Dynamics of the family of complex maps
F (z)  z 
n

z
n
, n2
(why the case n = 2 is

)
with:
Paul Blanchard
Toni Garijo
Matt Holzer
U. Hoomiforgot
Dan Look
Sebastian Marotta
Mark Morabito
Monica Moreno Rocha
Kevin Pilgrim
Elizabeth Russell
Yakov Shapiro
David Uminsky
F (z)  z 
n
The case n > 2 is great because:


zn
F (z)  z 
n

zn
The case n > 2 is great because:
There exists a McMullen domain around  = 0 ....

Parameter plane for n=3

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F (z)  z 
n

zn
The case n > 2 is great because:
There exists a McMullen domain around  = 0 ....
 by infinitely many “Mandelpinski” necklaces...
... surrounded

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F (z)  z 
n

zn
The case n > 2 is great because:
There exists a McMullen domain around  = 0 ....
 by infinitely many “Mandelpinski” necklaces...
... surrounded
... and the Julia sets behave nicely as   0


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  .01
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  .0001
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  .000001
F (z)  z 
n

zn
The case n = 2 is crazy because:
There is no McMullen domain....

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F (z)  z 
n

zn
The case n = 2 is crazy because:
There is no McMullen domain....
... and no 
“Mandelpinski” necklaces...
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F (z)  z 
n

zn
The case n = 2 is crazy because:
There is no McMullen domain....
... and no 
“Mandelpinski” necklaces...
... and the Julia sets “go crazy” as   0

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  .000001
Some definitions:
Julia set of F (z)  z 
n

zn
J = boundary of {orbits that escape to
= closure {repelling periodic orbits}

= {chaotic set}
Fatou set
= complement of J
= predictable set

}
F (z)  z 
3
Computation of J:
Color points that escape to
infinity shades of
red
orange
yellow
green
blue
violet
Black points do not escape.
J = boundary of the black
region.

z3

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  .08i
F (z)  z 
3
Easy computations:
 is superattracting, so
have immediate basin B
mapped n-to-1 to itself.


z3
B
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  .08i
F (z)  z 
3
Easy computations:
 is superattracting, so
have immediate basin B
mapped n-to-1 to itself.

z3
B
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T
0 is a pole, so have
trap door T mapped
n-to-1 to B.

  .08i
F (z)  z 
3
Easy computations:
 is superattracting, so
have immediate basin B
mapped n-to-1 to itself.

z3
B
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T
0 is a pole, so have
trap door T mapped
n-to-1 to B.

The Julia set has
2n-fold symmetry.
  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n


z3

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  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n


z3

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  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n

z3

Only 2 critical values

v  2 
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
  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n

z3

Only 2 critical values

v  2 
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
  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n

z3

Only 2 critical values

v  2 
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
  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n

z3

Only 2 critical values


v  2 
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But really only 1 free
critical orbit
  .08i
F (z)  z 
3
Easy computations:
2n free critical points
c  1/2n

z3

Only 2 critical values


v  2 
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But really only 1 free
critical orbit
And 2n prepoles
p  (  )1/2n
  .08i
The Escape Trichotomy
(with D. Look & D Uminsky)
There are three possible ways that the
critical orbits can escape to infinity,
and each yields a different type
of Julia set.
The Escape Trichotomy
(with D. Look & D Uminsky)
v  B




J(F ) is a Cantor set
The Escape Trichotomy
(with D. Look & D Uminsky)


v  B

v  T




J(F ) is a Cantor set
J(F ) is a Cantor set of
simple closed curves
(n > 2)

(McMullen)

The Escape Trichotomy
(with D. Look & D Uminsky)


v  B

v  T


J(F ) is a Cantor set
J(F ) is a Cantor set of
simple closed curves
(n > 2)

(McMullen)
 cases J(F ) is a connected set, and if
In all other
F (v )  T
k




J(F ) is a Sierpinski curve
Case 1:
v  B



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parameter plane
when n = 3

J(F ) is a Cantor set
v  B
Case 1:




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
parameter plane
when n = 3

J(F ) is a Cantor set
v  B
Case 1:


J(F ) is a Cantor set


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
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
parameter plane
when n = 3
J is a Cantor set
v  B
Case 1:


J(F ) is a Cantor set


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v

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c


parameter plane
when n = 3

J is a Cantor set
Case 2:
v  T

J(F ) is a Cantor set of
simple closed curves



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parameter plane
when n = 3
Case 2:
v  T

J(F ) is a Cantor set of
simple closed curves




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
parameter plane
when n = 3
Case 2:
v  T

J(F ) is a Cantor set of
simple closed curves




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
The central disk is
the McMullen domain
Case 2:
v  T

J(F ) is a Cantor set of
simple closed curves

B



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T

parameter plane
when n = 3
J is a Cantor set of
simple closed curves
Case 2:
v  T

J(F ) is a Cantor set of
simple closed curves




v


parameter plane
when n = 3
c
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
J is a Cantor set of
simple closed curves
k
Case 3: F (v )  T



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parameter plane
when n = 3

J(F ) is a Sierpinski
curve
k
Case 3: F (v )  T


J(F ) is a Sierpinski
curve


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parameter plane
when n = 3
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A Sierpinski curve is a planar
set homeomorphic to the
Sierpinski carpet fractal
Sierpinski curves are important for two reasons:
1. There is a “topological characterization” of the carpet
2. A Sierpinski curve is a “universal plane continuum”
Topological Characterization
Any planar set that is:
1. compact
2. connected
3. locally connected
4. nowhere dense
5. any two complementary
domains are bounded by
simple closed curves
that are pairwise disjoint
is a Sierpinski curve.
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The Sierpinski Carpet
Universal Plane Continuum
Any planar, one-dimensional, compact, connected set can
be homeomorphically embedded in a Sierpinski curve.
For example....
This set
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This set
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can be embedded
inside
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k
Case 3: F (v )  T



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parameter plane
when n = 3

J(F ) is a Sierpinski
curve
k
Case 3: F (v )  T




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
A Sierpinski “hole”

J(F ) is a Sierpinski
curve
k
Case 3: F (v )  T



J(F ) is a Sierpinski
curve


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
A Sierpinski “hole”
A Sierpinski curve
k
Case 3: F (v )  T





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
J(F ) is a Sierpinski
curve
v
c

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F (v )


A Sierpinski “hole”
A Sierpinski curve
Escape time 3
k
Case 3: F (v )  T


J(F ) is a Sierpinski
curve



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
Another Sierpinski “hole”
A Sierpinski curve
k
Case 3: F (v )  T


J(F ) is a Sierpinski
curve



v
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F (v )



Another Sierpinski “hole”

A Sierpinski curve
Escape time 4
c
k
Case 3: F (v )  T



J(F ) is a Sierpinski
curve


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
Another Sierpinski “hole”
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A Sierpinski curve
Escape time 7
k
Case 3: F (v )  T



J(F ) is a Sierpinski
curve


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
Another Sierpinski “hole”
A Sierpinski curve
Escape time 5
So to show that
is homeomorphic to
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
Fatou set is the union of
the preimages of B; all
disjoint, open disks.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
Fatou set is the union of
the preimages of B; all
disjoint, open disks.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
If J contains an open
set, then J = C.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
If J contains an open
set, then J = C.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
No recurrent critical orbits
and no parabolic points.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
No recurrent critical orbits
and no parabolic points.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
J locally connected, so the
boundaries are locally
connected. Need to show
they are s.c.c.’s. Can only
meet at (preimages of) critical
points, hence disjoint.
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compact
connected
nowhere dense
locally connected
bounded by disjoint
s.c.c.’s
So J is a Sierpinski curve.
Remark: All Julia sets drawn from Sierpinski holes are
homeomorphic, but only those in symmetrically located
Sierpinski holes have the same dynamics.
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The maps on all these Julia sets are dynamically different.
Remark: All Julia sets drawn from Sierpinski holes are
homeomorphic, but only those in symmetrically located
Sierpinski holes have the same dynamics.
In fact, there are exactly (n-1)(2n)k-3 Sierpinski holes with
escape time k, and (2n)k-3 different conjugacy classes (n odd).
(with K.Pilgrim)
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The maps on all these Julia sets are dynamically different.
Topics
1. The McMullen domain
2. Mandelpinski necklaces
3. Julia sets near 0
Part 1: The McMullen Domain
Why is there no McMullen domain when n = 2?
First suppose v  T:
What is the preimage of T?

Why is there no McMullen domain when n = 2?
First suppose v  T:
What is the preimage of T?
Can 
the preimage be 2n
disjoint disks, each of which
contains a critical point?
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Why is there no McMullen domain when n = 2?
First suppose v  T:
What is the preimage of T?
Can 
the preimage be 2n
disjoint disks, each of which
contains a critical point?
No --- there would then be 4n
preimages of any point in T,
but the map has degree 2n.
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Why is there no McMullen domain when n = 2?
So some of the preimages of T
must overlap, and by 2n-fold
symmetry, all must intersect.
Why is there no McMullen domain when n = 2?
So some of the preimages of T
must overlap, and by 2n-fold
symmetry, all must intersect.
v
By Riemann-Hurwitz, the
preimage of T must then
be an annulus.
c
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

Why is there no McMullen domain when n = 2?
So here is the picture:
B
A is the annulus
separating B and T
A
T
Why is there no McMullen domain when n = 2?
So here is the picture:
B
A is the annulus
separating B and T
X
F maps X 2n-to-1
onto T
A
T
Why is there no McMullen domain when n = 2?
So here is the picture:
B
A0
A is the annulus
separating B and T
X
F maps X 2n-to-1
onto T
A
F maps both A0
and A1 as an n-to-1
covering onto A
T
A1
Why is there no McMullen domain when n = 2?
So mod(A0) = 1/n mod(A)
And mod(A1) = 1/n mod(A)
B
A0
X
A
T
A1
Why is there no McMullen domain when n = 2?
So mod(A0) = 1/n mod(A)
And mod(A1) = 1/n mod(A)
B
A0
X
When n = 2,
mod(A0) + mod(A1) =
mod(A)
A
T
A1
Why is there no McMullen domain when n = 2?
So mod(A0) = 1/n mod(A)
And mod(A1) = 1/n mod(A)
B
A0
X
When n = 2,
mod(A0) + mod(A1) =
mod(A)
A
T
A1
So there is no room
for X, i.e., v does
not lie in T
Why is there no McMullen domain when n = 2?
Here is another reason:
v  2 
F (v )  2 
n


n /2

1
n n /2
 n n /2  2   n (n /21)
2 
2 
Why is there no McMullen domain when n = 2?
Here is another reason:
v  2 
F (v )  2 
n

n /2

1
n n /2
 n n /2  2   n (n /21)
2 
2 
n  2  F (v )   as   0

so v lies in T when n > 2


Why is there no McMullen domain when n = 2?
Here is another reason:
v  2 
F (v )  2 
n

n /2

1
n n /2
 n n /2  2   n (n /21)
2 
2 
n  2  F (v )   as   0

so v lies in T when n > 2


1
n  2  F (v )  4  
4
Why is there no McMullen domain when n = 2?
Here is another reason:
v  2 
F (v )  2 
n

n /2

1
n n /2
 n n /2  2   n (n /21)
2 
2 
n  2  F (v )   as   0

so v lies in T when n > 2


1
n  2  F (v )  4  
4
so F (v ) 1/ 4 (???) as   0
Part 2: Mandelpinski Necklaces
A “Mandelpinski necklace” is a simple closed curve in the
parameter plane that passes alternately through k centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
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TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
A “Mandelpinski necklace” is a simple closed curve in the
parameter plane that passes alternately through k centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
C1 passes through the
centers of 2 M-sets
and 2 S-holes
A “Mandelpinski necklace” is a simple closed curve in the
parameter plane that passes alternately through k centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
A “Mandelpinski necklace” is a simple closed curve in the
parameter plane that passes alternately through k centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
C2 passes through the
centers of 4 M-sets*
and 4 S-holes
* only exception:
2 centers of period
2 bulbs, not M-sets
Parameter plane for n = 3
A “Mandelpinski necklace” is a simple closed curve in the
parameter plane that passes alternately through k centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
C3 passes through the
centers of 10 M-sets
and 10 S-holes
A “Mandelpinski necklace” is a simple closed curve in the
parameter plane that passes alternately through k centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
C4 passes through the
centers 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 centers
of baby Mandelbrot sets and k centers of Sierpinski holes.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
C5 passes through the
centers of 82 M-sets
and 82 S-holes
Theorem: There exist closed curves Cj, j  1,..., 
surrounding the McMullen domain. Each Cj passes
alternately through (n-2)nj-1 +1 centers of baby
Mandelbrot sets and centers of Sierpinski holes.

Theorem: There exist closed curves Cj, j  1,..., 
surrounding the McMullen domain. Each Cj passes
alternately through (n-2)nj-1 +1 centers of baby
Mandelbrot sets and centers of Sierpinski holes.

QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane for n = 3
C14 passes through the
centers of 4,782,969
M-sets and S-holes
Theorem: There exist closed curves Cj, j  1,..., 
surrounding the McMullen domain. Each Cj passes
alternately through (n-2)nj-1 +1 centers of baby
Mandelbrot sets and centers of Sierpinski holes.

C1: 3 holes and M-sets
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Parameter plane for n = 4
Theorem: There exist closed curves Cj, j  1,..., 
surrounding the McMullen domain. Each Cj passes
alternately through (n-2)nj-1 +1 centers of baby
Mandelbrot sets and centers of Sierpinski holes.

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are needed to see this picture.
Parameter plane for n = 4
C2: 9 holes and M-sets*
C3: 33 holes and M-sets
F (z)  z 
3
Easy computations:
All of the critical
points and prepoles
lie on the “critical
circle”  0 : |z| = | | 1/2n

z3
0



QuickTime™ and a
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are needed to see this picture.

  .08i
F (z)  z 
3
Easy computations:
All of the critical
points and prepoles
lie on the “critical
circle”  0 : |z| = | | 1/2n


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and a

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are needed to see this picture.
v



z3
0
v
which is mapped 2n-to-1
onto
value line”

 the “critical
connecting v

  .08i
F (z)  z 
3
Easy computations:
Any other circle around 0
is mapped n-to-1 to an ellipse 
whose foci are 2 
z3
0
v



QuickTime™
and a

TIFF (LZW) decompressor
are needed to see this picture.
v


  .08i
F (z)  z 
3
Easy computations:
Any other circle around 0
is mapped n-to-1 to an ellipse 
whose foci are 2 
z3
0
v



QuickTime™
and a

TIFF (LZW) decompressor
are needed to see this picture.
v


  .08i
F (z)  z 
3
Easy computations:
Any other circle around 0
is mapped n-to-1 to an ellipse 
whose foci are 2 
So the exterior of  0 is mapped
as an n-to-1
 covering of the
exterior of the critical value line.


z3
0
v

QuickTime™
and a

TIFF (LZW) decompressor
are needed to see this picture.
v


  .08i
F (z)  z 
3
Easy computations:
Any other circle around 0
is mapped n-to-1 to an ellipse 
whose foci are 2 
So the exterior of  0 is mapped
as an n-to-1
 covering of the
exterior of the critical value line.
 0.
Same with
the
interior
of

z3
0
v

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and a

TIFF (LZW) decompressor
are needed to see this picture.
v




  .08i
Simplest case: C1
Assume that v both sit on the critical circle.
 2 |  |1/2  |  |1/2n
(i.e., | v |  | c |  | p | )


p
v

v


0
Simplest case: C1
Assume that v both sit on the critical circle.
 2 |  |1/2  |  |1/2n
 2 2n |  |n  |  |

 || 2

 2n 
 n1 
p
v

v


0
Simplest case: C1
Assume that v both sit on the critical circle.
 2 |  |1/2  |  |1/2n
 2 2n |  |n  |  |

 || 2

 2n 
 n1 
This is the “dividing circle”
in the parameter plane
r = 2-3
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Parameter plane n = 3
Simplest case: C1
Assume that v both sit on the critical circle.
 2 |  |1/2  |  |1/2n
 2 2n |  |n  |  |

 || 2
 2n 

 n1 
This is the “dividing circle”
in the parameter plane
r = 2-8/3
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TIFF (LZW) decompressor
are needed to see this picture.
Parameter plane n = 4
Simplest case: C1
Assume that  lies on the dividing circle,
so v both sit on the critical circle.


p
v

In this picture,
 is real and n = 3
v


0
Simplest case: C1
Assume that  lies on the dividing circle,
so v both sit on the critical circle.


As  rotates around the dividing circle,
v 
rotates a half-turn, while p and c rotate
1/2n of a turn. So each v meets exactly
n - 1 prepoles and critical points.



p
v

In this picture,
 is real and n = 3
v


0
Simplest case: C1
Assume that  lies on the dividing circle,
so v both sit on the critical circle.


As  rotates around the dividing circle,
v 
rotates a half-turn, while p and c rotate
1/2n of a turn. So each v meets exactly
n - 1 prepoles and critical points.


So the dividing 
circle is C1
p
v

In this picture,
 is real and n = 3
v


0
The dividing circle when n = 5
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
n-1 = 4 centers of Sierpinski holes;
n-1 = 4 centers of baby Mandelbrot sets
Now assume that v lies inside the critical circle:

v
v
0



Now assume that v lies inside the critical circle:
The exterior of  0 is mapped n-to-1
onto the exterior of the critical value
 is a preimage  1
ray, so there
mappedn-to-1 to  0 ,

v

v
0



Now assume that v lies inside the critical circle:
The exterior of  0 is mapped n-to-1
onto the exterior of the critical value
 is a preimage  1
ray, so there
mappedn-to-1 to  0 ,

v

v
0




1
Now assume that v lies inside the critical circle:
The exterior of  0 is mapped n-to-1
onto the exterior of the critical value
 is a preimage  1
ray, so there
mappedn-to-1 to  0 ,
then  2 is mapped
n-to-1 to  1 ,


v


v
0



 
1 2
Now assume that v lies inside the critical circle:
The exterior of  0 is mapped n-to-1
onto the exterior of the critical value
 is a preimage  1
ray, so there
mappedn-to-1 to  0 ,
then  2 is mapped
n-to-1 to  1 ,

 on
and on and
 out to B



v
v
0
1 2


 
B
 0 contains 2n critical points and 2n prepoles, so
 1 contains 2n2 pre-critical points and pre-prepoles
v
v
0




1
 0 contains 2n critical points and 2n prepoles, so
 1 contains 2n2 pre-critical points and pre-prepoles
 k contains 2nk+1 points that
map to the critical points
and pre-prepoles
under Fk
v
v

0

1 2


 
B
As  rotates by one turn, these 2nk+1 points on  k
each rotate by 1/2nk+1 of a turn.


v
v
0




k

As  rotates by one turn, these 2nk+1 points on  k
each rotate by 1/2nk+1 of a turn.
Since


1 1n /2
F (v )  n n /2  n 
2 
2
the second iterate of the critical
points rotate by 1 - n/2 of
a turn
v
v
0




k

As  rotates by one turn, these 2nk+1 points on  k
each rotate by 1/2nk+1 of a turn.
Since


1 1n /2
F (v )  n n /2  n 
2 
2
the second iterate of the critical
points rotate by 1 - n/2 of
a turn, so this point hits
exactly
(n / 2  1)(2n
k 1
)  1  (n  2)n
k 1
v
v
1
0

preimages of the critical points
and prepoles on  k



k
The real proof involves the Schwarz Lemma:
There is a natural parametrization  k ( ) of each  k


 k ( )
v
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TIFF (LZW) decompressor
are needed to see this picture.
v

0




k
The real proof involves the Schwarz Lemma:
There is a natural parametrization  k ( ) of each  k


 k ( )
v
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
v

Best to restrict to a “symmetry
region” inside the dividing


circle, so that  k ( ) is well-defined.
0




k
Then we have a second map from the parameter plane to the
dynamical plane, namely G( )  F (v ) which is invertible
on the symmetry sector

G 1
 k ( )
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v

v

Best to restrict to a “symmetry
region” inside the dividing


circle, so that  k ( ) is well-defined.
0




k
Then we have a second map from the parameter plane to the
dynamical plane, namely G( )  F (v ) which is invertible
on the symmetry sector

G 1
 k ( )
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v

v
So consider the composition
G 1 ( k ( ) )
a map from a “disk” to itself.

0




k
Schwarz implies that G 1 ( k ( ) ) has a unique fixed point,
i.e., a parameter for which the second iterate of the critical

point lands on the point  k ( ), so this proves the
existence of 
the centers of the S-holes and M-sets.

G 1
 k ( )
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v

v
So consider the composition
G 1 ( k ( ) )
a map from a “disk” to itself.

0




k
Remarks: This proves the existence of centers of
Sierpinski holes and Mandelbrot sets. Producing
the entire M-set involves polynomial-like maps;
while the entire S-hole involves qc-surgery.
Part 3: Behavior of the Julia sets

Part 3: Behavior of the Julia sets
n = 2: When   0, the Julia set of F is the unit circle. But,
as   0 , the Julia set of F converges to the closed unit disk




Part 3: Behavior of the Julia sets
n = 2: When   0, the Julia set of F is the unit circle. But,
as   0 , the Julia set of F converges to the closed unit disk



n > 2: J is always a Cantor set of “circles” when  is small.


Part 3: Behavior of the Julia sets
n = 2: When   0, the Julia set of F is the unit circle. But,
as   0 , the Julia set of F converges to the closed unit disk



n > 2: J is always a Cantor set of “circles” when  is small.
Moreover, there is a   0 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.


Theorem: When n = 2, the Julia sets converge to
the unit disk as   0

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n=2
Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as   0

Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as   0
Then there exists   0 and a sequence i  0 such that, for each i,
there is a point zi D such that B (zi ) lies in the Fatou set.





Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as   0
Then there exists   0 and a sequence i  0 such that, for each i,
there is a point zi D such that B (zi ) lies in the Fatou set.





B (z1 )
B (z3 )

B (z2 )



B (z4 )
Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as   0
Then there exists   0 and a sequence i  0 such that, for each i,
there is a point zi D such that B (zi ) lies in the Fatou set.


The zi must accumulate on

some nonzero point, say z*, 
may assume that 
so we
B (z* )
lies in the Fatou set for all i.


B (z1 )
B (z3 )

B (z2 )



B (z4 )
Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as   0
Then there exists   0 and a sequence i  0 such that, for each i,
there is a point zi D such that B (zi ) lies in the Fatou set.

The zi must accumulate on

some nonzero point, say z*, 
may assume that 
so we
B (z* )
lies in the Fatou set for all i.



B (z* )

Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as   0
Then there exists   0 and a sequence i  0 such that, for each i,
there is a point zi D such that B (zi ) lies in the Fatou set.



The zi must accumulate on

some nonzero point, say z*, 
may assume that 
so we
B (z* )
lies in the Fatou set for all i.
2
F

z
But for large i,  i
k
F
B (z* )
so  i stretches

into an “annulus” that
surrounds the origin, so this

disconnects
the Julia set.

F k


B (z* )
So the Fatou components must become arbitrarily small:
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  0.0001
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  0.0000001
n > 2: Note the “round” annuli in the Fatou set; there is always
such an annulus of some fixed width for |  | small.

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  .01
  .0001
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  .00000001
  .000001



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  .0000000001
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  .000000000001
Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
B
A0

T
Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

B
A0
T
Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

~
A1 and A1 mapped to A0;
X1 is mapped to T
B
A1
A0
~
 mod A1 = mod A
1 = mod X1 = m/3;
~ T XX
11
A1
A1  S1
Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

~
A1 and A1 mapped to A0;
X1 is mapped to T
B
A1
A2
T X1 X2
A2 is mapped to A1;
X2 is mapped to X1
~
 mod A1 = mod A
A1  S1
1 = mod X1 = m/3;
 mod A2 = mod X2 = m/32; A2  S1
Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

~
A1 and A1 mapped to A0;
X1 is mapped to T
A2 is mapped to A1;
X2 is mapped to X1
B
Ak-1
Ak
N
Xk

M
Ak is mapped to Ak-1; Xk to Xk-1
~
 mod A1 = mod A1 = mod X1 = m/3;
A1  S1
 mod A2 = mod X2 = m/32; A2  S1
 mod Ak = mod Xk = m/3k; Ak  S1
Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

B
Ak-1
Ak
Xk
Eventually find k so
that 1  mod Ak   < 3
and Ak  S1




Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

B
Ak-1
Ak
Xk
Eventually find k so
that 1  mod Ak   < 3
and Ak  S1




 Ak must contain a round annulus of modulus  1/2 1/2
(Ble, Douady, and Henriksen)

Say n = 3:
|  | small  T is tiny
mod A0 = m is huge
and the boundary of A0
is very close to S1

B
Ak-1
Ak
Xk
Eventually find k so
that 1  mod Ak   < 3
and Ak  S1




 Ak must contain a round annulus of modulus  1/2 1/2
(Ble, Douady, and Henriksen)
But does this annulus have definite “thickness?”

 Ak in here

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 Ak in here

mod Ak  
says that the inner
boundary of Ak
cannot be inside  0

or outside  1 ,
so the round annulus
in Ak is
“thick”

Ak
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 
 Ak in here
Same argument
says that Ak  Xk
is twice as thick


Ak  Xk

and a
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 
So Xk must
have definite
thickness as well
Xk
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Part 4: A major application
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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