Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta with: (why the case n = 2 is )

Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta with: (why the case n = 2 is ) Mark Morabito Monica Moreno Rocha Kevin Pilgrim Elizabeth Russell Yakov Shapiro David Uminsky C R A Z Y

The case n > 2 is great because:

There exists a McMullen domain around = 0.... Parameter plane for n=3

The case n > 2 is great because:... surrounded by infinitely many “Mandelpinski” necklaces... There exists a McMullen domain around = 0....

The case n > 2 is great because:... surrounded by infinitely many “Mandelpinski” necklaces...... and the Julia sets behave nicely as There exists a McMullen domain around = 0....

There is no McMullen domain.... The case n = 2 is crazy because:

There is no McMullen domain....... and no “Mandelpinski” necklaces... The case n = 2 is crazy because:

There is no McMullen domain....... and the Julia sets “go crazy” as... and no “Mandelpinski” necklaces... The case n = 2 is crazy because:

Some definitions: Julia set of J = boundary of {orbits that escape to } = closure {repelling periodic orbits} = {chaotic set} Fatou set = complement of J = predictable set

Computation of J: Color points that escape to infinity shades of red orange yellow green blue violet Black points do not escape. J = boundary of the black region.

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. 0 is a pole, so have trap door T mapped n-to-1 to B. The Julia set has 2n-fold symmetry. B T

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 But really only 1 free critical orbit

Easy computations: 2n free critical points Only 2 critical values But really only 1 free critical orbit And 2n prepoles

The Escape Trichotomy There are three possible ways that the critical orbits can escape to infinity, and each yields a different type of Julia set. (with D. Look & D Uminsky)

The Escape Trichotomy B is a Cantor set (with D. Look & D Uminsky)

The Escape Trichotomy B is a Cantor set T is a Cantor set of simple closed curves (McMullen) (with D. Look & D Uminsky) (n > 2)

The Escape Trichotomy B is a Cantor set T is a Cantor set of simple closed curves T is a Sierpinski curve (McMullen) (with D. Look & D Uminsky) (n > 2) In all other cases is a connected set, and if

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

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

parameter plane when n = 3 Case 2: T is a Cantor set of simple closed curves

The central disk is the McMullen domain Case 2: T is a Cantor set of simple closed curves

parameter plane when n = 3 Case 2: T is a Cantor set of simple closed curves J is a Cantor set of simple closed curves T B

parameter plane when n = 3 Case 2: T is a Cantor set of simple closed curves J is a Cantor set of simple closed curves

parameter plane when n = 3 Case 3: T is a Sierpinski curve

parameter plane when n = 3 Case 3: T A Sierpinski curve is a planar set homeomorphic to the Sierpinski carpet fractal is a Sierpinski curve

Sierpinski curves are important for two reasons: 1.There is a “topological characterization” of the carpet 2. A Sierpinski curve is a “universal plane continuum”

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.

Universal Plane Continuum Any planar, one-dimensional, compact, connected set can be homeomorphically embedded in a Sierpinski curve. For example....

This set

can be embedded inside

parameter plane when n = 3 Case 3: T is a Sierpinski curve

Case 3: T is a Sierpinski curve A Sierpinski “hole”

Case 3: T A Sierpinski curve is a Sierpinski curve A Sierpinski “hole”

Case 3: T A Sierpinski curve is a Sierpinski curve A Sierpinski “hole” Escape time 3

Case 3: T A Sierpinski curve is a Sierpinski curve Another Sierpinski “hole”

Case 3: T A Sierpinski curve is a Sierpinski curve Another Sierpinski “hole” Escape time 4

So to show that is homeomorphic to

Need to show: compact connected nowhere dense locally connected bounded by disjoint s.c.c.’s

Need to show: compact connected nowhere dense locally connected bounded by disjoint s.c.c.’s Fatou set is the union of the preimages of B; all disjoint, open disks.

Need to show: compact connected nowhere dense locally connected bounded by disjoint s.c.c.’s If J contains an open set, then J = C.

Need to show: compact connected nowhere dense locally connected bounded by disjoint s.c.c.’s No recurrent critical orbits and no parabolic points.

Need to show: compact connected nowhere dense locally connected bounded by disjoint s.c.c.’s J locally connected, so the boundaries are locally connected. Need to show they are s.c.c.’s. Can only meet at (preimages of) critical points, hence disjoint.

Need to show: compact connected nowhere dense locally connected bounded by disjoint s.c.c.’s So J is a Sierpinski curve.

Remark: All Julia sets drawn from Sierpinski holes are homeomorphic, but only those in symmetrically located Sierpinski holes have the same dynamics. The maps on all these Julia sets are dynamically different.

Remark: All Julia sets drawn from Sierpinski holes are homeomorphic, but only those in symmetrically located Sierpinski holes have the same dynamics. In fact, there are exactly (n-1)(2n) k-3 Sierpinski holes with escape time k, and (2n) k-3 different conjugacy classes (n odd). The maps on all these Julia sets are dynamically different. (with K.Pilgrim)

1. The McMullen domain 2. Mandelpinski necklaces 3. Julia sets near 0 Topics

Part 1: The McMullen Domain

Why is there no McMullen domain when n = 2? What is the preimage of T? First suppose T:

Why is there no McMullen domain when n = 2? First suppose What is the preimage of T? T: Can the preimage be 2n disjoint disks, each of which contains a critical point?

Why is there no McMullen domain when n = 2? First suppose What is the preimage of T? T: Can the preimage be 2n disjoint disks, each of which contains a critical point? No --- there would then be 4n preimages of any point in T, but the map has degree 2n.

Why is there no McMullen domain when n = 2? So some of the preimages of T must overlap, and by 2n-fold symmetry, all must intersect.

Why is there no McMullen domain when n = 2? So some of the preimages of T must overlap, and by 2n-fold symmetry, all must intersect. By Riemann-Hurwitz, the preimage of T must then be an annulus.

Why is there no McMullen domain when n = 2? So here is the picture: T B A A is the annulus separating B and T

Why is there no McMullen domain when n = 2? So here is the picture: B T A A is the annulus separating B and T X F maps X 2n-to-1 onto T

Why is there no McMullen domain when n = 2? So here is the picture: B T A A is the annulus separating B and T X A A 0 1 F maps bothA 0 1 and A as an n-to-1 covering onto A F maps X 2n-to-1 onto T

Why is there no McMullen domain when n = 2? B T A X A A 0 1 So mod(A 0 ) = 1/n mod(A) And mod(A 1 ) = 1/n mod(A)

Why is there no McMullen domain when n = 2? B T A X A A 0 1 So mod(A 0 ) = 1/n mod(A) And mod(A 1 ) = 1/n mod(A) When n = 2, mod(A 0 ) + mod(A 1 ) = mod(A)

Why is there no McMullen domain when n = 2? B T A X A A 0 1 So mod(A 0 ) = 1/n mod(A) And mod(A 1 ) = 1/n mod(A) When n = 2, mod(A 0 ) + mod(A 1 ) = mod(A) So there is no room for X, i.e., does not lie in T

Why is there no McMullen domain when n = 2? Here is another reason:

Why is there no McMullen domain when n = 2? Here is another reason: so lies in T when n > 2

Why is there no McMullen domain when n = 2? Here is another reason: (???) so lies in T when n > 2

Part 2: Mandelpinski Necklaces

Parameter plane for n = 3 A “Mandelpinski necklace” is a simple closed curve in the parameter plane that passes alternately through k centers of baby Mandelbrot sets and k centers of Sierpinski holes.

Parameter plane for n = 3 A “Mandelpinski necklace” is a simple closed curve in the parameter plane that passes alternately through k centers of baby Mandelbrot sets and k centers of Sierpinski holes. C 1 passes through the centers of 2 M-sets and 2 S-holes

Parameter plane for n = 3 A “Mandelpinski necklace” is a simple closed curve in the parameter plane that passes alternately through k centers of baby Mandelbrot sets and k centers of Sierpinski holes.

Parameter plane for n = 3 A “Mandelpinski necklace” is a simple closed curve in the parameter plane that passes alternately through k centers of baby Mandelbrot sets and k centers of Sierpinski holes. C 2 passes through the centers of 4 M-sets and 4 S-holes * * only exception: 2 centers of period 2 bulbs, not M-sets

A “Mandelpinski necklace” is a simple closed curve in the parameter plane that passes alternately through k centers of baby Mandelbrot sets and k centers of Sierpinski holes. C 3 passes through the centers of 10 M-sets and 10 S-holes Parameter plane for n = 3

Theorem: There exist closed curves C j, surrounding the McMullen domain. Each C j passes alternately through (n-2)n j-1 +1 centers of baby Mandelbrot sets and centers of Sierpinski holes.

C 14 passes through the centers of 4,782,969 M-sets and S-holes Parameter plane for n = 3 Theorem: There exist closed curves C j, surrounding the McMullen domain. Each C j passes alternately through (n-2)n j-1 +1 centers of baby Mandelbrot sets and centers of Sierpinski holes.

Parameter plane for n = 4 C 1 : 3 holes and M-sets Theorem: There exist closed curves C j, surrounding the McMullen domain. Each C j passes alternately through (n-2)n j-1 +1 centers of baby Mandelbrot sets and centers of Sierpinski holes.

Parameter plane for n = 4 C 2 : 9 holes and M-sets C 3 : 33 holes and M-sets Theorem: There exist closed curves C j, surrounding the McMullen domain. Each C j passes alternately through (n-2)n j-1 +1 centers of baby Mandelbrot sets and centers of Sierpinski holes. *

Easy computations: All of the critical points and prepoles lie on the “critical circle” : |z| = | | 1/2n

All of the critical points and prepoles lie on the “critical circle” : |z| = | | 1/2n which is mapped 2n-to-1 onto the “critical value line” connecting Easy computations:

Any other circle around 0 is mapped n-to-1 to an ellipse whose foci are Easy computations:

Any other circle around 0 is mapped n-to-1 to an ellipse whose foci are

So the exterior of is mapped as an n-to-1 covering of the exterior of the critical value line. Easy computations: Any other circle around 0 is mapped n-to-1 to an ellipse whose foci are

So the exterior of is mapped as an n-to-1 covering of the exterior of the critical value line. Same with the interior of. Easy computations: Any other circle around 0 is mapped n-to-1 to an ellipse whose foci are

Simplest case: C 1 Assume that both sit on the critical circle. (i.e., )

Assume that both sit on the critical circle. Simplest case: C 1

Assume that both sit on the critical circle. This is the “dividing circle” in the parameter plane Parameter plane n = 3 r = 2 -3 Simplest case: C 1

Assume that both sit on the critical circle. This is the “dividing circle” in the parameter plane Parameter plane n = 4 r = 2 -8/3 Simplest case: C 1

Assume that lies on the dividing circle, so both sit on the critical circle. In this picture, is real and n = 3 Simplest case: C 1

Assume that lies on the dividing circle, so both sit on the critical circle. As rotates around the dividing circle, rotates a half-turn, while and rotate 1/2n of a turn. So each meets exactly n - 1 prepoles and critical points. In this picture, is real and n = 3 Simplest case: C 1

Assume that lies on the dividing circle, so both sit on the critical circle. As rotates around the dividing circle, rotates a half-turn, while and rotate 1/2n of a turn. So each meets exactly n - 1 prepoles and critical points. In this picture, is real and n = 3 Simplest case: C 1 So the dividing circle is C 1

The dividing circle when n = 5 n-1 = 4 centers of Sierpinski holes; n-1 = 4 centers of baby Mandelbrot sets

Now assume that lies inside the critical circle:

The exterior of is mapped n-to-1 onto the exterior of the critical value ray, so there is a preimage mapped n-to-1 to,

Now assume that lies inside the critical circle: The exterior of is mapped n-to-1 onto the exterior of the critical value ray, so there is a preimage mapped n-to-1 to,

then is mapped n-to-1 to, Now assume that lies inside the critical circle: The exterior of is mapped n-to-1 onto the exterior of the critical value ray, so there is a preimage mapped n-to-1 to,

and on out to Now assume that lies inside the critical circle: then is mapped n-to-1 to, B The exterior of is mapped n-to-1 onto the exterior of the critical value ray, so there is a preimage mapped n-to-1 to,

contains 2n critical points and 2n prepoles, so contains 2n 2 pre-critical points and pre-prepoles

contains 2n critical points and 2n prepoles, so contains 2n 2 pre-critical points and pre-prepoles B contains 2n k+1 points that map to the critical points and pre-prepoles under

As rotates by one turn, these 2n k+1 points on each rotate by 1/2n k+1 of a turn.

Since the second iterate of the critical points rotate by 1 - n/2 of a turn As rotates by one turn, these 2n k+1 points on each rotate by 1/2n k+1 of a turn.

Since the second iterate of the critical points rotate by 1 - n/2 of a turn, so this point hits exactly preimages of the critical points and prepoles on As rotates by one turn, these 2n k+1 points on each rotate by 1/2n k+1 of a turn.

There is a natural parametrization of each The real proof involves the Schwarz Lemma:

There is a natural parametrization of each The real proof involves the Schwarz Lemma: Best to restrict to a “symmetry region” inside the dividing circle, so that is well-defined.

Then we have a second map from the parameter plane to the dynamical plane, namely which is invertible on the symmetry sector Best to restrict to a “symmetry region” inside the dividing circle, so that is well-defined.

Then we have a second map from the parameter plane to the dynamical plane, namely which is invertible on the symmetry sector a map from a “disk” to itself. So consider the composition

a map from a “disk” to itself. So consider the composition Schwarz implies that has a unique fixed point, i.e., a parameter for which the second iterate of the critical point lands on the point, so this proves the existence of the centers of the S-holes and M-sets.

Remarks: This proves the existence of centers of Sierpinski holes and Mandelbrot sets. Producing the entire M-set involves polynomial-like maps; while the entire S-hole involves qc-surgery.

Part 3: Behavior of the Julia sets

n = 2: When, the Julia set of is the unit circle. But, as, the Julia set of converges to the closed unit disk Part 3: Behavior of the Julia sets

n > 2: J is always a Cantor set of “circles” when is small. n = 2: When, the Julia set of is the unit circle. But, as, the Julia set of converges to the closed unit disk Part 3: Behavior of the Julia sets

n > 2: J is always a Cantor set of “circles” when is small. n = 2: When, the Julia set of is the unit circle. But, as, the Julia set of converges to the closed unit disk 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 3: Behavior of the Julia sets

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. 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

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

Part 4: A major application

Here’s the parameter plane when n = 2:

Rotate it by 90 degrees: and this object appears everywhere.....

Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta with: (why the case n = 2 is )

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Dynamics of the family of complex maps Paul Blanchard Toni Garijo Matt Holzer U. Hoomiforgot Dan Look Sebastian Marotta with: (why the case n = 2 is )

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