Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer Dan Look Sebastian Marotta Mark Morabito with: Monica Moreno Rocha Kevin Pilgrim Elizabeth Russell Yakov Shapiro David Uminsky Sum Wun Ellce Dynamics of Singularly Perturbed Rational Maps

1. The Escape Trichotomy Cantor setSierpinski curveCantor set of circles Dynamics of Singularly Perturbed Rational Maps

2. Classification of escape time Sierpinski curve Julia sets All are Sierpinski curve Julia sets, but with very different dynamics Dynamics of Singularly Perturbed Rational Maps

3. Julia sets converging to the unit disk Dynamics of Singularly Perturbed Rational Maps = -.01 = =

4. Cantor necklaces and webs Dynamics of Singularly Perturbed Rational Maps

These lectures will deal with the dynamics of the family of complex maps where c is the center of a hyperbolic component of the Multibrot set

These lectures will deal with the dynamics of the family of complex maps where c is the center of a hyperbolic component of the Multibrot set But for simplicity, we’ll concentrate for the most part on the easier family

Why the interest in these maps?

First, these are singular perturbations of.

Why the interest in these maps? First, these are singular perturbations of. When but when, the dynamical behavior “explodes.” we completely understand the dynamics of

Second, how do you solve the equation ? Why the interest in these maps? First, these are singular perturbations of.

Why the interest in these maps? First, these are singular perturbations of. Second, how do you solve the equation ? You use Newton’s method (of course!):

Why the interest in these maps? First, these are singular perturbations of. Second, how do you solve the equation ? You use Newton’s method (of course!): Iterate:

Why the interest in these maps? First, these are singular perturbations of. Second, how do you solve the equation ? You use Newton’s method (of course!): Iterate: a singular perturbation of z/2

Why the interest in these maps? First, these are singular perturbations of. Second, how do you solve the equation ? You use Newton’s method (of course!): Whenever the equation has a multiple root, the corresponding Newton’s method involves a singular perturbation.

Why the interest in these maps? First, these are singular perturbations of. Second, how do you solve the equation ? Third, we are looking at maps on the boundary of the set of rational maps of degree 2n --- a very interesting topic of contemporary research.

Dynamics of complex and A rational map of degree 2n.

Dynamics of complex and The Julia set is: The closure of the set of repelling periodic points; The boundary of the escaping orbits; The chaotic set. The Fatou set is the complement of. A rational map of degree 2n.

When, the Julia set is the unit circle

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Colored points have orbits that escape to infinity: Escape time: red (fastest) orange yellow green blue violet (slowest)

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle Black points have orbits that do not escape to infinity:

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

When, the Julia set is the unit circle The Julia set is the boundary of the black & colored regions.

But when, the Julia set explodes When, the Julia set is the unit circle

But when, the Julia set explodes A Sierpinski curve When, the Julia set is the unit circle

But when, the Julia set explodes Another Sierpinski curve When, the Julia set is the unit circle

But when, the Julia set explodes Also a Sierpinski curve When, the Julia set is the unit circle

Easy computations: J(F ) has 2n-fold symmetry

Easy computations: 2n free critical points J(F ) has 2n-fold symmetry

Easy computations: 2n free critical points J(F ) has 2n-fold symmetry

Easy computations: 2n free critical points Only 2 critical values J(F ) has 2n-fold symmetry

Easy computations: 2n free critical points Only 2 critical values J(F ) has 2n-fold symmetry

Easy computations: 2n free critical points Only 2 critical values J(F ) has 2n-fold symmetry

Easy computations: 2n free critical points Only 2 critical values But really only 1 free critical orbit since the map has 2n-fold symmetry J(F ) has 2n-fold symmetry

Easy computations: is superattracting, so have immediate basin B mapped n-to-1 to itself. B

Easy computations: is superattracting, so have immediate basin B mapped n-to-1 to itself. B T 0 is a pole, so have trap door T mapped n-to-1 to B.

Easy computations: is superattracting, so have immediate basin B mapped n-to-1 to itself. B T So any orbit that eventually enters B must do so by passing through T. 0 is a pole, so have trap door T mapped n-to-1 to B.

Part 1. The Escape Trichotomy Dynamics of the family of complex maps Paul Blanchard Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta Monica Moreno Rocha Yakov Shapiro David Uminsky with:

There are three distinct ways the critical orbit can enter B: 1. The Escape Trichotomy

B is a Cantor set There are three distinct ways the critical orbit can enter B: 1. The Escape Trichotomy

B is a Cantor set T is a Cantor set of simple closed curves There are three distinct ways the critical orbit can enter B: (this case does not occur if n = 2) 1. The Escape Trichotomy

B is a Cantor set T is a Cantor set of simple closed curves T is a Sierpinski curve There are three distinct ways the critical orbit can enter B: (this case does not occur if n = 2) 1. The Escape Trichotomy

B is a Cantor set parameter plane when n = 3 Case 1:

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set parameter plane when n = 3 J is a Cantor set

B is a Cantor set Draw symmetric curves connecting the two critical values to in B v1v1 v2v2

B is a Cantor set The preimages are curves passing through the critical points and connecting c to v1v1 v2v2

B is a Cantor set The preimages are curves passing through the critical points and connecting c to and to 0

B is a Cantor set Choose a large circle in B

B is a Cantor set and locate its two preimages

B is a Cantor set Construct the regions I 0,... I 2n-1 I0I0 I1I1 I3I3 I2I2 I4I4 I5I5

B is a Cantor set I0I0 I1I1 I3I3 I2I2 I4I4 I5I5 Each of the regions I 0,... I 2n-1 is mapped 1-1 over the union of the I j

B is a Cantor set I0I0 I1I1 I3I3 I2I2 I4I4 I5I5 Each of the regions I 0,... I 2n-1 is mapped 1-1 over the union of the I j

B is a Cantor set I0I0 I1I1 I3I3 I2I2 I4I4 I5I5 Each of the regions I 0,... I 2n-1 is mapped 1-1 over the union of the I j

B is a Cantor set I0I0 I1I1 I3I3 I2I2 I4I4 I5I5 Each of the regions I 0,... I 2n-1 is mapped 1-1 over the union of the I j

B is a Cantor set I0I0 I1I1 I3I3 I2I2 I4I4 I5I5 Each of the regions I 0,... I 2n-1 is mapped 1-1 over the union of the I j is a Cantor set

parameter plane when n = 3 Case 2: the critical values lie in T, not B

T parameter plane when n = 3 lies in the McMullen domain

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain Remark: There is no McMullen domain in the case n = 2.

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain

T parameter plane when n = 3 J is a Cantor set of simple closed curves lies in the McMullen domain

T J is a Cantor set of “circles” Why is the preimage of T an annulus? c v

Could it be that each critical point lies in a disjoint preimage of T?

No. The map would then be 4n to 1 on the preimage of T.

By 2n-fold symmetry, there can then be only one preimage of T Could it be that each critical point lies in a disjoint preimage of T?

Riemann-Hurwitz formula: F: D R holomorphic branched covering n(D), n(R) = number of boundary components of D, R

Riemann-Hurwitz formula: F: D R holomorphic branched covering n(D), n(R) = number of boundary components of D, R then: n(D) - 2 = (deg F) (n(R) - 2) + (# of critical points in D) *

Riemann-Hurwitz formula: then: n(D) - 2 = (deg F) (n(R) - 2) + (# of critical points in D) * In our case, R is the trap door and D = ??? is the preimage of T F: D R holomorphic branched covering n(D), n(R) = number of boundary components of D, R

Riemann-Hurwitz formula: then: n(D) - 2 = (deg F) (n(R) - 2) + (# of critical points in D) * We have deg F = 2n on D, so F: D R holomorphic branched covering n(D) - 2 = ( 2n ) (n(R) - 2) + (# of critical points in D) * n(D), n(R) = number of boundary components of D, R

Riemann-Hurwitz formula: then: n(D) - 2 = (deg F) (n(R) - 2) + (# of critical points in D) * and R = T is a disk, so n(R) = 1 F: D R holomorphic branched covering n(D) - 2 = ( 2n ) (1 - 2) + (# of critical points in D) * n(D), n(R) = number of boundary components of D, R

Riemann-Hurwitz formula: then: n(D) - 2 = (deg F) (n(R) - 2) + (# of critical points in D) * F: D R holomorphic branched covering and there are 2n critical points in D n(D) - 2 = ( 2n ) (1 - 2) + ( 2n ) = 0 * n(D), n(R) = number of boundary components of D, R

Riemann-Hurwitz formula: then: n(D) - 2 = (deg F) (n(R) - 2) + (# of critical points in D) * F: D R holomorphic branched covering and there are 2n critical points in D n(D) - 2 = ( 2n ) (1 - 2) + ( 2n ) = 0 * n(D), n(R) = number of boundary components of D, R so n(D) = 2 and the preimage of T is an annulus.

T J is a Cantor set of “circles” So the preimage of T is an annulus. c v

T J is a Cantor set of “circles” B and T are mapped n-to-1 to B B T

T J is a Cantor set of “circles” The white annulus is mapped 2n-to-1 to T B T

T J is a Cantor set of “circles” So all that’s left are the blue annuli, and each are mapped n-to-1 to the union of the blue and white annuli. B T

T So there are sub-annuli in the blue annuli that are mapped onto the white annulus. B T J is a Cantor set of “circles”

T 2n to 1 J is a Cantor set of “circles”

T 2n to 1 n to 1

T J is a Cantor set of “circles” then all other preimages of F -1 (T) contain no critical points, and F is an n - to -1 covering on each, so the remaining preimages of T are all annuli. 2n to 1 n to 1

T J is a Cantor set of “circles” 2n to 1 n to 1 These annuli fill out the Fatou set; removing all of them leaves us with a Cantor set of simple closed curves (McMullen)

Curiously, this cannot happen when n = 2. One reason involves the moduli of the annuli in the preimages of T.

A is mapped 4-to-1 to T B T A Case n = 2

A 0 and A 1 are mapped 2-to-1 to A  A 0  A 1 B T A Case n = 2 A0A0 A1A1

So mod(A 0 ) = mod( A  A 0  A 1 ) and mod(A 1 ) = mod( A  A 0  A 1 ) B T A Case n = 2 A0A0 A1A1

B T A A0A0 A1A1 So mod(A 0 ) = mod( A  A 0  A 1 ) and mod(A 1 ) = mod( A  A 0  A 1 ) so there is no room in the middle for A

Another reason this does not happen: The critical values are When n > 2 we have

Another reason this does not happen: The critical values are so the critical values lie in T when is small. When n > 2 we have

Another reason this does not happen: The critical values are But when n = 2 we have

Another reason this does not happen: The critical values are So, as, and 1/4 is nowhere near the basin of when is small. But when n = 2 we have

Parameter planes n = 2 n = 3 No McMullen domainMcMullen domain

Parameter planes n = 2 n = 3 No McMullen domainMcMullen domain

n = 2 n = 3 No McMullen domainMcMullen domain

n = 2 n = 3 No McMullen domainMcMullen domain

There is a lot of structure around the McMullen domain when n > 2 but a very different structure when n = 2.

Case 3: the critical orbit eventually lands in the trap door is a “Sierpinski curve.”

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

The Sierpinski Carpet 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.

T parameter plane when n = 3 lies in a Sierpinski hole Case 3: the critical orbit eventually lands in the trap door.

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

T parameter plane when n = 3 J is a Sierpinski curve lies in a Sierpinski hole

Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k. Have an exact count of the number of Sierpinski holes:

Reason: The equation reduces to a polynomial of degree (n-1)(2n) (k-3), and it can be shown that all the roots of this polynomial are distinct. (You can put a Böttcher coordinate on each Sierpinski hole). Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: Reason: The equation reduces to a polynomial of degree (n-1)(2n) (k-3), and it can be shown that all the roots of this polynomial are distinct. (You can put a Böttcher coordinate on each Sierpinski hole). So we have exactly that many “centers” of Sierpinski holes, i.e., parameters for which the critical points all land on 0 and then. Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 3 escape time 3 2 Sierpinski holes parameter plane n = 3 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 3 escape time 3 2 Sierpinski holes parameter plane n = 3 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 3 escape time 4 12 Sierpinski holes parameter plane n = 3 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 3 escape time 4 12 Sierpinski holes parameter plane n = 3 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 4 escape time 3 3 Sierpinski holes parameter plane n = 4 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 4 escape time 4 24 Sierpinski holes parameter plane n = 4 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

Have an exact count of the number of Sierpinski holes: n = 4 escape time ,653,184 Sierpinski holes Sorry. I forgot to indicate their locations. parameter plane n = 4 Theorem (Roesch): Given n, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k.

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: These sets are homeomorphic, but are the dynamics on them the same?

Part 2. Dynamic Classification of Escape Time Sierpinski Curve Julia Sets J ( F ) J ( F  ) When do F and F  have the same (conjugate) dynamics?

#1: If and are drawn from the same Sierpinski hole, then the corresponding maps have the same dynamics, i.e., they are topologically conjugate on their Julia sets. parameter plane n = 4

#1: If and are drawn from the same Sierpinski hole, then the corresponding maps have the same dynamics, i.e., they are topologically conjugate on their Julia sets. So all these parameters have the same dynamics on their Julia sets. parameter plane n = 4

#1: If and are drawn from the same Sierpinski hole, then the corresponding maps have the same dynamics, i.e., they are topologically conjugate on their Julia sets. parameter plane n = 4 This uses quasiconformal surgery techniques

#1: If and are drawn from the same Sierpinski hole, then the corresponding maps have the same dynamics, i.e., they are topologically conjugate on their Julia sets. #2: If these parameters come from Sierpinski holes with different “escape times,” then the maps cannot be conjugate.

Two Sierpinski curve Julia sets, so they are homeomorphic.

escape time 3escape time 4 So these maps cannot be topologically conjugate.

is the only invariant boundary of an escape component, so must be preserved by any conjugacy.

is the only preimage of, so this curve must also be preserved by a conjugacy.

If a boundary component is mapped to after k iterations, its image under the conjugacy must also have this property, and so forth.....

The curves around c are special; they are the only other ones in J mapped 2-1 onto their images. c

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. c

For this it suffices to consider the centers of the Sierpinski holes; i.e., parameter values for which for some k  3. #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).

#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 0  0 under the conjugacy, the Mobius conjugacy must be of the form z   z.

then: If we have a conjugacy

then: Comparing coefficients:

then: Comparing coefficients: If we have a conjugacy

then: Comparing coefficients: Easy check --- for the orientation reversing case: is conjugate to via If we have a conjugacy

Theorem. If and are centers of Sierpinski holes, then iff or where is a primitive root of unity; then any two parameters drawn from these holes have the same dynamics. n = 3:Only and are conjugate centers since,,,,,n = 4:Only are conjugate centers where.

n = 3, escape time 4, 12 Sierpinski holes, but only six conjugacy classes conjugate centers:,

n = 3, escape time 4, 12 Sierpinski holes, but only six conjugacy classes conjugate centers:,

n = 3, escape time 4, 12 Sierpinski holes, but only six conjugacy classes conjugate centers:,

n = 3, escape time 4, 12 Sierpinski holes, but only six conjugacy classes conjugate centers:,

,,,,,where n = 4, escape time 4, 24 Sierpinski holes, but only five conjugacy classes conjugate centers:

,,,,,where n = 4, escape time 4, 24 Sierpinski holes, but only five conjugacy classes conjugate centers:

,,,,,where n = 4, escape time 4, 24 Sierpinski holes, but only five conjugacy classes conjugate centers:

,,,,,where n = 4, escape time 4, 24 Sierpinski holes, but only five conjugacy classes conjugate centers:

,,,,,where 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 of distinct conjugacy classes is given by: k-3 a. (2n) when n is odd; k-3 b. (2n) /2 + 2 when n is even. k-3k-4

For n odd, there are no Sierpinski holes along the real axis, so there are exactly n - 1 conjugate Sierpinski holes. n = 3n = 5

For n even, there is a “Cantor necklace” along the negative axis, so there are some “real” Sierpinski holes, n = 4

For n even, there is a “Cantor necklace” along the negative axis, so there are some “real” Sierpinski holes, n = 4

For n even, there is a “Cantor necklace” along the negative axis, so there are some “real” Sierpinski holes, n = 4magnification M

For n even, there is a “Cantor necklace” along the negative axis, so there are some “real” Sierpinski holes, n = 4magnification M

For n even, there is a “Cantor necklace” along the negative axis, so we can count the number of “real” Sierpinski holes, and there are exactly n - 1 conjugate holes in this case: n = 4magnification M

For n even, there are also 2(n - 1) “complex” Sierpinski holes that have conjugate dynamics: n = 4magnification M

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.

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.

Part 3: Julia sets converging to the unit disk With Toni Garijo, Mark Morabito, and Robert T. Kozma

n = 2: When, the Julia set of is the unit circle. But, as, the Julia set of converges to the closed unit disk With Toni Garijo, Mark Morabito, and Robert T. Kozma Part 3: Julia sets converging to the unit disk

n = 2: When, the Julia set of is the unit circle. But, as, the Julia set of converges to the closed unit disk n > 2: J is always a Cantor set of “circles” when is small. With Toni Garijo, Mark Morabito, and Robert T. Kozma Part 3: Julia sets converging to the unit disk

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 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. With Toni Garijo, Mark Morabito, and Robert T. Kozma Part 3: Julia sets converging to the unit disk

n = 2 Theorem: When n = 2, the Julia sets converge to the unit disk as

Suppose the Julia sets do not converge to the unit disk D as Sketch of the proof:

Suppose the Julia sets do not converge to the unit disk D as 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.

Suppose the Julia sets do not converge to the unit disk D as 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.

Suppose the Julia sets do not converge to the unit disk D as 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.

Suppose the Julia sets do not converge to the unit disk D as 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.

Suppose the Julia sets do not converge to the unit disk D as 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.

Suppose the Julia sets do not converge to the unit disk D as 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. These disks cannot lie in the trap door since T vanishes as. (Remember is never in the trap door when n = 2.)

Suppose the Julia sets do not converge to the unit disk D as 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. The must accumulate on some nonzero point, say, so we may assume that lies in the Fatou set for all i.

Suppose the Julia sets do not converge to the unit disk D as 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. The must accumulate on some nonzero point, say, so we may assume that lies in the Fatou set for all i.

Suppose the Julia sets do not converge to the unit disk D as 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. The must accumulate on some nonzero point, say, so we may assume that lies in the Fatou set for all i. But for large i, so stretches into an “annulus” that surrounds the origin, so this disconnects the Julia set.

So the Fatou components must become arbitrarily small:

For the family the Julia sets again converge to the unit disk, but only if  0 along n - 1 special rays. (with M. Morabito) n = 6 n = 4

Things are much different in the family when 1/n + 1/d < 1, i.e. n, d  2 (but not both = 2)

n = d = 3: Note the “round” annuli in the Fatou set

It can be shown that, when n, d ≥ 2 and not both equal to 2 and lies in the McMullen domain, the Fatou set always contains a round annulus of some fixed width, so the Julia sets do not converge to the unit disk in this case.

Consider the family of maps where c is the center of a hyperbolic component of the Mandelbrot set. c = 0

Consider the family of maps where c is the center of a hyperbolic component of the Mandelbrot set. c = -1

Consider the family of maps where c is the center of a hyperbolic component of the Mandelbrot set. c = i

When, the Julia set again expodes and converges to the filled Julia set for z 2 + c. (with R. Kozma)

When, the Julia set again expodes and converges to the filled Julia set for z 2 + c. (with R. Kozma)

When, the Julia set again expodes and converges to the filled Julia set for z 2 + c. (with R. Kozma)

When, the Julia set again expodes and converges to the filled Julia set for z 2 + c. (with R. Kozma)

When, the Julia set again expodes and converges to the filled Julia set for z 2 + c. (with R. Kozma) An inverted Douady rabbit

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 also very different: (E. Russell)

When is small, the Julia set contains a Cantor set of “circles” surrounding the origin.....

infinitely many of which are “decorated” and there are also Cantor sets of buried points

Part 4: Cantor necklaces and webs A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals.

A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals. Do you see a necklace in the carpet? Part 4: Cantor necklaces and webs

A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals. Do you see a necklace in the carpet? Part 4: Cantor necklaces and webs

A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals. There are lots of necklaces in the carpet Part 4: Cantor necklaces and webs

A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals. There are lots of necklaces in the carpet Part 4: Cantor necklaces and webs

a Julia set with n = 2 There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet.

There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

n = 2 Even if we choose a parameter from the Mandelbrot set, there are Cantor necklaces in the Julia set

n = 2 Even if we choose a parameter from the Mandelbrot set, there are Cantor necklaces in the Julia set

And there are Cantor necklaces in the parameter planes. n = 2

The critical circle & ray

Critical points

The critical circle & ray Critical points n = 2 0

The critical circle & ray Critical points Critical values n = 2 0

The critical circle & ray Critical points Critical values n = 2 0

The critical circle & ray Critical points Prepoles Critical values n = 2 0

The critical circle & ray Critical points Prepoles Critical values n = 2 0

The critical circle & ray Critical points Prepoles Critical points & prepoles lie on the critical circle Critical values n = 2 0

The critical circle & ray Critical points Prepoles Critical values Critical points & prepoles lie on the critical circle n = 2 0

The critical circle & ray Critical points Prepoles Critical values Critical points & prepoles lie on the critical circle Which is mapped 2n to 1 onto the critical line n = 2 0

The critical circle & ray Critical points Prepoles Which is mapped 2n to 1 onto the critical line Critical values Critical points & prepoles lie on the critical circle n = 2 0

Critical points Critical values Critical point rays n = 2

Critical points Critical values Critical point rays are mapped 2 to 1 to a ray external to the critical line, a critical value ray. n = 2

Suppose is not positive real. n = 2

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

I0I0 I1I1 There are 4 prepole sectors when n = 2 and the critical values always lie in I 1 and I 3. I2I2 I3I3 Suppose is not positive real.

I0I0 I1I1 I2I2 I3I3 And the interior of each I j is mapped one-to-one over the entire plane minus the critical value rays Suppose is not positive real. There are 4 prepole sectors when n = 2 and the critical values always lie in I 1 and I 3.

I0I0 I1I1 I2I2 I3I3 In particular, maps I 0 and I 2 one-to-one over I 0 I 2 Suppose is not positive real. And the interior of each I j is mapped one-to-one over the entire plane minus the critical value rays There are 4 prepole sectors when n = 2 and the critical values always lie in I 1 and I 3.

Choose a circle in B that is mapped strictly outside itself I0I0 I2I2

Choose a circle in B that is mapped strictly outside itself I0I0 I2I2

Choose a circle in B that is mapped strictly outside itself Then there is another circle in the trap door that is also mapped to

Consider the portions of I 0 and I 2 that lie between and, say U 0 and U 2 U2U2 U0U0

U2U2 U0U0 Consider the portions of I 0 and I 2 that lie between and, say U 0 and U 2 U 0 and U 2 are mapped one-to-one over U 0 U 2

U0U0 Consider the portions of I 0 and I 2 that lie between and, say U 0 and U 2 U 0 and U 2 are mapped one-to-one over U 0 U 2 So the set of points whose orbits lie for all iterations in U 0 U 2 is an invariant Cantor set U2U2

Consider the portions of I 0 and I 2 that lie between and, say U 0 and U 2 U 0 and U 2 are mapped one-to-one over U 0 U 2 So the set of points whose orbits lie for all iterations in U 0 U 2 is an invariant Cantor set

The Cantor set in U 0  U 2 contains: 2 points on  B

The Cantor set in U 0  U 2 contains: 2 points on  B BB BB

2 points on  T The Cantor set in U 0  U 2 contains: 2 points on  B TT

4 points on the 2 preimages of  T 2 points on  T The Cantor set in U 0  U 2 contains: 2 points on  B

Add in the appropriate preimages of  T points on the 2 preimages of  T 2 points on  T The Cantor set in U 0  U 2 contains: 2 points on  B

Add in the appropriate preimages of  T..... to get an “invariant” Cantor necklace in the dynamical plane 4 points on the 2 preimages of  T 2 points on  T The Cantor set in U 0  U 2 contains: 2 points on  B

Cantor necklaces in the dynamical plane when n = 2

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

n = 3 When n > 2, get Cantor “webs” in the dynamical plane: Start with an open disk....

n = 3 When n > 2, get Cantor “webs” in the dynamical plane: Then surround it by 4 smaller disks

n = 3 When n > 2, get Cantor “webs” in the dynamical plane: Then do it again....

When n > 2, get Cantor “webs” in the dynamical plane: n = 3 and so forth, joining the open disks by a Cantor set of points

n = 4 When n > 2, get Cantor “webs” in the dynamical plane: n = 3

When n > 2, get Cantor “webs” in the dynamical plane: n = 3n = 4

When n > 2, get Cantor “webs” in the dynamical plane: n = 3

Same argument: we have 2n prepole sectors I 0,...,I 2n-1 I0I0 I1I1 I2I2 I3I3 I4I4 I5I5 n = 3

Same argument: we have 2n prepole sectors I 0,...,I 2n-1 I0I0 I1I1 I2I2 I3I3 I4I4 I5I5 If is not positive real, then the critical value ray always lies in I 0 and I n as long as lies in a symmetry region n = 3

Same argument: we have 2n prepole sectors I 0,...,I 2n-1 I0I0 I1I1 I2I2 I3I3 I4I4 I5I5 If is not positive real, then the critical value ray always lies in I 0 and I n as long as lies in a symmetry region n = 3 A symmetry region when n = 3

Same argument: we have 2n prepole sectors I 0,...,I 2n-1 I0I0 I1I1 I2I2 I3I3 I4I4 I5I5 If is not positive real, then the critical value ray always lies in I 0 and I n as long as lies in a symmetry region n = 3 A symmetry region when n = 4

Same argument: we have 2n prepole sectors I 0,...,I 2n-1 U1U1 U2U2 U4U4 U5U5 So consider the corresponding regions U j not including U 0 and U n which contain n = 3 If is not positive real, then the critical value ray always lies in I 0 and I n as long as lies in a symmetry region

n = 3 U1U1 U2U2 U4U4 U5U5 Each of these U j are mapped univalently over all the others, excluding U 0 and U n, so we get an invariant Cantor set in these regions.

n = 3 Each of these U j are mapped univalently over all the others, excluding U 0 and U n, so we get an invariant Cantor set in these regions. Then join in the preimages of T in these regions....

n = 3 Each of these U j are mapped univalently over all the others, excluding U 0 and U n, so we get an invariant Cantor set in these regions. Then join in the preimages of T in these regions.... and their preimages....

n = 3 Each of these U j are mapped univalently over all the others, excluding U 0 and U n, so we get an invariant Cantor set in these regions. Then join in the preimages of T in these regions.... and their preimages.... etc., etc. to get the Cantor web

Other Cantor webs n = 4n = 5

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