€

λ



n = 4


€

λ



n = 4




€

λ



n = 4



A Sierpinski curve, but very different dynamically from the earlier ones.


€

λ

n = 3

3. If n > 2, there are uncountably many simpleclosed curves surrounding the McMullen domain;

all these parameters are (non-escape) Sierpinski curves.



n = 3







n = 3





All non-symmetric parameters on these curves have non-conjugate dynamics.




are needed to see this picture.€

γ2

10 centers

n = 3

If n > 2, there are also countably many different simple closed curves accumulating on the McMullen domain. Each alternately passes through centers of Sierpinski holes and centers of baby Mandelbrot sets (j > 1).

€

γj

€

(n − 2)n j +1

€

γj



€

γ3

28 centers

n = 3


€

γj

€

(n − 2)n j +1

€

γj



€

γ4

82 centers

n = 3


€

γj

€

(n − 2)n j +1

€

γj



n = 3

€

γ13 passes through1,594,324 centers.

€

γ13


€

γj

€

(n − 2)n j +1

€

γj



As before, all non-symmetrically locatedcenters havedifferent dynamics.

n = 3


€

γj

€

(n − 2)n j +1

€

γj

€

γ13

4. Consider the family of maps

where c is the center of a hyperbolic component of the Mandelbrot set.

€

Fλ (z ) = z 2 + c +λ

z 2



€

λ =0

€

c = −1



€

Fλ (z ) = z 2 + c +λ

z 2

€

c = −.12 +.75i



4. Consider the family of maps

where c is the center of a hyperbolic component of the Mandelbrot set.



€

λ =0





€

λ ≠0, the Julia set again expodes.When



€




€






A doubly-invertedDouady rabbit.





If you chop off the “ears” of each internal rabbit in each component of the original Fatou set, then what’s left is another Sierpinski curve (provided that both of the critical orbits eventually escape).

The case n = 2 is very different from (and much more difficult than) the case n > 2.

n = 3 n = 2





One difference: there is a McMullen domain whenn > 2, but no McMullen domain when n = 2





n = 3 n = 2

There is lots of structure when n > 2, but what is going on when n = 2?

n = 3 n = 2





Another difference: as

n = 3 n = 2





€

λ → 0,

€

Fλ2(cλ ) →∞

€

Fλ2(cλ ) →1/ 4

when n > 2when n = 2

Also, not much is happening for the Julia sets near when n > 2

n = 3



€

λ =.01



€

λ =0

The Julia set is always aCantor set of circles.

n = 3





€

λ =.0001

n = 3






€

λ =.000001




€

λ =.000001

There is always a round annulus of some fixed width in the Fatou set,

so the Fatou set is “large.”

n = 2

But when n = 2, lots of things happen near the origin;in fact, the Julia sets converge to the unit disk as



€

λ → 0

disk-converge

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.....

dynamic classification of escape time sierpinski curve julia sets dynamics of the family of complex...

Documents

b slide

mcmullen slide

unit circle slide

planar set

n free critical points

critical values

free critical orbit

easy computations