Topological Structures in the Julia Setsof Rational Maps

Dynamics of the family of complex maps

Paul Blanchard Mark MorabitoToni Garijo Monica Moreno RochaMatt Holzer Kevin PilgrimU. Hoomiforgot Elizabeth RussellDan Look Yakov ShapiroSebastian Marotta David Uminsky

with:

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Fλ (z) = z n + λz n

Three different types of topological objects:

1. Cantor Necklaces

A Cantor necklace is a planar set that ishomeomorphic to the Cantor middle thirds

set with open disks replacing removed intervals.

Three different types of topological objects:

1. Cantor Necklaces

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dynamical planen = 2

Three different types of topological objects:

1. Cantor Necklaces

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dynamical planen = 2

Three different types of topological objects:

1. Cantor Necklaces

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dynamical planen = 2

Three different types of topological objects:

1. Cantor Necklaces

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dynamical plane parameter plane

n = 2

Three different types of topological objects:

2. Mandelpinski Necklaces

Infinitely many simple closed curves in the parameter planethat pass alternately through centers of “Sierpinski holes”

and centers of baby Mandelbrot sets.

Three different types of topological objects:

2. Mandelpinski Necklaces

parameter plane zoom inn = 3

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Three different types of topological objects:

3. CanManPinski Trees

A tree of Cantor necklaces with Mandelbrot sets interspersed between each branch

Three different types of topological objects:

3. CanManPinski Trees

parameter plane parameter plane

n = 2

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Three different types of topological objects:

3. CanManPinski Trees

parameter plane parameter plane

n = 2

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Dynamics of

complex and

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J ( Fλ

)The Julia set is:The closure of the set of repelling periodic points;The boundary of the escaping orbits;The chaotic set.

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Fλ (z) = z n + λz n

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λ,z

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n ≥ 2

The Fatou set is the complement of .

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J ( Fλ

)

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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are needed to see this picture.

€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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are needed to see this picture.

€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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are needed to see this picture.

€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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are needed to see this picture.

€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

€

λ = 0

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€

λ = 0

€

F λ ( z ) = z2

+λ

z2

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

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λ ≠ 0

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λ = −1 / 16

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But when , theJulia set explodes

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

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λ ≠ 0

€

λ = −1 / 16

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But when , theJulia set explodes

A Sierpinski curve

When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

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λ ≠ 0But when , theJulia set explodes

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λ = −0 . 01

Another Sierpinski curve

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When , the Julia set is the unit circle

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λ = 0

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λ = 0

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F λ ( z ) = z2

+λ

z2

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λ ≠ 0But when , theJulia set explodes

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λ = −0 . 2

Also a Sierpinski curve

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When , the Julia set is the unit circle

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λ = 0

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

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The Sierpinski Carpet

Sierpinski Curve

Easy computations:

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Fλ (z ) = z 3 +λz 3

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λ=.036+.026i

2n free critical points

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cλ = λ1/2n

Easy computations:

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λ=.036+.026i

2n free critical points

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cλ = λ1/2n€

Fλ (z ) = z 3 +λz 3

Easy computations:

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λ=.036+.026i

2n free critical points

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cλ = λ1/2n

Only 2 critical values

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vλ = ±2 λ

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Fλ (z ) = z 3 +λz 3

Easy computations:

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€

λ=.036+.026i

2n free critical points

€

cλ = λ1/2n

Only 2 critical values

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vλ = ±2 λ

€

Fλ (z ) = z 3 +λz 3

Easy computations:

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λ=.036+.026i

2n free critical points

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cλ = λ1/2n

Only 2 critical values

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vλ = ±2 λ

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Fλ (z ) = z 3 +λz 3

Easy computations:

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λ=.036+.026i

2n free critical points

€

cλ = λ1/2n

Only 2 critical values

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vλ = ±2 λ

But really only 1 freecritical orbit since

the map has 2n-foldsymmetry

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Fλ (z ) = z 3 +λz 3

Easy computations:

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λ=.036+.026i

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

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∞ B€

Fλ (z ) = z 3 +λz 3

Easy computations:

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λ=.036+.026i

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

B

T

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Fλ (z ) = z 3 +λz 3

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∞

0 is a pole, so havetrap door T mapped

n-to-1 to B.

Easy computations:

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λ=.036+.026i

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

B

0 is a pole, so havetrap door T mapped

n-to-1 to B.

T

€

Fλ (z ) = z 3 +λz 3

€

∞

So any orbit that eventuallyenters B must do so by

passing through T.

The Escape Trichotomy

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v λ ∈

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J ( Fλ

)

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⇒B

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⇒

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⇒

is a Cantor set

T

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v λ ∈ is a Cantor set ofsimple closed curves

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J ( Fλ

)

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Fλk(v λ ) ∈ T

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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)

(with Dan Look & David Uminsky)

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v λ ∈

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J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

Case 1:

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v λ ∈

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J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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€

v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

€

⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

€

λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

€

λ

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v λ ∈

€

J ( Fλ

)

€

⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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v λ ∈

€

J ( Fλ

)

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⇒B is a Cantor set

parameter planewhen n = 3

J is a Cantor set

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λ

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parameter planewhen n = 3

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

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v λ ∈

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⇒T

parameter planewhen n = 3

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λ lies in the McMullen domain

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λ

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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λQuickTime™ and a

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

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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λQuickTime™ and a

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QuickTime™ and aTIFF (LZW) decompressor

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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λQuickTime™ and a

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QuickTime™ and aTIFF (LZW) decompressor

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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λQuickTime™ and a

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QuickTime™ and aTIFF (LZW) decompressor

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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v λ ∈

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⇒T

parameter planewhen n = 3

J is a Cantor set of simple closed curves

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λ lies in the McMullen domain

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⇒T

parameter planewhen n = 3

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λ lies in a Sierpinski hole

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Fλk(v λ ) ∈

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

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€

⇒T

parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

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λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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parameter planewhen n = 3

J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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J is a Sierpinski curve

€

λ lies in a Sierpinski hole

€

λ

€

Fλk(v λ ) ∈

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1. Cantor necklaces in thedynamical and parameter plane

The Cantor necklace is homeomorphic to the Cantor middle thirds set with

open disks replacing removed intervals.

1. Cantor necklaces in thedynamical and parameter plane

Julia set n = 2λ = -0.23

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The Cantor necklace is homeomorphic to the Cantor middle thirds set with

open disks replacing removed intervals.

1. Cantor necklaces in thedynamical and parameter plane

parameter plane n = 4

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The Cantor necklace is homeomorphic to the Cantor middle thirds set with

open disks replacing removed intervals.

Dynamical plane: n = 2

Suppose B and T are disjoint.

B

T

Dynamical plane: n = 2

Four critical points

λ1/4

Dynamical plane: n = 2

And two critical valuesthat do not lie in T

2λ1/2

Dynamical plane: n = 2

The critical lines...

Dynamical plane: n = 2

are mapped two-to-one toone of two critical value rays

So the sectors S0 and S1 are mappedone-to-one to C - {critical value rays)

S1 S0

And the regions I0 - T and I1 - T are mapped

one-to-one to C - B - {critical value rays)

Dynamical plane: n = 2

I1 I0

And the regions I0 - T and I1 - T are mapped

one-to-one to C - B - {critical value rays)

Dynamical plane: n = 2

Dynamical plane: n = 2

I1 I0

So consider the bow-tie I0 T I1

T

Dynamical plane: n = 2

I1 I0

Both I0 and I1 are mapped one-to-oneover the entire bow-tie I0 T I1

T

Dynamical plane: n = 2

So we have a preimage of the bow-tie inside each of I0 and I1

T

Dynamical plane: n = 2

Then a second preimage, etc., etc.

T

Dynamical plane: n = 2

The points whose orbits stay in I0 I1 form a Cantor set, and the preimages of T give the adjoined disks.

T

Dynamical plane: n = 2

The points whose orbits stay in I0 I1 form a Cantor set, and the preimages of T give the adjoined disks.

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Cantor Necklaces in the Parameter Plane

€

Fλ (z) = z2 +λz2

cλ = λ1/4 vλ = 2 λ1/2 Fλ(vλ) = 1/4 + 4λ

D = {λ| |λ| < 1, Re(λ) < 0}

D

For each λ D, have a Cantor set of points inside I1

I1

I0T. . . .. ...:. ..

I1

I0T

For each λ D, have a Cantor set of points inside I1

. . . .. ...:. ..

Let zs(λ) be the point in the Cantor set with itinerary s

zs(λ)

I1

I0T

For each λ D, have a Cantor set of points inside I1

. . . .. ...:. ..

Let zs(λ) be the point in the Cantor set with itinerary s

zs(λ)

zs(λ) depends analytically on λand continuously on s

I1

I0T

For each λ D, have a Cantor set of points inside I1

. . . .. ...:. ..

Let zs(λ) be the point in the Cantor set with itinerary s

zs(λ)

and zs(λ) lies in the half-diskH given by |z| < 2, Re(z) < 0

H

zs(λ) depends analytically on λand continuously on s

So have an analytic map λ zs(λ) that takes D into H

zs(λ)

DH

So have an analytic map λ zs(λ) that takes D into H

zs(λ)

DH

Have another map λ G(λ) = Fλ(vλ) = 1/4 + 4λ which maps D over a larger half disk containing H

G(λ)

So have an analytic map λ zs(λ) that takes D into H

zs(λ)

DH

Have another map λ G(λ) = Fλ(vλ) = 1/4 + 4λ which maps D over a larger half disk containing H

G-1

But G is invertible.

So have an analytic map λ zs(λ) that takes D into H

zs(λ)

DH

Have another map λ G(λ) = Fλ(vλ) = 1/4 + 4λ which maps D over a larger half disk containing H

G-1

But G is invertible. So G-1(zs(λ)) maps D strictly inside itself.

zs(λ)

DH

G-1

By the Schwarz Lemma, for each itinerary s there is a unique fixed point λs for the map G-1(zs(λ)).

λs

zs(λ)

DH

This is a parameter for which G(λs) = zs(λs),i.e., the second iterate of the critical points lands

on a point in the Cantor set portion of the Cantor necklace.

G-1

By the Schwarz Lemma, for each itinerary s there is a unique fixed point λs for the map G-1(zs(λ)).

λs

zs(λs)

zs(λ)

DH

G-1

So the points λs for each s give aCantor set of points in the parameter plane.

λs

zs(λs)

zs(λ)

DH

G-1

So the points λs for each s give aCantor set of points in the parameter plane.

λs

Similar arguments involving Böttcher coordinates onand itineraries of preimages of the trap door

then append the Sierpinski holes to the necklace.

zs(λs)

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This necklace lies along the negative real axis.

parameter plane n = 2

parameter plane n = 2

There are lots of other Cantornecklaces in the parameter planes.

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parameter plane n = 2

There are lots of other Cantornecklaces in the parameter planes.

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parameter plane n = 2

There are lots of other Cantornecklaces in the parameter planes.

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parameter plane n = 2

There are lots of other Cantornecklaces in the parameter planes.

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When n > 2, get more complicated Cantor webscase n = 3:

When n > 2, get more complicated Cantor webscase n = 3:

When n > 2, get more complicated Cantor webscase n = 3:

When n > 2, get more complicated Cantor webscase n = 3:

When n > 2, get more complicated Cantor webscase n = 3:

Continue in this wayand then adjoin

Cantor sets

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Parameter plane n = 3Dynamical plane

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Parameter plane n = 3Dynamical plane

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Parameter plane n = 3

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Dynamical plane

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Parameter plane n = 3

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Dynamical plane

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Parameter plane n = 3Dynamical plane

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Parameter plane n = 3

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Cantor webs in the parameter plane

Parameter plane n = 3

Cantor webs in the parameter plane

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Parameter plane n = 3

Cantor webs in the parameter plane

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Parameter plane

Parameter plane n = 3

Cantor webs in the parameter plane

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Parameter plane

Parameter plane n = 3

Cantor webs in the parameter plane

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Parameter plane

Parameter plane n = 4

Different Cantor webs when n = 4

Dynamical plane

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Part 2: Mandelpinski Necklaces

Parameter plane for n = 3

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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.

Parameter plane for n = 3

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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

Easy check: C1 is the circle r = 2-2n/n-1

Parameter plane for n = 3

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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.

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.

C2 passes through thecenters of 4 M-sets

and 4 S-holesQuickTime™ and a

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*

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

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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.

Parameter plane for n = 3

C3 passes through thecenters of 10 M-sets

and 10 S-holes

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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.

Parameter plane for n = 3

C4 passes through thecenters of 28 M-sets

and 28 S-holes

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.

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Parameter plane for n = 3

C5 passes through thecenters of 82 M-sets

and 82 S-holes

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

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Parameter plane for n = 3

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,...,∞

Parameter plane for n = 4

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C1: 3 holes and M-sets

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,...,∞

Parameter plane for n = 4

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

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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,...,∞

*

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Easy computations:

€

Fλ (z ) = z 3 +λz 3

€

λ =.08i

Critical points: λ1/2n

Prepoles: (-λ)1/2n

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Easy computations:

€

Fλ (z ) = z 3 +λz 3

€

λ =.08i

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

€

λ 1/2n

€

γ0

€

γ0

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€

Fλ (z ) = z 3 +λz 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:

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€

Fλ (z ) = z 3 +λz 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

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€

Fλ (z ) = z 3 +λz 3

€

λ =.08i

€

γ0

€

vλ

€

±2 λ

Easy computations:

€

−vλ

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

whose foci are

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€

Fλ (z ) = z 3 +λz 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

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

whose foci are

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€

Fλ (z ) = z 3 +λz 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

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

whose foci are

Now assume that lies inside the critical circle:

€

±vλ

€

γ0€

vλ

€

−vλ

Warning: this is not a real proof....

€

γ0

Now assume that lies inside the critical circle:

€

±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

Now assume that lies inside the critical circle:

€

±vλ

€

γ0€

vλ

€

−vλ

€

γ1

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

then is mapped n-to-1 to ,

€

γ1

Now assume that lies inside the critical circle:

€

±vλ

€

γ2

€

γ0€

vλ

€

−vλ

€

γ1

€

γ2

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

and on and onout to

€

∂B

Now assume that lies inside the critical circle:

€

±vλ

then is mapped n-to-1 to ,

€

γ1

€

γ0€

vλ

€

−vλ

€

γ1

€

γ2

B

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

€

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 contains 2n critical points and 2n prepoles, so

€

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

€

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

€

Fλk

n = 3

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€

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

€

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

€

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

€

Fλk

n = 3

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€

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

€

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

€

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

€

Fλk

n = 3

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€

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

Since

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

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

€

γk

€

λ

€

Fλ (vλ ) = 2nλn /2 +λ

2nλn /2 ≈12nλ1−n /2

€

γ0€

vλ

€

−vλ

€

γk

€

Fλ (vλ ) = 2nλn /2 +λ

2nλn /2 ≈12nλ1−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

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

€

γk

€

λ

There is a natural parametrization of each

€

γk

€

γkλ (θ )

€

γ0€

vλ

€

−vλ

€

γk

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€

γkλ (θ )

The real proof involves the Schwarz Lemma (as before):

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There is a natural parametrization of each

€

γk

€

γkλ (θ )

The real proof involves the Schwarz Lemma (as before):

€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

Best to restrict to a “symmetry region” inside the circle C1, so that is well-defined.

€

γkλ (θ )

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€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

Best to restrict to a “symmetry region” inside the circle C1, so that is well-defined.

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

€

γkλ (θ )

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€

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(γ kλ (θ ) )

a map from a “disk” to itself.

So consider the composition€

G−1

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€

γ0€

vλ

€

−vλ

€

γk

€

γkλ (θ )

€

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 lots of parameters for which the critical orbits are periodic and land on 0.

€

G−1(γ kλ (θ ) )

€

γkλ (θ )

€

G−1

Remarks:

1. 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.

Remarks:

1. 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.

2. It is known that each S-hole in the Mandelpinskinecklace is also surrounded by infinitely many sub-necklaces, which in turn are surrounded bysub-sub-necklaces, etc.

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n = 3

3. CanManPinski Trees

A tree of Cantor necklaces with Mandelbrot sets interspersed between each branch

parameter plane parameter plane

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n = 2

Dynamical plane: n = 2

I1

I0

Recall that we have a Cantor necklace in the dynamical plane lying in I0 T I1

Dynamical plane: n = 2

I1

I0

The regions I2 and I3 are mapped one-to-one overI0 T I1, so there are Cantor necklaces in I2 and I3

I2

I3

Dynamical plane: n = 2

I1

I0

I2

I3

The regions I2 and I3 are mapped one-to-one overI0 T I1, so there are Cantor necklaces in I2 and I3

This necklace is mapped one-to-one ontothe original necklace.

This necklace is mapped one-to-one ontothe original necklace.

And so is the bottom necklace

S2

S0

S1

S3

Now consider the regions Sj.

S2

S0

S1

S3

Now consider the regions Sj.

S0 is mapped two-to-one onto S0 S1

S2

S0

S1

S3

Now consider the regions Sj.

S0 is mapped two-to-one onto S0 S1

Similarly, S1 S2 S3,S2 S0 S1

and S3 S2 S3

S2

S0

S1

S3

Assuming λ lies in the upper half plane, the critical values ±vλ lie in S0 and S2 (easy check)

vλ

-vλ

S2

S0

S1

S3

Assuming λ lies in the upper half plane, the critical values ±vλ lie in S0 and S2 (easy check)So there is a region in S3 mapped one-to-one ontoS3.

vλ

-vλ

S2

S0

S1

S3

So there is a preimage of this Cantor necklace in S3

vλ

-vλ

S2

S0

S1

S3

So there is a preimage of this Cantor necklace in S3,

vλ

-vλ

S2

S0

S1

S3

So there is a preimage of this Cantor necklace in S3,and then another preimage,

vλ

-vλ

S2

S0

S1

S3

So there is a preimage of this Cantor necklace in S3,and then another preimage, and so on, yielding infinitelymany necklaces eventually mapping to the original necklace. Looking like branchesof a tree....

vλ

-vλ

S2

S0

S1

S3

By symmetry, we have similar branches in S1

vλ

-vλ

S2

S0

S1

S3

By symmetry, we have similar branches in S1, S0,

S2

S0

S1

S3

By symmetry, we have similar branches in S1, S0, and S2

This produces trees of Cantor necklaces in the dynamical plane

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This produces trees of Cantor necklaces in the dynamical plane

Assuming λ is in the upper half plane, we can again use G(λ) = 1/4 + 4 λ and an appropriate coding of points in the necklace, and then the Schwarz Lemma produces a similar

tree in the upper half of the parameter plane.

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Symmetry under complex conjugation yields a similar tree in the lower half-plane.

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Then polynomial-like map theory produces a Mandelbrot set in each region in between the branches.

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Open problems:

Can we use these trees to give a complete map of the Sierpinski-hole regions and buried parameters in the parameter planes?

Same question for the baby Mandelbrot sets.

Using “Yoccoz puzzles,” do these trees allow us tosee that the boundary of the parameter plane locusis a simple closed curve?

Open problems:

Can we use these trees to give a complete map of the Sierpinski-hole regions and buried parameters in the parameter planes?

Same question for the baby Mandelbrot sets.

Using “Yoccoz puzzles,” do these trees allow us tosee that the boundary of the parameter plane locusis a simple closed curve?

Would it be better to call these thingsCantormandelbrotsierpinski trees?

Open problems:

Can we use these trees to give a complete map of the Sierpinski-hole regions and buried parameters in the parameter planes?

Same question for the baby Mandelbrot sets.

Using “Yoccoz puzzles,” do these trees allow us tosee that the boundary of the parameter plane locusis a simple closed curve?

Would it be better to call these thingsCantormandelbrotsierpinski trees?

Who the hell is this?

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when n = 2

Other topics:

Main cardioid of a buried baby M-set

Perturbed rabbit

Convergence to the unit disk

Major application

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

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n = 4

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n = 4

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

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n = 4

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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

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n = 4

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

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n = 4

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

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

Consider the family of maps

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

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Fλ (z) = zn + c+ λzn

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λ =0

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c = −1

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Consider the family of maps

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

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Fλ (z) = zn + c+ λzn

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c = −.12 +.75i

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λ =0

Consider the family of maps

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

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Fλ (z) = zn + c+ λzn

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λ ≠0, the Julia set again expodes.When

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λ ≠0, the Julia set again expodes.When

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λ ≠0, the Julia set again expodes.When

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λ ≠0, the Julia set again expodes.When

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λ ≠0, the Julia set again expodes.When

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A doubly-invertedDouady rabbit.

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If you chop off the “ears” of each internal rabbit in each component of the original Julia 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

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

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n = 3 n = 2

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

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n = 3 n = 2

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

n = 3 n = 2

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

n = 3 n = 2

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

n = 3 n = 2

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Also, not much is happening for the Julia sets near 0 when n > 2

n = 3

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λ =.01

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The Julia set is always aCantor set of circles.

n = 3

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λ =.0001

n = 3

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The Julia set is always aCantor set of circles.

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λ =.000001

The Julia set is always aCantor set of circles.

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λ =.000001

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

so the Julia set does not convergeto the unit disk.

n = 2

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

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λ → 0

disk-converge

Here’s the parameter plane when n = 2:

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Rotate it by 90 degrees:

and this object appears everywhere.....

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