Structures in the Dynamical Plane 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, n > 1

1.Dynamic classsification of escape time Sierpinski curve Julia sets 3. Cantor necklaces in the dynamical plane 2. Julia sets converging to the unit disk 4. Internal rays

Sierpinski curve Julia sets occur for many different parameters in this family 1. When critical points “eventually” escape 2. As buried points in Cantor necklaces 3. In the main cardioid of buried Mandelbrot sets 4. On Cantor sets of circles surrounding the McMullen domain 5. On “Mandelpinski” necklaces And almost all have very different dynamical behavior

Sierpinski curve Julia sets occur for many different parameters in this family All these sets are homeomorphic, but the dynamics on each are very different

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

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.

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

Easy computations: 2n free critical points

Easy computations: 2n free critical points

Easy computations: 2n free critical points Only 2 critical values

Easy computations: 2n free critical points Only 2 critical values

Easy computations: 2n free critical points Only 2 critical values

Easy computations: 2n free critical points Only 2 critical values

Easy computations: 2n free critical points Only 2 critical values But really only one free critical orbit

Easy computations: 2n free critical points Only 2 critical values And J has 2n-fold symmetry But really only one free critical orbit

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.

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) (McMullen)

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

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 lies in a Sierpinski hole J is an escape time Sierpinski curve c

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: n = 3 escape time 3 2 Sierpinski holes parameter plane n = 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 = 3 escape time 4 12 Sierpinski holes parameter plane n = 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 = 4 escape time 3 3 Sierpinski holes parameter plane n = 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 = 4 escape time 4 24 Sierpinski holes parameter plane n = 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 = 4 escape time ,653,184 Sierpinski holes Guess what? I again 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: Part 1: Dynamic Classification of Escape Time Sierpinski Curve Julia sets

#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 boundary curves 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.

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 under the conjugacy, the Mobius conjugacy must be of the form.

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. (with K. Pilgrim) 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

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

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 3

For n even, there is a “Cantor necklace” along the negative axis, so there are some “real” Sierpinski holes, n = 4magnification M 344

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. Will someone please remind me to indicate their locations?

n = 4: 402,653,184 Sierpinski holes with escape time 12; 67,108,832 distinct conjugacy classes. Problem: use symbolic dynamics to describe these conjugacy classes.

Part 2: Julia sets converging to the unit disk With Toni Garijo

n = 2: When, the Julia set of is the unit circle. But, as, the Julia set of converges to the closed unit disk Part 2: Julia sets converging to the unit disk With Toni Garijo

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. Part 2: Julia sets converging to the unit disk With Toni Garijo

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 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. Part 2: Julia sets converging to the unit disk With Toni Garijo

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. 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:

n > 2: Note the “round” annuli in the Fatou set; there is always such an annulus of some fixed width for small.

T B small T is tiny A0A0 mod A 0 = m is huge Say n = 3:

T B small T is tiny A0A0 mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3:

T B small T is tiny mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3: X1X1 A1A1 A 1 and A 1 mapped to A 0 ; X 1 is mapped to T X1X1 A1A1 A1A1 ~ ~ A0A0 mod A 1 = mod A 1 = mod X 1 = m/3; A 1 S 1 ~

T B small T is tiny mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3: X1X1 A1A1 A 1 and A 1 mapped to A 0 ; X 1 is mapped to T X2X2 A1A1 A2A2 A 2 is mapped to A 1 ; X 2 is mapped to X 1 mod A 2 = mod X 2 = m/3 2 ; A 2 S 1 mod A 1 = mod A 1 = mod X 1 = m/3; A 1 S 1 ~ ~

B small T is tiny mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3: A 1 and A 1 mapped to A 0 ; X 1 is mapped to T XkXk AkAk A k-1 A k is mapped to A k-1 ; X k to X k-1 mod A 2 = mod X 2 = m/3 2 ; A 2 S 1 mod A 1 = mod A 1 = mod X 1 = m/3; A 1 S 1 ~ mod A k = mod X k = m/3 k ; A k S 1 A 2 is mapped to A 1 ; X 2 is mapped to X 1 ~

B small T is tiny mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3: XkXk AkAk A k-1 1 mod A k < 3 Eventually find k so that and A k S 1

B small T is tiny mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3: XkXk AkAk A k-1 1 mod A k < 3 Eventually find k so that and A k S 1 A k must contain a round annulus of modulus (Ble, Douady, and Henriksen)

B small T is tiny mod A 0 = m is huge and the boundary of A 0 is very close to S 1 Say n = 3: XkXk AkAk A k-1 A k must contain a round annulus of modulus (Ble, Douady, and Henriksen) But does this annulus have definite “thickness?” 1 mod A k < 3 Eventually find k so that and A k S 1

A k in here

Construct an outer round annulus with modulus

A k in here Construct an outer round annulus with modulus and a similar inner annulus

A k in here mod A k says that the inner boundary of A k cannot be inside or outside, so the round annulus in A k is “thick” AkAk

A k in here AkAk Same argument says that A k X k is twice as thick XkXk

XkXk So X k must have definite thickness as well

And that’s what we saw earlier: there is always an annulus X k far out with definite thickness.

Part 3: Cantor necklaces and webs A Cantor necklace is the Cantor middle thirds set with open disks replacing the removed intervals.

Part 3: 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 3: 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 3: 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 3: 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 3: 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.

Part 3: Cantor necklaces and webs There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

Part 3: Cantor necklaces and webs There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

Part 3: Cantor necklaces and webs There are lots of Cantor necklaces in these Julia sets, just as in the Sierpinski carpet. another Julia set with n = 2

Part 3: Cantor necklaces and webs 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

Part 3: Cantor necklaces and webs 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

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

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

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

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

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

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

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.

Part 4: Internal Rays These straight rays are preserved by z n 1/4 0 1/8 We can put a Böttcher coordinate on B to produce external rays in the dynamical plane.

Part 4: Internal Rays We can put a Böttcher coordinate on B to produce external rays in the dynamical plane. 1/4 0 1/8 and are mapped by to external rays in B These straight rays are preserved by z n

And using the Cantor necklace/web, these can be extended to “internal rays” connecting the external rays to the origin and passing through the Julia set in interesting ways:

And using the Cantor necklace/web, these can be extended to “internal rays” connecting the external rays to the origin and passing through the Julia set in interesting ways:

These two internal rays pass through all the points in the Cantor set of points in the Cantor necklace as well as through the appropriate preimages of T:

In the necklace construction, you just pull back appropriate images of the external rays in B;

In the necklace construction, you just pull back appropriate images of the external rays in B; 0 1/2

In the necklace construction, you just pull back appropriate images of the external rays in B; 0 1/2

In the necklace construction, you just pull back appropriate images of the external rays in B; 0 1/2

In the necklace construction, you just pull back appropriate images of the external rays in B; which then connect up with the appropriate endpoints of the Cantor set to produce the internal ray. 0 1/2

Some facts: When n > 2, there exists an easy-to-construct Cantor set of such internal rays, and each crosses infinitely many others.

Some facts: And the regions in between these crossing rays give disks on which the map is polynomial- like in the sense of Douady and Hubbard. is polynomial- like in this disk.

Some facts: So this enables us to prove the existence of the infinitely many baby Mandelbrot sets you see in the parameter planes:

Some facts: So this enables us to prove the existence of the infinitely many baby Mandelbrot sets you see in the parameter planes:

Some facts: So this enables us to prove the existence of the infinitely many baby Mandelbrot sets you see in the parameter planes:

Some facts: So this enables us to prove the existence of the infinitely many baby Mandelbrot sets you see in the parameter planes:

Some facts: So this enables us to prove the existence of the infinitely many baby Mandelbrot sets you see in the parameter planes:

Some facts: And there is a piecewise linear model for these internal rays that you can embed in the Sierpinski carpet:

Some facts: And there is a piecewise linear model for these internal rays that you can embed in the Sierpinski carpet:

Some facts: And finally, this should provide a mechanism (Yoccoz puzzles) to prove what is “obviously” true: that the boundaries of the parameter planes for these maps are simple closed curves (like Milnor’s cubic family). n = 4 n = 7