dynamic classification of escape time sierpinski curve julia sets dynamics of the family of complex...
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Dynamic Classification of Escape TimeSierpinski Curve Julia Sets
Dynamics of the family of complex maps
Paul BlanchardToni GarijoMatt HolzerU. HoomiforgotDan LookSebastian Marotta
with:
€
Fλ (z) = zn +λ
zn
Mark MorabitoMonica Moreno RochaKevin PilgrimElizabeth RussellYakov ShapiroDavid Uminsky
, n > 1
A Sierpinski curve is any planar set that is homeomorphic to the Sierpinski carpet fractal.
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The Sierpinski Carpet
Sierpinski Curve
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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.
Any planar, one-dimensional, compact, connected set can be homeomorphically embedded in a Sierpinski curve.
More importantly....
A Sierpinski curve is a universal plane continuum:
For example....
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This set
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can be embedded inside
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This set
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can be embedded inside
Moreover, Sierpinski curves occur all the time as Julia sets.
Dynamics of
complex and
€
Fλ (z) = z n +λ
z n
€
λ,z
€
n ≥ 2
A rational map of degree 2n.
Also a “singular perturbation” of zn.
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€
λ = 0
€
Fλ
( z ) = z2
+λ
z2
When , the Julia set is the unit circle
€
λ = 0
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€
λ = 0
€
Fλ
( z ) = z2
+λ
z2
€
λ ≠ 0
€
λ = −1 / 16
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But when , theJulia set explodes
When , the Julia set is the unit circle
€
λ = 0
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€
λ = 0
€
Fλ
( z ) = z2
+λ
z2
€
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
€
λ = 0
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€
λ = 0
€
Fλ
( z ) = z2
+λ
z2
€
λ ≠ 0But when , theJulia set explodes
€
λ = −0 . 01
Another Sierpinski curve
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When , the Julia set is the unit circle
€
λ = 0
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€
λ = 0
€
Fλ
( z ) = z2
+λ
z2
€
λ ≠ 0But when , theJulia set explodes
€
λ = −0 . 2
Also a Sierpinski curve
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When , the Julia set is the unit circle
€
λ = 0
Easy computations:
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€
Fλ (z ) = z 3 +λ
z 3
€
λ=.036+.026i
2n free critical points
€
cλ = λ1/2n
Easy computations:
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€
λ=.036+.026i
2n free critical points
€
cλ = λ1/2n€
Fλ (z ) = z 3 +λ
z 3
Easy computations:
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€
λ=.036+.026i
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
€
Fλ (z ) = z 3 +λ
z 3
Easy computations:
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€
λ=.036+.026i
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
€
Fλ (z ) = z 3 +λ
z 3
Easy computations:
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€
λ=.036+.026i
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
€
Fλ (z ) = z 3 +λ
z 3
Easy computations:
€
λ=.036+.026i
2n free critical points
€
cλ = λ1/2n
Only 2 critical values
€
vλ = ±2 λ
But really only 1 freecritical orbit since
the map is symmetricunder
€
Fλ (z ) = z 3 +λ
z 3
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€
z → −z
Easy computations:
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€
λ=.036+.026i
is superattracting, so have immediate basin Bmapped n-to-1 to itself.
€
∞ B€
Fλ (z ) = z 3 +λ
z 3
Easy computations:
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€
λ=.036+.026i
is superattracting, so have immediate basin Bmapped n-to-1 to itself.
B
T
€
Fλ (z ) = z 3 +λ
z 3
€
∞
0 is a pole, so havetrap door T mapped
n-to-1 to B.
Easy computations:
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€
λ=.036+.026i
is superattracting, so have immediate basin Bmapped n-to-1 to itself.
B
T
€
Fλ (z ) = z 3 +λ
z 3
€
∞
So any orbit that eventuallyenters B must do so by
passing through T.
0 is a pole, so havetrap door T mapped
n-to-1 to B.
The Escape Trichotomy
€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
There are three distinct ways the critical orbit can enter B:
The Escape Trichotomy
€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
T
€
vλ∈ is a Cantor set of
simple closed curves
€
J ( Fλ
)
There are three distinct ways the critical orbit can enter B:
(this case does not occur if n = 2)
€
⇒
(McMullen)
The Escape Trichotomy
€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
T
€
vλ∈ is a Cantor set of
simple closed curves
€
J ( Fλ
)
€
Fλ
k(v
λ) ∈ T
€
J ( Fλ
) is a Sierpinski curve
There are three distinct ways the critical orbit can enter B:
(this case does not occur if n = 2)
€
⇒
€
⇒
(McMullen)
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vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
Case 1:
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
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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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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€
vλ∈
€
J ( Fλ
)
€
⇒B is a Cantor set
parameter planewhen n = 3
J is a Cantor set
€
λ
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parameter planewhen n = 3
Case 2: the critical values lie in T, not B
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€
vλ∈
€
⇒T
parameter planewhen n = 3
€
λ lies in the McMullen domain
€
λ
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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Remark: There is no McMullen domain in the case n = 2.
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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€
vλ∈
€
⇒T
parameter planewhen n = 3
J is a Cantor set of simple closed curves
€
λ lies in the McMullen domain
€
λQuickTime™ and a
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
Fλ
k(v
λ) ∈
Case 3: the critical orbit eventually lands in the trap door.
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€
⇒T
parameter planewhen n = 3
J is an escape time Sierpinski curve
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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€
⇒T
parameter planewhen n = 3
€
λ lies in a Sierpinski hole
€
λ
€
Fλ
k(v
λ) ∈
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J is an escape time Sierpinski curve
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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.
Have an exact count of the number of Sierpinski holes:
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
Reason: The equation reduces to a polynomial of degree (n-1)(2n)(k-3) ;we have a Bottcher coordinate on each Sierpinski hole;and so all the roots of this polynomial are distinct. So we have exactly that many “centers” of Sierpinski holes, i.e., parameters for which the critical points all land on 0 and then on .
€
Fλk−1(λ1/2n ) = 0
€
∞
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
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n = 3escape time 32 Sierpinski holes
parameter planen = 3
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
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n = 3escape time 32 Sierpinski holes
parameter planen = 3
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
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n = 3escape time 412 Sierpinski holes
parameter planen = 3
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
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n = 3escape time 412 Sierpinski holes
parameter planen = 3
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
n = 4escape time 33 Sierpinski holes
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parameter planen = 4
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
n = 4escape time 424 Sierpinski holes
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parameter planen = 4
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Have an exact count of the number of Sierpinski holes:
n = 4escape time 12 402,653,184 Sierpinski holes
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Sorry. I forgot to indicate their locations. parameter plane
n = 4
Theorem (Roesch): Given n, there are exactly (n-1)(2n) Sierpinski holes with escape time k. (k-3)
Given two Sierpinski curve Julia sets, when do we know that the dynamics on them are the same, i.e., the maps are conjugate on the Julia sets?
Main Question:
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These sets are homeomorphic, but are the dynamics on them the same?
#1: If and are drawn from the same Sierpinski hole, then the correspondingmaps have the same dynamics, i.e., they are topologically conjugate on their Julia sets.€
λ
€
μ
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parameter planen = 4
#1: If and are drawn from the same Sierpinski hole, then the correspondingmaps have the same dynamics, i.e., they are topologically conjugate on their Julia sets.
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parameter planen = 4
€
λ
€
μ
#1: If and are drawn from the same Sierpinski hole, then the correspondingmaps have the same dynamics, i.e., they are topologically conjugate on their Julia sets.€
λ
€
μ
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parameter planen = 4
This uses quasiconformalsurgery techniques
#1: If and are drawn from the same Sierpinski hole, then the correspondingmaps have the same dynamics, i.e., they are topologically conjugate on their Julia sets.€
λ
€
μ
#2: If these parameters come from Sierpinski holes with different “escape times,” then the maps cannot be conjugate.
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€
Fλ (z ) = z 3 +λ
z 3
Two Sierpinski curve Julia sets, so they are homeomorphic.
€
cλ
€
cμ
€
Fμ (z ) = z 3 +μ
z 3
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escape time 3 escape time 4
So these maps cannot be topologically conjugate.
€
cλ
€
cμ
€
Fλ (z ) = z 3 +λ
z 3
€
Fμ (z ) = z 3 +μ
z 3
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is the only invariant boundary of an escapecomponent, so must be preserved by any conjugacy.
€
∂B
€
Fλ (z ) = z 3 +λ
z 3
€
Fμ (z ) = z 3 +μ
z 3
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is the only preimage of , so this curve must also be preserved by a conjugacy.
€
∂B
€
∂T
€
Fλ (z ) = z 3 +λ
z 3
€
Fμ (z ) = z 3 +μ
z 3
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If a boundary component is mapped toafter k iterations, its image under the conjugacy must also have this property,and so forth.....
€
∂T
€
Fλ (z ) = z 3 +λ
z 3
€
Fμ (z ) = z 3 +μ
z 3
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2-11-1
3-1
The curves around c are special; they are the only other ones in Jmapped 2-1 onto their images.
c
€
Fλ (z ) = z 3 +λ
z 3
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2-11-1
3-1
2-1
3-11-11-1
This bounding region takes 3 iterates to land on the
boundary of B.
But this bounding region takes 4 iterates to land, so
these maps are not conjugate.
€
Fλ (z ) = z 3 +λ
z 3
€
Fμ (z ) = z 3 +μ
z 3
For this it suffices to consider the centers of the Sierpinski holes; i.e., parameter values
for which for some k 3.
€
Fλk(cλ ) = ∞
#3: What if two maps lie in different Sierpinski holes that have the same escape time?
#3: What if two maps lie in different Sierpinski holes that have the same escape time?
For this it suffices to consider the centers of the Sierpinski holes; i.e., parameter values
for which for some k 3.
Two such centers of Sierpinski holes are “critically finite” maps, so by Thurston’s Theorem, if they are topologically
conjugate in the plane, they can be conjugated by a Mobius transformation (in the orientation preserving case).
€
Fλk(cλ ) = ∞
#3: What if two maps lie in different Sierpinski holes that have the same escape time?
For this it suffices to consider the centers of the Sierpinski holes; i.e., parameter values
for which for some k 3.
Two such centers of Sierpinski holes are “critically finite” maps, so by Thurston’s Theorem, if they are topologically
conjugate in the plane, they can be conjugated by a Mobius transformation (in the orientation preserving case).
Since and under the conjugacy,the Mobius conjugacy must be of the form .
€
∞ a ∞
€
0 a 0
€
z a αz
€
Fλk(cλ ) = ∞
€
h(Fλ (z)) = Fμ (h(z))then:
€
α zn +λ
zn
⎛
⎝ ⎜
⎞
⎠ ⎟=αzn +
αλ
zn=α nzn +
μ
α nzn
If we have a conjugacy
€
h(z) =αz
If we have a conjugacy
€
h(Fλ (z)) = Fμ (h(z))
€
h(z) =αz
then:
€
α zn +λ
zn
⎛
⎝ ⎜
⎞
⎠ ⎟=αzn +
αλ
zn=α nzn +
μ
α nzn
Comparing coefficients:
€
αn−1 =1
€
h(Fλ (z)) = Fμ (h(z))then:
€
α zn +λ
zn
⎛
⎝ ⎜
⎞
⎠ ⎟=αzn +
αλ
zn=α nzn +
μ
α nzn
Comparing coefficients:
€
αn−1 =1
€
μ =α2λ
If we have a conjugacy
€
h(z) =αz
€
h(Fλ (z)) = Fμ (h(z))then:
€
α zn +λ
zn
⎛
⎝ ⎜
⎞
⎠ ⎟=αzn +
αλ
zn=α nzn +
μ
α nzn
Comparing coefficients:
€
αn−1 =1
€
μ =α2λ
Easy check --- for the orientation reversing case:
is conjugate to via
€
Fλ
€
Fλ
€
h(z) = z
If we have a conjugacy
€
h(z) =αz
Theorem. If and are centers of Sierpinski holes, then iff or whereis a primitive root of unity; then any twoparameters drawn from these holes have the samedynamics.
€
Fλ ≈ Fμ
€
α
€
(n −1)st
€
μ
€
λ
n = 3: Only and are conjugatecenters since
€
λ
€
λ
€
αλ
€
α2λ
€
λ
€
λ
€
αλ
€
α2λ, , , , ,n = 4: Only
are conjugate centers where .
€
α 3 =1
€
μ =α2 jλ
€
μ =α2 jλ
€
α =−1
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n = 3, escape time 4, 12 Sierpinski holes,but only six conjugacy classes
€
λ
€
λconjugate centers: ,
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n = 3, escape time 4, 12 Sierpinski holes,but only six conjugacy classes
€
λ
€
λconjugate centers: ,
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n = 3, escape time 4, 12 Sierpinski holes,but only six conjugacy classes
€
λ
€
λconjugate centers: ,
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n = 3, escape time 4, 12 Sierpinski holes,but only six conjugacy classes
€
λ
€
λconjugate centers: ,
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€
αλ
€
α2λ
€
λ
€
λ
€
αλ
€
α2λ, , , , , where
€
α 3 =1
n = 4, escape time 4, 24 Sierpinski holes,but only five conjugacy classes
conjugate centers:
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€
αλ
€
α2λ
€
λ
€
λ
€
αλ
€
α2λ, , , , , where
€
α 3 =1
n = 4, escape time 4, 24 Sierpinski holes,but only five conjugacy classes
conjugate centers:
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€
αλ
€
α2λ
€
λ
€
λ
€
αλ
€
α2λ, , , , , where
€
α 3 =1
n = 4, escape time 4, 24 Sierpinski holes,but only five conjugacy classes
conjugate centers:
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€
αλ
€
α2λ
€
λ
€
λ
€
αλ
€
α2λ, , , , , where
€
α 3 =1
n = 4, escape time 4, 24 Sierpinski holes,but only five conjugacy classes
conjugate centers:
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€
αλ
€
α2λ
€
λ
€
λ
€
αλ
€
α2λ, , , , , where
€
α 3 =1
n = 4, escape time 4, 24 Sierpinski holes,but only five conjugacy classes
conjugate centers:
Theorem: For any n there are exactly (n-1) (2n) Sierpinski holes with escape time k. The number ofdistinct conjugacy classes is given by:
k-3
a. (2n) when n is odd;k-3
b. (2n) /2 + 2 when n is even.k-3 k-4
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For n odd, there are no Sierpinski holes along the real axis,so there are exactly n - 1 conjugate Sierpinski holes.
n = 3 n = 5
For n even, there is a “Cantor necklace” along the negativeaxis, so there are some “real” Sierpinski holes,
n = 4
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For n even, there is a “Cantor necklace” along the negativeaxis, so there are some “real” Sierpinski holes,
n = 4
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For n even, there is a “Cantor necklace” along the negativeaxis, so there are some “real” Sierpinski holes,
n = 4 magnification
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For n even, there is a “Cantor necklace” along the negativeaxis, so there are some “real” Sierpinski holes,
n = 4 magnification
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5 5
For n even, there is a “Cantor necklace” along the negativeaxis, so we can count the number of “real” Sierpinski holes,
and there are exactly n - 1 conjugate holes in this case:
n = 4 magnification
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5 5
For n even, there are also 2(n - 1) “complex” Sierpinski holes that have conjugate dynamics:
n = 4 magnification
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n = 4: 402,653,184 Sierpinski holes with escape time 12; 67,108,832 distinct conjugacy classes.
Sorry. I again forgot to indicate their locations.
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n = 4: 402,653,184 Sierpinski holes with escape time 12; 67,108,832 distinct conjugacy classes.
Problem: Describe the dynamics on these different conjugacy classes.
Other ways that Sierpinski curve Julia sets arise:
1. Buried points in Cantor necklaces;
2. Main cardioids in buried Mandelbrot sets;
3. Structure around the McMullen domain;
4. Other families of rational maps;
5. The difference between n = 2 and n >2 ;
6. Major applications
1. Cantor necklaces in the parameter plane
parameter plane n = 4
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1. Cantor necklaces in the parameter plane
parameter plane n = 4
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parameter plane n = 4
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The open disks are Sierpinski holes
34 4
5 5
1. Cantor necklaces in the parameter plane
parameter plane n = 4
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The open disks are Sierpinski holes;the buried points in the Cantor setalso correspond to Sierpinski curves;
34 4
5 5
1. Cantor necklaces in the parameter plane
parameter plane n = 4
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The open disks are Sierpinski holes;the buried points in the Cantor setalso correspond to Sierpinski curves;and all are dynamically different.
34 4
5 5
1. Cantor necklaces in the parameter plane
The “endpoints” in the Cantor set (parameters on the boundaries ofthe Sierpinski holes) do not correspond to Sierpinski curves.
1. Cantor necklaces in the parameter plane
A “hybrid” Sierpinski curve;some boundary curves meet.
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parameter plane n = 2
There are lots of other Cantornecklaces in the parameter planes.
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1. Cantor necklaces in the parameter plane
parameter plane n = 2
There are lots of other Cantornecklaces in the parameter planes.
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1. Cantor necklaces in the parameter plane
2. If 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
€
λ
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n = 4
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
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n = 4
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A Sierpinski curve, but very different dynamically from the earlier ones.
2. If lies in the main cardioid of a buried Mandelbrot set, then again the Julia set is a Sierpinski curve.
€
λ
n = 3
3. If n > 2, there are uncountably many simpleclosed curves surrounding the McMullen domain;
all these parameters are (non-escape) Sierpinski curves.
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n = 3
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3. If n > 2, there are uncountably many simpleclosed curves surrounding the McMullen domain;
all these parameters are (non-escape) Sierpinski curves.
n = 3
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All non-symmetric parameters on these curves have non-conjugate dynamics.
3. If n > 2, there are uncountably many simpleclosed curves surrounding the McMullen domain;
all these parameters are (non-escape) Sierpinski curves.
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γ2
10 centers
n = 3
If n > 2, there are also countably many different simple closed curves accumulating on the McMullen domain. Each alternately passes through centers of Sierpinski holes and centers of baby Mandelbrot sets (j > 1).
€
γj
€
(n − 2)n j +1
€
γj
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€
γ3
28 centers
n = 3
If n > 2, there are also countably many different simple closed curves accumulating on the McMullen domain. Each alternately passes through centers of Sierpinski holes and centers of baby Mandelbrot sets (j > 1).
€
γj
€
(n − 2)n j +1
€
γj
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€
γ4
82 centers
n = 3
If n > 2, there are also countably many different simple closed curves accumulating on the McMullen domain. Each alternately passes through centers of Sierpinski holes and centers of baby Mandelbrot sets (j > 1).
€
γj
€
(n − 2)n j +1
€
γj
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n = 3
€
γ13 passes through1,594,324 centers.
€
γ13
If n > 2, there are also countably many different simple closed curves accumulating on the McMullen domain. Each alternately passes through centers of Sierpinski holes and centers of baby Mandelbrot sets (j > 1).
€
γj
€
(n − 2)n j +1
€
γj
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As before, all non-symmetrically locatedcenters havedifferent dynamics.
n = 3
If n > 2, there are also countably many different simple closed curves accumulating on the McMullen domain. Each alternately passes through centers of Sierpinski holes and centers of baby Mandelbrot sets (j > 1).
€
γj
€
(n − 2)n j +1
€
γj
€
γ13
4. Consider the family of maps
where c is the center of a hyperbolic component of the Mandelbrot set.
€
Fλ (z ) = z 2 + c +λ
z 2
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λ =0
€
c = −1
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€
Fλ (z ) = z 2 + c +λ
z 2
€
c = −.12 +.75i
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4. Consider the family of maps
where c is the center of a hyperbolic component of the Mandelbrot set.
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λ =0
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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 Fatou set, then what’s left is another Sierpinski curve (provided that both of the critical orbits eventually escape).
The case n = 2 is very different from (and much more difficult than) the case n > 2.
n = 3 n = 2
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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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Another difference: as
n = 3 n = 2
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€
λ → 0,
€
Fλ2(cλ ) →∞
€
Fλ2(cλ ) →1/ 4
when n > 2when n = 2
Also, not much is happening for the Julia sets near when n > 2
n = 3
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λ =.01
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λ =0
The Julia set is always aCantor set of circles.
n = 3
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λ =.0001
n = 3
QuickTime™ and aTIFF (LZW) decompressor
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The Julia set is always aCantor set of circles.
QuickTime™ and aTIFF (LZW) decompressor
are needed to see this picture.
€
λ =.000001
The Julia set is always aCantor set of circles.
QuickTime™ and aTIFF (LZW) decompressor
are needed to see this picture.
€
λ =.000001
There is always a round annulus of some fixed width in the Fatou set,
so the Fatou set is “large.”
n = 2
But when n = 2, lots of things happen near the origin;in fact, the Julia sets converge to the unit disk as
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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.....