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Dynamics of the family of complex maps
Paul BlanchardToni GarijoMatt HolzerU. HoomiforgotDan LookSebastian Marotta
with:€
Fλ (z) = zn +λ
zn, n ≥ 2
(why the case n = 2 is )
Mark MorabitoMonica Moreno RochaKevin PilgrimElizabeth RussellYakov ShapiroDavid Uminsky
CR AZ Y
€
Fλ (z) = z n +λ
z n
The case n > 2 is great because:
€
Fλ (z) = z n +λ
z n
The case n > 2 is great because:
There exists a McMullen domain around = 0 ....
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Parameter plane for n=3
€
λ
€
Fλ (z) = z n +λ
z n
The case n > 2 is great because:
... surrounded by infinitely many “Mandelpinski” necklaces...
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There exists a McMullen domain around = 0 ....
€
λ
€
Fλ (z) = z n +λ
z n
The case n > 2 is great because:
... surrounded by infinitely many “Mandelpinski” necklaces...
... and the Julia sets behave nicely as
€
λ → 0
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€
λ =.01
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€
λ =.0001
€
λ =.000001
There exists a McMullen domain around = 0 ....
€
λ
€
Fλ (z) = z n +λ
z n
There is no McMullen domain....
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The case n = 2 is crazy because:
€
Fλ (z) = z n +λ
z n
There is no McMullen domain....
... and no “Mandelpinski” necklaces...
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The case n = 2 is crazy because:
€
Fλ (z) = z n +λ
z n
There is no McMullen domain....
... and the Julia sets “go crazy” as
€
λ → 0
... and no “Mandelpinski” necklaces...
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€
λ =−.000001
€
λ =−.0001
€
λ =−.01
The case n = 2 is crazy because:
Some definitions:
Julia set of
J = boundary of {orbits that escape to }
€
∞
= closure {repelling periodic orbits}
= {chaotic set}
Fatou set
= complement of J
= predictable set
€
Fλ (z) = zn +λ
zn
Computation of J:
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Color points that escape toinfinity shades of red orange yellow green blue violet Black points do not escape.J = boundary of the black region.
€
λ =.08i
€
Fλ (z ) = z 3 +λ
z 3
Easy computations:
€
λ =.08i
is superattracting, so have immediate basin Bmapped n-to-1 to itself.
€
∞€
Fλ (z ) = z 3 +λ
z 3
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B
Easy computations:
is superattracting, so have immediate basin Bmapped n-to-1 to itself.
€
∞€
Fλ (z ) = z 3 +λ
z 3
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B
T
€
λ =.08i
0 is a pole, so havetrap door T mapped
n-to-1 to B.
Easy computations:
is superattracting, so have immediate basin Bmapped n-to-1 to itself.
€
∞€
Fλ (z ) = z 3 +λ
z 3
0 is a pole, so havetrap door T mapped
n-to-1 to B.
The Julia set has 2n-fold symmetry.
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B
T
€
λ =.08i
Easy computations:
€
Fλ (z ) = z 3 +λ
z 3
2n free critical points
€
cλ = λ1/2n
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€
λ =.08i
Easy computations:
2n free critical points
€
cλ = λ1/2n
€
Fλ (z ) = z 3 +λ
z 3
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€
λ =.08i
Easy computations:
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
€
Fλ (z ) = z 3 +λ
z 3
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€
λ =.08i
Easy computations:
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
€
Fλ (z ) = z 3 +λ
z 3
€
λ =.08i
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Easy computations:
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
€
Fλ (z ) = z 3 +λ
z 3
€
λ =.08i
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Easy computations:
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
But really only 1 freecritical orbit
€
Fλ (z ) = z 3 +λ
z 3
€
λ =.08i
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Easy computations:
2n free critical points
€
pλ = (−λ )1/2n
Only 2 critical values
€
vλ = ±2 λ
But really only 1 freecritical orbit
€
Fλ (z ) = z 3 +λ
z 3
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€
λ =.08iAnd 2n prepoles
€
cλ = λ1/2n
The Escape Trichotomy
There are three possible ways that thecritical orbits can escape to infinity,
and each yields a different typeof Julia set.
(with D. Look & D Uminsky)
The Escape Trichotomy
€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
(with D. Look & D Uminsky)
The Escape Trichotomy
€
vλ∈
€
J ( Fλ
)
€
⇒B
€
⇒
is a Cantor set
T
€
vλ∈ is a Cantor set of
simple closed curves
€
J ( Fλ
)
(McMullen)
(with D. Look & D Uminsky)
(n > 2)
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
(McMullen)
(with D. Look & D Uminsky)
(n > 2)
In all other cases is a connected set, and if
€
J ( Fλ
)
€
vλ∈
€
J ( Fλ
)
€
⇒BCase 1:
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parameter planewhen n = 3
is a Cantor set
€
vλ∈
€
J ( Fλ
)
€
⇒BCase 1:
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parameter planewhen n = 3
€
λ
is a Cantor set
€
vλ∈
€
J ( Fλ
)
€
⇒BCase 1:
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parameter planewhen n = 3
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J is a Cantor set
€
λ
is a Cantor set
€
vλ∈
€
J ( Fλ
)
€
⇒BCase 1:
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parameter planewhen n = 3
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J is a Cantor set
€
λ
€
cλ
€
vλ
is a Cantor set
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parameter planewhen n = 3
Case 2: T
€
vλ∈
€
⇒
€
J ( Fλ
) is a Cantor set ofsimple closed curves
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parameter planewhen n = 3
€
λ
Case 2: T
€
vλ∈
€
⇒
€
J ( Fλ
) is a Cantor set ofsimple closed curves
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The central disk isthe McMullen domain
€
λ
Case 2: T
€
vλ∈ is a Cantor set of
simple closed curves
€
⇒
€
J ( Fλ
)
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parameter planewhen n = 3
€
λ
Case 2: T
€
vλ∈ is a Cantor set of
simple closed curves
J is a Cantor set of simple closed curves
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€
⇒
€
J ( Fλ
)
T
B
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parameter planewhen n = 3
€
λ
Case 2: T
€
vλ∈ is a Cantor set of
simple closed curves
J is a Cantor set of simple closed curves
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€
⇒
€
J ( Fλ
)
€
cλ
€
vλ
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parameter planewhen n = 3
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
) is a Sierpinski curve
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parameter planewhen n = 3
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
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A Sierpinski curve is a planarset homeomorphic to theSierpinski carpet fractal
is a Sierpinski curve
Sierpinski curves are important for two reasons:
1. There is a “topological characterization” of the carpet
2. A Sierpinski curve is a “universal plane continuum”
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The Sierpinski Carpet
Topological Characterization
Any planar set that is:1. compact2. connected3. locally connected4. nowhere dense5. any two complementary domains are bounded by simple closed curves that are pairwise disjoint is a Sierpinski curve.
Universal Plane Continuum
Any planar, one-dimensional, compact, connected set can be homeomorphically embedded in a Sierpinski curve.
For example....
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This set
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This set
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can be embedded inside
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parameter planewhen n = 3
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
) is a Sierpinski curve
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€
λ
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
) is a Sierpinski curve
A Sierpinski “hole”
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€
λ
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
A Sierpinski curve
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is a Sierpinski curve
A Sierpinski “hole”
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€
λ
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
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€
cλ
€
vλ
€
Fλ (vλ )
A Sierpinski curve
is a Sierpinski curve
A Sierpinski “hole”Escape time 3
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€
λ
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
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A Sierpinski curve
is a Sierpinski curve
Another Sierpinski “hole”
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€
λ
Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
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A Sierpinski curve
is a Sierpinski curve
Another Sierpinski “hole”
€
cλ
€
vλ
€
Fλ2(vλ )
Escape time 4
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Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
€
λ
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A Sierpinski curve
is a Sierpinski curve
Another Sierpinski “hole”Escape time 7
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Case 3:
€
⇒
€
Fλk(v λ ) ∈ T
€
J ( Fλ
)
€
λ
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A Sierpinski curve
is a Sierpinski curve
Another Sierpinski “hole”Escape time 5
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So to show that is homeomorphic to
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
Fatou set is the union of the preimages of B; all disjoint, open disks.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
Fatou set is the union of the preimages of B; all disjoint, open disks.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
If J contains an open set, then J = C.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
If J contains an open set, then J = C.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
No recurrent critical orbits and no parabolic points.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
No recurrent critical orbits and no parabolic points.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
J locally connected, so theboundaries are locally connected. Need to show they are s.c.c.’s. Can only meet at (preimages of) critical points, hence disjoint.
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Need to show:
compactconnectednowhere denselocally connectedbounded by disjoint s.c.c.’s
So J is a Sierpinski curve.
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Remark: All Julia sets drawn from Sierpinski holes are homeomorphic, but only those in symmetrically locatedSierpinski holes have the same dynamics.
The maps on all these Julia sets are dynamically different.
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Remark: All Julia sets drawn from Sierpinski holes are homeomorphic, but only those in symmetrically locatedSierpinski holes have the same dynamics.
In fact, there are exactly (n-1)(2n)k-3 Sierpinski holes withescape time k, and (2n)k-3 different conjugacy classes (n odd).
The maps on all these Julia sets are dynamically different.
(with K.Pilgrim)
1. The McMullen domain
2. Mandelpinski necklaces
3. Julia sets near 0
Topics
Part 1: The McMullen Domain
Why is there no McMullen domain when n = 2?
What is the preimage of T?
First suppose
€
vλ∈ T:
Why is there no McMullen domain when n = 2?
First suppose
What is the preimage of T?
€
vλ∈ 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
What is the preimage of T?
€
vλ∈ T:
Can the preimage be 2n disjoint disks, each of which contains a critical point?
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No --- there would then be 4n preimages of any point in T, but the map has degree 2n.
Why is there no McMullen domain when n = 2?
So some of the preimages of Tmust overlap, and by 2n-foldsymmetry, all must intersect.
Why is there no McMullen domain when n = 2?
So some of the preimages of Tmust overlap, and by 2n-foldsymmetry, all must intersect.
By Riemann-Hurwitz, the preimage of T must then
be an annulus.
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€
cλ
€
vλ
Why is there no McMullen domain when n = 2?
So here is the picture:
T
B
A
A is the annulus separating B and T
Why is there no McMullen domain when n = 2?
So here is the picture:B
TA
A is the annulus separating B and T
XF maps X 2n-to-1
onto T
Why is there no McMullen domain when n = 2?
So here is the picture:B
TA
A is the annulus separating B and T
X
A
A
0
1
F maps both A0
1and A as an n-to-1
covering onto A
F maps X 2n-to-1onto T
Why is there no McMullen domain when n = 2?
B
TA
X
A
A
0
1
So mod(A0) = 1/n mod(A)
And mod(A1) = 1/n mod(A)
Why is there no McMullen domain when n = 2?
B
TA
X
A
A
0
1
So mod(A0) = 1/n mod(A)
And mod(A1) = 1/n mod(A)
When n = 2,mod(A0) + mod(A1) =
mod(A)
Why is there no McMullen domain when n = 2?
B
TA
X
A
A
0
1
So mod(A0) = 1/n mod(A)
And mod(A1) = 1/n mod(A)
When n = 2,mod(A0) + mod(A1) =
mod(A)
So there is no room for X, i.e., does
not lie in T
€
vλ
€
vλ = ±2 λ
€
Fλ (vλ ) = 2nλn /2 +λ
2nλn /2 = 2nλn /2 +1
2nλ (n /2−1)
Why is there no McMullen domain when n = 2?
Here is another reason:
€
Fλ (vλ ) = 2nλn /2 +λ
2nλn /2 = 2nλn /2 +1
2nλ (n /2−1)
Why is there no McMullen domain when n = 2?
Here is another reason:
€
n > 2⇒ Fλ (vλ ) → ∞ as λ → 0
€
vλso lies in T when n > 2
€
vλ = ±2 λ
Why is there no McMullen domain when n = 2?
Here is another reason:
€
n > 2⇒ Fλ (vλ ) → ∞ as λ → 0
€
n = 2⇒ Fλ (vλ ) = 4λ +1
4€
vλso lies in T when n > 2
€
vλ = ±2 λ
€
Fλ (vλ ) = 2nλn /2 +λ
2nλn /2 = 2nλn /2 +1
2nλ (n /2−1)
€
Fλ (vλ ) = 2nλn /2 +λ
2nλn /2 = 2nλn /2 +1
2nλ (n /2−1)
Why is there no McMullen domain when n = 2?
Here is another reason:
€
n > 2⇒ Fλ (vλ ) → ∞ as λ → 0
€
so Fλ (vλ ) →1/ 4 as λ → 0(???)€
vλso lies in T when n > 2
€
vλ = ±2 λ
€
n = 2⇒ Fλ (vλ ) = 4λ +1
4
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
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.
C3 passes through thecenters of 10 M-sets
and 10 S-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.
C4 passes through thecenters of 28 M-sets
and 28 S-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.
C5 passes through thecenters of 82 M-sets
and 82 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,...,∞
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 +λ
z 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 +λ
z 3
€
λ =.08i
All of the criticalpoints and prepoleslie on the “criticalcircle” : |z| = | |
€
λ 1/2n
€
γ0
€
γ0
which is mapped 2n-to-1onto the “critical value line”
connecting
€
±vλ €
vλ
€
−vλ
Easy computations:
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€
Fλ (z ) = z 3 +λ
z 3
€
λ =.08i
€
γ0
€
vλ
Any other circle around 0is mapped n-to-1 to an ellipse
whose foci are
€
±2 λ
Easy computations:
€
−vλ
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€
Fλ (z ) = z 3 +λ
z 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 +λ
z 3
€
λ =.08i
€
γ0
€
vλ
€
±2 λ
So the exterior of is mapped as an n-to-1 covering of the
exterior of the critical value line. €
γ0
Easy computations:
€
−vλ
Any other circle around 0is mapped n-to-1 to an ellipse
whose foci are
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€
Fλ (z ) = z 3 +λ
z 3
€
λ =.08i
€
γ0
€
vλ
€
±2 λ
So the exterior of is mapped as an n-to-1 covering of the
exterior of the critical value line. Same with the interior of . €
γ0
€
γ0
Easy computations:
€
−vλ
Any other circle around 0is mapped n-to-1 to an ellipse
whose foci are
Simplest case: C1
Assume that both sit on the critical circle.
€
±vλ
€
vλ
€
−vλ
€
γ0
€
pλ
(i.e., )
€
| vλ | = | cλ | = | pλ |
€
⇒ 2 | λ |1/2 = | λ |1/2n
Assume that both sit on the critical circle.
€
±vλ
€
vλ
€
−vλ
€
γ0
€
pλ
Simplest case: C1
€
⇒ 2 | λ |1/2 = | λ |1/2n
€
⇒ 22n | λ |n = | λ |
€
⇒ | λ | = 2−
2n
n−1 ⎛ ⎝
⎞ ⎠
Assume that both sit on the critical circle.
€
±vλ
This is the “dividing circle” in the parameter plane
Parameter plane n = 3
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r = 2-3
Simplest case: C1
€
⇒ 2 | λ |1/2 = | λ |1/2n
€
⇒ 22n | λ |n = | λ |
€
⇒ | λ | = 2−
2n
n−1 ⎛ ⎝
⎞ ⎠
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Assume that both sit on the critical circle.
€
±vλ
€
⇒ 2 | λ |1/2 = | λ |1/2n
€
⇒ 22n | λ |n = | λ |
This is the “dividing circle” in the parameter plane
Parameter plane n = 4
€
⇒ | λ | = 2−
2n
n−1 ⎛ ⎝
⎞ ⎠
r = 2-8/3
Simplest case: C1
Assume that lies on the dividing circle,so both sit on the critical circle.
€
±vλ
€
vλ
€
−vλ
€
γ0
€
λ
€
pλ
In this picture, is real and n = 3
€
λ
Simplest case: C1
Assume that lies on the dividing circle,so both sit on the critical circle.
€
±vλ
€
vλ
€
−vλ
€
γ0
As rotates around the dividing circle, rotates a half-turn, while and rotate1/2n of a turn. So each meets exactlyn - 1 prepoles and critical points.€
λ
€
±vλ
€
pλ
€
vλ
€
λ
€
pλ€
cλ
In this picture, is real and n = 3
€
λ
Simplest case: C1
Assume that lies on the dividing circle,so both sit on the critical circle.
€
±vλ
€
vλ
€
−vλ
€
γ0
As rotates around the dividing circle, rotates a half-turn, while and rotate1/2n of a turn. So each meets exactlyn - 1 prepoles and critical points.€
λ
€
±vλ
€
pλ
€
vλ
€
λ
€
pλ€
cλ
In this picture, is real and n = 3
€
λ
Simplest case: C1
So the dividing circle is C1
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The dividing circle when n = 5
n-1 = 4 centers of Sierpinski holes;n-1 = 4 centers of baby Mandelbrot sets
Now assume that lies inside the critical circle:
€
±vλ
€
γ0€
vλ
€
−vλ
€
γ0
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€
vλ
€
−vλ
€
γk
As rotates by one turn, these 2nk+1 points on each rotate by 1/2nk+1 of a turn.
€
γk
€
λ
€
γ0€
vλ
€
−vλ
€
γk
€
Fλ (vλ ) ≈λ
2nλn /2 =1
2nλ1−n /2
Since
the second iterate of the criticalpoints rotate by 1 - n/2 ofa turn
As rotates by one turn, these 2nk+1 points on each rotate by 1/2nk+1 of a turn.
€
γk
€
λ
€
γ0€
vλ
€
−vλ
€
γk
€
Fλ (vλ ) ≈λ
2nλn /2 =1
2nλ1−n /2
Since
the second iterate of the criticalpoints rotate by 1 - n/2 ofa turn, 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 )
The real proof involves the Schwarz Lemma:
€
γ0€
vλ
€
−vλ
€
γk
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€
γkλ (θ )
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There is a natural parametrization of each
€
γk
€
γkλ (θ )
The real proof involves the Schwarz Lemma:
€
γ0€
vλ
€
−vλ
€
γk
€
γkλ (θ )
Best to restrict to a “symmetry region” inside the dividingcircle, so that is well-defined.
€
γkλ (θ )
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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
Best to restrict to a “symmetry region” inside the dividingcircle, so that is well-defined.
€
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
€
G−1(γ kλ (θ ) )
a map from a “disk” to itself.
So consider the composition
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€
γ0€
vλ
€
−vλ
€
γk
€
γkλ (θ )
€
G−1
€
G−1(γ kλ (θ ) )
a map from a “disk” to itself.
So consider the composition
Schwarz implies that has a unique fixed point,i.e., a parameter for which the second iterate of the criticalpoint lands on the point , so this proves theexistence of the centers of the S-holes and M-sets.€
G−1(γ kλ (θ ) )
€
γkλ (θ )
Remarks: This proves the existence of centers of Sierpinski holes and Mandelbrot sets. Producingthe entire M-set involves polynomial-like maps;while the entire S-hole involves qc-surgery.
Part 3: Behavior of the Julia sets
n = 2: When , the Julia set of is the unit circle. But, as , the Julia set of converges to the closed unit disk
€
Fλ
€
λ → 0
€
λ =0
€
Fλ
Part 3: Behavior of the Julia sets
n > 2: J is always a Cantor set of “circles” when is small.
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
€
λ
Part 3: Behavior of the Julia sets
n > 2: J is always a Cantor set of “circles” when is small.
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
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
Part 3: Behavior of the Julia sets
n = 2
Theorem: When n = 2, the Julia sets converge to the unit disk as
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€
λ → 0
Suppose the Julia sets do not converge to the unit disk D as
€
λ → 0
Sketch of the proof:
Suppose the Julia sets do not converge to the unit disk D as
€
λ → 0
Sketch of the proof:
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
Suppose the Julia sets do not converge to the unit disk D as
€
λ → 0
Sketch of the proof:
Then there exists and a sequence such that, for each i,there is a point such that lies in the Fatou set.
€
Bδ (zi )
€
λi → 0
€
zi ∈D
€
Bδ (z1)
€
Bδ (z2 )
€
Bδ (z3)
€
Bδ (z4 )
€
δ > 0
Suppose the Julia sets do not converge to the unit disk D as
€
λ → 0
Sketch of the proof:
Then there exists and a sequence such that, for each i,there is a point such that lies in the Fatou set.
€
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
Suppose the Julia sets do not converge to the unit disk D as
€
λ → 0
Sketch of the proof:
Then there exists and a sequence such that, for each i,there is a point such that lies in the Fatou set.
€
Bδ (zi )
€
λi → 0
€
zi ∈D
€
Bδ (z*)
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*)
€
δ > 0
Suppose the Julia sets do not converge to the unit disk D as
€
λ → 0
Sketch of the proof:
Then there exists and a sequence such that, for each i,there is a point such that lies in the Fatou set.
€
Bδ (zi )
€
λi → 0
€
zi ∈D
€
Bδ (z*)
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*)
But for large i, so stretchesinto an “annulus” that surrounds the origin, so thisdisconnects the Julia set.€
Fλ i ≈ z2
€
Fλ ik
€
Bδ (z*)€
Fλk
€
δ > 0
So the Fatou components must become arbitrarily small:
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€
λ =−0.0001
€
λ =−0.0000001
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€
λ =.01
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€
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.
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€
λ =.00000001
€
λ =.0000000001
€
| λ |
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€
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
€
⇒ 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
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
€
⇒ T is tinymod A0 = m is hugeand the boundary of A0
is very close to S1
Say n = 3:
A1 and A1 mapped to A0;X1 is mapped to T
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
€
⇒ T is tinymod A0 = m is hugeand the boundary of A0
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
€
⇒ T is tinymod A0 = m is hugeand the boundary of A0
is very close to S1
Say n = 3:
Xk
Ak
Ak-1
1 mod Ak < 3
€
≤Eventually find k sothatand Ak S1
€
≈
€
∂
€
⇒ Ak must contain a round annulus of modulus
(Ble, Douady, and Henriksen)
€
=α
€
α −1/2 >1/2
B
€
| λ | small
€
⇒ T is tinymod A0 = m is hugeand the boundary of A0
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
€
≤Eventually find k sothatand Ak S1
€
≈
€
∂
€
=α
€
α −1/2 >1/2
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€
∂Ak in here
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€
∂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
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€
∂Ak in here
€
μ1
€
μ0
Ak
Same argumentsays that Ak Xk
is twice as thick
€
∪
€
∪ Xk
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Xk
So Xk musthave definite thickness as well
Part 4: A major application
Here’s the parameter plane when n = 2:
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Qu
ickTim
e™
an
d a
TIF
F (
LZ
W)
decom
pre
sso
rare
nee
de
d t
o s
ee t
his
pic
ture
.
Rotate it by 90 degrees:
and this object appears everywhere.....
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