Structures in the Dynamical Plane Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer Dan Look Sebastian Marotta Mark Morabito

Structures in the Dynamical Plane

Dynamics of the family of complex maps

Paul BlanchardToni GarijoMatt HolzerDan LookSebastian Marotta Mark Morabito

with:

€

Fλ (z) = zn +λ

zn

Monica Moreno RochaKevin PilgrimElizabeth RussellYakov ShapiroDavid UminskySum Wun Ellce

, n > 1

1. Dynamic classsification of escape time Sierpinski curve Julia sets

3. Cantor necklaces in the dynamical plane

2. Julia sets converging to the unit disk

4. Internal rays

Sierpinski curve Julia sets occur formany different parameters in this family

1. When critical points “eventually” escape

2. As buried points in Cantor necklaces

3. In the main cardioid of buried Mandelbrot sets

4. On Cantor sets of circles surrounding the McMullen domain

5. On “Mandelpinski” necklaces

And almost all have very different dynamical behavior

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.





Sierpinski curve Julia sets occur formany different parameters in this family

All these sets are homeomorphic, but the dynamics on each are very different

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



The Sierpinski Carpet

Recall:



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.



€

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

A Sierpinski curve

When , the Julia set is the unit circle

€

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


€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3

Easy computations:



€

λ=.036+.026i


€

cλ = λ1/2n


€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3

Easy computations:



€

λ=.036+.026i


€

cλ = λ1/2n


€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3

Easy computations:

€

λ=.036+.026i


€

cλ = λ1/2n


€

vλ = ±2 λ

But really only one free critical orbit

€

Fλ (z ) = z 3 +λ

z 3



Easy computations:

€

λ=.036+.026i


€

cλ = λ1/2n


€

vλ = ±2 λ

€

Fλ (z ) = z 3 +λ

z 3



And J has 2n-fold symmetry

But really only one free critical orbit

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

€

vλ∈

€

J ( Fλ

)

€

⇒B is a Cantor set

T

€

vλ∈ is a Cantor set of

simple closed curves

€

J ( Fλ

)

€

Fλ

k(v

λ) ∈ T

€

J ( Fλ

) is a Sierpinski curve

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

(this case does not occur if n = 2)

€

⇒

€

⇒

(McMullen)





€

vλ∈

€

J ( Fλ

)

€

⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

€

λ



€

vλ∈

€

⇒T


J is a Cantor set of simple closed curves

€

λ lies in the McMullen domain

€

λQuickTime™ and a

TIFF (LZW) decompressorare needed to see this picture.



€

⇒T


€

λ lies in a Sierpinski hole

€

λ

€

Fλ

k(v

λ) ∈



J is an escape time Sierpinski curve

c

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)




n = 3escape time 32 Sierpinski holes

parameter planen = 3





















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



Guess what? I again 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:





Part 1: Dynamic Classification of Escape Time Sierpinski Curve Julia sets

#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 boundary curvesin J mapped 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 .

€

∞→ ∞

€

0 → 0

€

z →α z

€

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) = α z


€


€

α zn +λ

zn

⎛

⎝ ⎜

⎞

⎠ ⎟=αzn +

αλ

zn=α nzn +

μ

α nzn

Comparing coefficients:

€

αn−1 =1

€

h(z) = α z

€


€

α zn +λ

zn

⎛

⎝ ⎜

⎞

⎠ ⎟=αzn +

αλ

zn=α nzn +

μ

α nzn


€

αn−1 =1

€

μ =α2λ


€

h(z) = α z

€


€

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) = α z

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. (with K. Pilgrim)

€

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: ,




€

λ

€





€

λ

€





€

λ

€




€

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:



€

αλ

€

α2λ

€

λ

€

λ

€

αλ

€

α2λ, , , , , where

€

α 3 =1


conjugate centers:



€

αλ

€

α2λ

€

λ

€

λ

€

αλ

€

α2λ, , , , , where

€

α 3 =1


conjugate centers:



€

αλ

€

α2λ

€

λ

€

λ

€

αλ

€

α2λ, , , , , where

€

α 3 =1


conjugate centers:



€

αλ

€

α2λ

€

λ

€

λ

€

αλ

€

α2λ, , , , , where

€

α 3 =1


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

n = 4



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. M3


n = 4 magnification




are needed to see this picture. M34 4


n = 4 magnification





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





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.

Will someone please remind me to

indicate their locations?



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

Problem: use symbolic dynamics to describe these conjugacy classes.

Part 2: Julia sets converging to the unit disk

With Toni Garijo

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


With Toni Garijo


€

Fλ

€

λ → 0

€

λ =0

€

Fλ

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

€

λ


With Toni Garijo

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


With Toni Garijo

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

€

Bδ (z1)

€

Bδ (z2 )

€

Bδ (z3)

€

Bδ (z4 )

€

δ > 0

These disks cannot lie in the trapdoor since T vanishes as .(Remember is never in the trap door when n = 2.)

€

λ → 0

€

vλ = 2 λ


€

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*€

δ > 0

€

Bδ (z*)


€

λ → 0



€

Bδ (zi )

€

λi → 0

€

zi ∈D

€

Bδ (z*)


€

zi

€

z*

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

Fλ i≈ z2

€

Fλ i

k

€

Bδ (z*)€

Fλk

€

δ > 0

€

Bδ (z*)

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

€

r = 1− ε

€

r = 1+ ε



€

∂Ak in here

€

r = 1+ εConstruct an outerround annulus withmodulus

€

α



€

∂Ak in here

€

r = 1+ εConstruct an outerround annulus withmodulus

€

α

and a similarinner annulus



€

∂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





€

λ =.01



€

λ =.0001

€

λ =.000001

And that’s what we saw earlier: there is always an annulus Xk far out with definite thickness.





€

λ =.00000001

€

λ =.0000000001



€

λ =.000000000001

Part 3: Cantor necklaces and webs

A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals.





Do you see a necklace in the carpet?





Do you see a necklace in the carpet?





There are lots of necklaces in the carpet





There are lots of necklaces in the carpet




a Julia set with n = 2

There are lots of Cantornecklaces in these Julia sets,

just as in the Sierpinski carpet.




another Julia set with n = 2

























n = 2

Even if we choose a parameter from the Mandelbrot set, there are Cantor necklaces in the Julia set





n = 2

Even if we choose a parameter from the Mandelbrot set, there are Cantor necklaces in the Julia set


And there are Cantor necklaces in the parameter planes.



n = 2

The critical circle & ray


Critical points

€

λ1/2n


Critical points

€

λ1/2n

n = 2

0


Critical points

€

λ1/2n

Critical values

€

±2 λ

n = 2

0


Critical points

€

λ1/2n

Critical values

€

±2 λ

n = 2

0


Critical points

€

λ1/2n

€

(−λ )1/2nPrepoles

Critical values

€

±2 λ

n = 2

0


Critical points

€

λ1/2n

€


Critical values

€

±2 λ

n = 2

0


Critical points

€

λ1/2n

€


Critical points &prepoles lie on the

critical circle

€

r = |λ |1/2n

Critical values

€

±2 λ

n = 2

0


Critical points

€

λ1/2n

€


Critical values

€

±2 λ


critical circle

€

r = |λ |1/2n

n = 2

0


Critical points

€

λ1/2n

€


Critical values

€

±2 λ


critical circle

€

r = |λ |1/2n

Which is mapped 2n to 1onto the critical line

n = 2

0


Critical points

€

λ1/2n

€


Which is mapped 2n to 1onto the critical line

Critical values

€

±2 λ


critical circle

€

r = |λ |1/2n

n = 2

0

Critical points

€

λ1/2n

Critical values

€

±2 λ

Critical point rays

Critical points

Critical values

€

±2 λ

Critical point raysare mapped 2 to 1 toa ray external to thecritical line, a criticalvalue ray.

€

λ1/2n

Suppose is not positive real.

€

λ

Suppose is not positive real. Then the critical values do not lie on the critical point rays. €

λ

€

±2 λ

€

−2 λ

€

+ 2 λ

€

λ1/4

I0

I1There are 4 prepole sectors when n = 2 andthe critical values alwayslie in I1 and I3.

I2

I3

Suppose is not positive real.

€

λ

I0

I1

I2

I3

And the interior ofeach Ij is mappedone-to-one over theentire plane minusthe critical value rays

€

λSuppose is not positive real.

There are 4 prepole sectors when n = 2 andthe critical values alwayslie in I1 and I3.

I0

I1

I2

I3In particular,maps I0 and I2

one-to-one overI0 I2 €

Fλ

€

U

€

λSuppose is not positive real.

And the interior ofeach Ij is mappedone-to-one over theentire plane minusthe critical value rays

There are 4 prepole sectors when n = 2 andthe critical values alwayslie in I1 and I3.

Choose a circle in B that is mapped strictly outside itself

€

γ

€

γ

I0

I2


€

γ

€

γ

€

Fλ (γ )

I0

I2


€

γ

€

γ

€

Fλ (γ )

Then there is anothercircle in the trapdoor that is alsomapped to

€

Fλ (γ )

€

υ

€

υ

€

=Fλ (υ )

€

γ

€

υ€

γ

€

υ

Consider the portions ofI0 and I2 that lie between and , say U0 and U2

U2

U0

€

γ

€

γ

€

υ

€

υ

U2

U0


U0 and U2 are mappedone-to-one over U0 U2

€

U

€

γ

€

υ

U0



€

U

So the set of pointswhose orbits lie forall iterations in U0 U2 is an invariant Cantorset

€

U

U2

€

γ

€

υ



€

U

So the set of pointswhose orbits lie forall iterations in U0 U2 is an invariant Cantorset

€

U

The Cantor set in U0 U2 contains:

€

U

2 points on B

€

∂


€

U

2 points on B

€

∂


€

U

2 points on B

€

∂

2 points on T

€

∂


€

U

2 points on B

€

∂

2 points on T

€

∂

4 points on the 2 preimages of T

€

∂


€

U

2 points on B

€

∂

2 points on T

€

∂

Add in the appropriatepreimages of T.....

€

∂


€

∂


€

U

2 points on B

€

∂

2 points on T

€

∂

Add in the appropriatepreimages of Tto get an “invariant”Cantor necklace in thedynamical plane€

∂


€

∂



Cantor necklaces in the dynamical plane when n = 2



When n > 2, get Cantor “webs” in the dynamical plane:

n = 3


Start with an open disk....

n = 3


Then surround it by 4smaller disks

n = 3


Then do it again....


n = 3

and so forth, joiningthe open disks by a Cantor set of points





n = 4


n = 3






n = 3 n = 4






n = 3 n = 3



Same argument: we have 2n prepole sectors I0,...,I2n-1

I0I1

I2

I3 I4

I5

n = 3

€

c0




I0I1

I2

I3 I4

I5

If is not positive real, then the critical value ray always lies in I0 and In as long as lies in a symmetry region

€

λ

€

vλ

€

−vλ

n = 3

€

c0

€

λ






I0I1

I2

I3 I4

I5


€

λ

€

vλ

€

−vλ

n = 3

€

c0

€

λ

A symmetry region when n = 3






I0I1

I2

I3 I4

I5


€

λ

€

vλ

€

−vλ

n = 3

€

c0

€

λ

A symmetry region when n = 4




U1

U2

U4

U5€

vλ

€

−vλ

So consider the correspondingregions Uj not including

U0 and Un which contain

€

±vλ

n = 3

€

c0


€

λ

€

λ



n = 3

U1

U2

U4

U5€

vλ

€

−vλ

Each of these Uj are mappedunivalently over all the others,

excluding U0 and Un, sowe get an invariant Cantor set

in these regions.



n = 3



in these regions.

Then join in the preimages of Tin these regions....



n = 3



in these regions.


and their preimages....



n = 3



in these regions.


and their preimages....

etc., etc. to get the Cantor web



Other Cantor webs

n = 4 n = 5



Other Cantor webs

Next time we’ll see how Cantor webs and necklacesalso appear in the parameter planes for these maps.





Part 4: Internal Rays



These straight rays are preserved by zn

1/4

0

1/8

We can put a Böttcher coordinate on B to produce external rays in the dynamical plane.

€

ϕ

Part 4: Internal Rays



We can put a Böttcher coordinate on B to produce external rays in the dynamical plane.



1/4

0

1/8€

ϕ

€

ϕ

and are mapped by to external rays in B

€

ϕThese straight rays are preserved by zn



And using the Cantor necklace/web, these can beextended to “internal rays” connecting the

external rays to the origin and passing throughthe Julia set in interesting ways:



And using the Cantor necklace/web, these can beextended to “internal rays” connecting the

external rays to the origin and passing throughthe Julia set in interesting ways:



These two internal rays pass through all the points in the Cantor set of points inthe Cantor necklace as well as through

the appropriate preimages of T:

In the necklace construction, you justpull back appropriate images of theexternal rays in B;


0

1/2


0

1/2


0

1/2

In the necklace construction, you justpull back appropriate images of theexternal rays in B; which then connectup with the appropriate endpoints ofthe Cantor set to produce the internal ray.

0

1/2

Some facts:



When n > 2, there exists an easy-to-constructCantor set of such internal rays, and each crossesinfinitely many others.

Some facts:



And the regions in between these crossingrays give disks on which the map is polynomial-like in the sense of Douady and Hubbard.

€

Fλ is polynomial-like in this disk.

Some facts:

So this enables us to prove the existence ofthe infinitely many baby Mandelbrot setsyou see in the parameter planes:



Some facts:




Some facts:




Some facts:




Some facts:




Some facts:

And there is a piecewise linear model forthese internal rays that you can embed in the Sierpinski carpet:



Some facts:

And there is a piecewise linear model forthese internal rays that you can embed in the Sierpinski carpet:


are needed to see this picture.QuickTime™ and a

TIFF (LZW) decompressorare needed to see this picture.

Some facts:

And finally, this should provide a mechanism (Yoccozpuzzles) to prove what is “obviously” true: that the boundaries of the parameter planes for these maps are simple closed curves (like Milnor’s cubic family).





n = 4 n = 7

Documents

Structures in the Dynamical Plane Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer Dan Look Sebastian Marotta Mark Morabito