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# Exponential Dynamics and (Crazy) Topology - PowerPoint PPT Presentation

Exponential Dynamics and (Crazy) Topology. Cantor bouquets. Indecomposable continua. Exponential Dynamics and (Crazy) Topology. Cantor bouquets. Indecomposable continua. These are Julia sets of. Example 1: Cantor Bouquets. with Clara Bodelon Michael Hayes Gareth Roberts

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Cantor bouquets

Indecomposable continua

Cantor bouquets

Indecomposable continua

These are Julia sets of

with

Clara Bodelon

Michael Hayes

Gareth Roberts

Ranjit Bhattacharjee

Lee DeVille

Monica Moreno Rocha

Kreso Josic

Alex Frumosu

Eileen Lee

Question: What is the fate of orbits?

*

J = closure of {orbits that escape to }

= closure {repelling periodic orbits}

= {chaotic set}

Fatou set

= complement of J

= predictable set

* not the boundary of {orbits that escape to }

critical points that determined everything.

But has no critical points.

critical points that determined everything.

But has no critical points.

But 0 is an asymptotic value; any far

left half-plane is wrapped infinitely often

around 0, just like a critical value.

So the orbit of 0 for the exponential plays

a similar role as in the quadratic family

(only what happens to the Julia sets

is very different in this case).

is a “Cantor bouquet”

is a “Cantor bouquet”

is a “Cantor bouquet”

attracting

fixed point

q

is a “Cantor bouquet”

The orbit of 0 always

tends this attracting

fixed point

attracting

fixed point

q

is a “Cantor bouquet”

q

p

repelling

fixed point

is a “Cantor bouquet”

q

x0

p

is a “Cantor bouquet”

And

for all

q

x0

p

in this half plane

Green points lie in the Fatou set

Green points lie in the Fatou set

Green points lie in the Fatou set

Green points lie in the Fatou set

Green points lie in the Fatou set

The Julia set is a collection of curves (hairs) in the right half plane, each with an endpoint

and a stem.

hairs

endpoints

stems

A “Cantor bouquet” half plane, each with an endpoint

q

p

Colored points escape to half plane, each with an endpoint

and so now are in the Julia set.

q

p

One such hair lies on the real axis. half plane, each with an endpoint

repelling

fixed point

stem

Orbits of points on the stems all tend to . half plane, each with an endpoint

hairs

So bounded orbits lie in the set of endpoints. half plane, each with an endpoint

hairs

So bounded orbits lie in the set of endpoints. half plane, each with an endpoint

Repelling cycles lie

in the set of endpoints.

hairs

So bounded orbits lie in the set of endpoints. half plane, each with an endpoint

Repelling cycles lie

in the set of endpoints.

hairs

So the endpoints are

dense in the bouquet.

So bounded orbits lie in the set of endpoints. half plane, each with an endpoint

Repelling cycles lie

in the set of endpoints.

hairs

So the endpoints are

dense in the bouquet.

S half plane, each with an endpoint

Some Facts:

S half plane, each with an endpoint

Some Facts:

The only accessible points in J from

the Fatou set are the endpoints;

you cannot touch the stems

S half plane, each with an endpoint

Some Crazy Facts:

The only accessible points in J from

the Fatou set are the endpoints;

you cannot touch the stems

The set of endpoints together with the

point at infinity is connected ...

S half plane, each with an endpoint

Some Crazy Facts:

The only accessible points in J from

the Fatou set are the endpoints;

you cannot touch the stems

The set of endpoints together with the

point at infinity is connected ...

but the set of endpoints is

totally disconnected (Mayer)

S half plane, each with an endpoint

Some Crazy Facts:

The only accessible points in J from

the Fatou set are the endpoints;

you cannot touch the stems

The set of endpoints together with the

point at infinity is connected ...

but the set of endpoints is

totally disconnected (Mayer)

Hausdorff dimension of {stems} = 1...

S half plane, each with an endpoint

Some Crazy Facts:

The only accessible points in J from

the Fatou set are the endpoints;

you cannot touch the stems

The set of endpoints together with the

point at infinity is connected ...

but the set of endpoints is

totally disconnected (Mayer)

Hausdorff dimension of {stems} = 1...

but the Hausdorff dimension of

{endpoints} = 2! (Karpinska)

Another example: half plane, each with an endpoint

Looks a little different, but still a pair

of Cantor bouquets

Another example: half plane, each with an endpoint

The interval [-, ] is

contracted inside itself,

and all these orbits tend to

0 (so are in the Fatou set)

Another example: half plane, each with an endpoint

The real line is contracted

onto the interval ,

and all these orbits tend to 0

(so are in the Fatou set)

Another example: half plane, each with an endpoint

-/2

/2

The vertical lines x = n + /2 are

mapped to either [, ∞) or (-∞, - ],

so these lines are in the Fatou set....

Another example: half plane, each with an endpoint

c

-c

The lines y = c are both

wrapped around an ellipse

with foci at

Another example: half plane, each with an endpoint

c

-c

The lines y = c are both

wrapped around an ellipse

with foci at , and all orbits

in the ellipse tend to 0 if

c is small enough

Another example: half plane, each with an endpoint

c

-c

So all points in the ellipse

lie in the Fatou set

Another example: half plane, each with an endpoint

c

-c

So do all points in the

strip

Another example: half plane, each with an endpoint

c

-c

The vertical lines given

by x = n + /2 are

also in the Fatou set.

And all points in the half plane, each with an endpoint

preimages of the strip

lie in the Fatou set...

And so on to get another half plane, each with an endpoint

Cantor bouquet.

The difference here is that the Cantor bouquet for half plane, each with an endpoint

the sine function has infinite Lebesgue measure,

while the exponential bouquet has zero measure.

Example 2: Indecomposable Continua half plane, each with an endpoint

with

Nuria Fagella

Xavier Jarque

Monica Moreno Rocha

When half plane, each with an endpoint

,

The two fixed points coalesce

and disappear from the real

axis when  goes above 1/e.

And now the orbit of 0 goes off to half plane, each with an endpoint∞ ....

And as increases through 1/e, explodes. half plane, each with an endpoint

Theorem: half plane, each with an endpoint If the orbit of 0 goes to ∞, then the

Julia set is the entire complex plane.

(Sullivan, Goldberg, Keen)

As half plane, each with an endpoint

increases through

,

; however:

As half plane, each with an endpoint

increases through

,

; however:

No new periodic cycles are born;

As half plane, each with an endpoint

increases through

,

; however:

No new periodic cycles are born;

All move continuously to fill in the plane densely;

As half plane, each with an endpoint

increases through

,

; however:

No new periodic cycles are born;

All move continuously to fill in the plane;

Infinitely many hairs suddenly become

“indecomposable continua.”

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

For example:

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

For example: indecomposable?

0

1

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

No, decomposable.

For example:

0

1

(subsets need not be disjoint)

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

For example: indecomposable?

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

No, decomposable.

For example:

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

For example: indecomposable?

An indecomposable continuum is a compact, connected half plane, each with an endpoint

set that cannot be broken into the union of two (proper)

compact, connected subsets.

No, decomposable.

For example:

Knaster continuum half plane, each with an endpoint

A well known example of an indecomposable continuum

Knaster continuum half plane, each with an endpoint

Connect symmetric points about 1/2 with semicircles

Knaster continuum half plane, each with an endpoint

Do the same below about 5/6

Knaster continuum half plane, each with an endpoint

And continue....

Knaster continuum half plane, each with an endpoint

Properties of K: half plane, each with an endpoint

There is one curve that

passes through all the

endpoints of the Cantor

set.

Properties of K: half plane, each with an endpoint

There is one curve that

passes through all the

endpoints of the Cantor

set.

It accumulates everywhere on

itself and on K.

Properties of K: half plane, each with an endpoint

There is one curve that

passes through all the

endpoints of the Cantor

set.

It accumulates everywhere on

itself and on K.

And is the only piece

of K that is accessible

from the outside.

Properties of K: half plane, each with an endpoint

There is one curve that

passes through all the

endpoints of the Cantor

set.

It accumulates everywhere on

itself and on K.

And is the only piece

of K that is accessible

from the outside.

But there are infinitely many other curves

in K, each of which is dense in K.

Properties of K: half plane, each with an endpoint

There is one curve that

passes through all the

endpoints of the Cantor

set.

It accumulates everywhere on

itself and on K.

And is the only piece

of K that is accessible

from the outside.

But there are infinitely many other curves

in K, each of which is dense in K.

So K is compact, connected, and....

Indecomposable! half plane, each with an endpoint

Try to write K as the union

of two compact, connected sets.

Indecomposable! half plane, each with an endpoint

Can’t divide it this way....

subsets are closed

but not connected.

Indecomposable! half plane, each with an endpoint

Or this way...

again closed but

not connected.

Indecomposable! half plane, each with an endpoint

Or the union of the outer curve and all

the inaccessible curves ... not closed.

How the hairs become indecomposable: half plane, each with an endpoint

repelling

fixed pt

.

.

.

.

... .

.

.

.

.

attracting

fixed pt

stem

How the hairs become indecomposable: half plane, each with an endpoint

.

.

.

.

... .

.

.

.

.

.

.

.

.

2 repelling

fixed points

.

.

.

.

.

.

.

.

Now all points in R escape,

so the hair is much longer

.

.

.

.

But the hair is even longer! half plane, each with an endpoint

0

But the hair is even longer! half plane, each with an endpoint

0

But the hair is even longer! And longer. half plane, each with an endpoint

0

But the hair is even longer! And longer... half plane, each with an endpoint

0

But the hair is even longer! And longer....... half plane, each with an endpoint

0

But the hair is even longer! And longer............. half plane, each with an endpoint

0

Compactify to get a single curve in a compact region half plane, each with an endpoint

in the plane that accumulates everywhere on itself.

The closure is then an indecomposable continuum.

0

The dynamics on this continuum is very simple: half plane, each with an endpoint

one repelling fixed point

all other orbits either tend to

or accumulate on the orbit of 0 and

But the topology is not at all understood:

Conjecture: the continuum for each

parameter is topologically distinct.

sin(z)

A pair of Cantor bouquets half plane, each with an endpoint

Julia set of sin(z)

A pair of Cantor bouquets half plane, each with an endpoint

A similar explosion occurs for

the sine family (1 + ci) sin(z)

Julia set of sin(z)

sin(z) half plane, each with an endpoint

sin(z) half plane, each with an endpoint

sin(z) half plane, each with an endpoint

(1+.2i) sin(z) half plane, each with an endpoint

(1+ ci) sin(z) half plane, each with an endpoint

Questions: half plane, each with an endpoint

Do the hairs become indecomposable

continua as in the exponential case?

If so, what is the topology of these sets?

Parameter plane for half plane, each with an endpoint

To plot the parameter plane (the analogue

of the Mandelbrot set), for each  plot the

corresponding orbit of 0.

If 0 escapes, the color ; J is the entire plane.

If 0 does not escape, leave  black; J is

usually a “pinched” Cantor bouquet.

Parameter plane for half plane, each with an endpoint

Parameter plane for half plane, each with an endpoint

Parameter plane for half plane, each with an endpoint

Parameter plane for half plane, each with an endpoint

has an

attracting fixed

point in this

cardioid

1

Parameter plane for half plane, each with an endpoint

2

Period 2 region

1

Parameter plane for half plane, each with an endpoint

4

3

5

2

1

5

3

4

Fixed point bifurcations half plane, each with an endpoint

2

Period 2 region

1

So undergoes

a period doubling

bifurcation along

this path in the

parameter plane

at

Parameter plane for

The Cantor bouquet half plane, each with an endpoint

The Cantor bouquet half plane, each with an endpoint

A repelling

2-cycle at

two endpoints

The hairs containing the 2-cycle meet at the half plane, each with an endpoint

neutral fixed point, and then remain attached.

Meanwhile an attracting 2 cycle emerges.

We get a “pinched” Cantor bouquet

Other bifurcations half plane, each with an endpoint

Period 3 region

3

2

1

Parameter plane for

Other bifurcations half plane, each with an endpoint

Period tripling

bifurcation

3

2

1

Parameter plane for

Three hairs containing a 3-cycle meet at the half plane, each with an endpoint

neutral fixed point, and then remain attached

We get a different “pinched” Cantor bouquet

Other bifurcations half plane, each with an endpoint

Period 5

bifurcation

5

3

2

1

Parameter plane for

Five hairs containing a 5-cycle meet at the half plane, each with an endpoint

neutral fixed point, and then remain attached

We get a different “pinched” Cantor bouquet

Lots of explosions occur.... half plane, each with an endpoint

3

1

Lots of explosions occur.... half plane, each with an endpoint

4 half plane, each with an endpoint

1

5

On a path like this, we pass through infinitely half plane, each with an endpoint

many regions where there is an attracting cycle,

so J is a pinched Cantor bouquet.....

and infinitely many hairs where J is the entire plane.

slower half plane, each with an endpoint

Example 3: Sierpinski Curves half plane, each with an endpoint

with:

Paul Blanchard

Toni Garijo

Matt Holzer

U. Hoomiforgot

Dan Look

Sebastian Marotta

Mark Morabito

Monica Moreno Rocha

Kevin Pilgrim

Elizabeth Russell

Yakov Shapiro

David Uminsky

Sierpinski Curve half plane, each with an endpoint

A Sierpinski curve is any planar

set that is homeomorphic to the Sierpinski carpet fractal.

The Sierpinski Carpet

Topological Characterization half plane, each with an endpoint

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.

The Sierpinski Carpet

More importantly.... half plane, each with an endpoint

A Sierpinski curve is a universal plane continuum:

Any planar, one-dimensional, compact, connected set can be homeomorphically embedded in a Sierpinski curve.

For example....

can be embedded inside half plane, each with an endpoint

The topologist’s sine curve

can be embedded inside half plane, each with an endpoint

The topologist’s sine curve

can be embedded inside half plane, each with an endpoint

The topologist’s sine curve

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

The Knaster continuum

can be embedded inside half plane, each with an endpoint

Even this “curve”

Some easy to verify facts: half plane, each with an endpoint

Some easy to verify facts: half plane, each with an endpoint

Have an immediate basin of infinity B

Some easy to verify facts: half plane, each with an endpoint

Have an immediate basin of infinity B

0 is a pole so have a “trap door” T

(the preimage of B)

Some easy to verify facts: half plane, each with an endpoint

Have an immediate basin of infinity B

0 is a pole so have a “trap door” T

(the preimage of B)

2n critical points given by but really only

one critical orbit due to symmetry

Some easy to verify facts: half plane, each with an endpoint

Have an immediate basin of infinity B

0 is a pole so have a “trap door” T

(the preimage of B)

2n critical points given by but really only

one critical orbit due to symmetry

J is now the boundary of the escaping orbits

(not the closure)

When , the Julia set half plane, each with an endpoint

is the unit circle

When , the Julia set half plane, each with an endpoint

is the unit circle

But when , the

Julia set explodes

A Sierpinskicurve

When , the Julia set half plane, each with an endpoint

is the unit circle

But when , the

Julia set explodes

B

T

A Sierpinskicurve

When , the Julia set half plane, each with an endpoint

is the unit circle

But when , the

Julia set explodes

Another Sierpinskicurve

When , the Julia set half plane, each with an endpoint

is the unit circle

But when , the

Julia set explodes

Also a Sierpinski curve

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

1. If the critical orbits

eventually fall into

the trap door (which

is disjoint from B),

then J is a Sierpinski

curve.

Lots of ways this happens: half plane, each with an endpoint

parameter plane

when n = 3

Lots of ways this happens: half plane, each with an endpoint

parameter plane

when n = 3

J is a Sierpinski curve

T

lies in a Sierpinski hole

Lots of ways this happens: half plane, each with an endpoint

parameter plane

when n = 3

J is a Sierpinski curve

T

lies in a Sierpinski hole

Lots of ways this happens: half plane, each with an endpoint

parameter plane

when n = 3

J is a Sierpinski curve

T

lies in a Sierpinski hole

Lots of ways this happens: half plane, each with an endpoint

parameter plane

when n = 3

J is a Sierpinski curve

T

lies in a Sierpinski hole

Theorem: half plane, each with an endpoint Two maps drawn from the same Sierpinski

hole have the same dynamics, but those drawn from

different holes are not conjugate (except in very

few symmetric cases).

n = 4, escape time 4, 24 Sierpinski holes,

Theorem: half plane, each with an endpoint Two maps drawn from the same Sierpinski

hole have the same dynamics, but those drawn from

different holes are not conjugate (except in very

few symmetric cases).

n = 4, escape time 4, 24 Sierpinski holes,

but only five conjugacy classes

Theorem: half plane, each with an endpoint Two maps drawn from the same Sierpinski

hole have the same dynamics, but those drawn from

different holes are not conjugate (except in very

few symmetric cases).

n = 4, escape time 12: 402,653,184 Sierpinski holes,

but only 67,108,832distinctconjugacy classes

Sorry. I forgot to

indicate their locations.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

2. If the parameter lies in

the main cardioid of a

buried baby Mandelbrot

set, J is again a

Sierpinski curve.

parameter plane

when n = 4

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

2. If the parameter lies in

the main cardioid of a

buried baby Mandelbrot

set, J is again a

Sierpinski curve.

parameter plane

when n = 4

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

2. If the parameter lies in

the main cardioid of a

buried baby Mandelbrot

set, J is again a

Sierpinski curve.

parameter plane

when n = 4

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

2. If the parameter lies in

the main cardioid of a

buried baby Mandelbrot

set, J is again a

Sierpinski curve.

Black regions are the basin

of an attracting cycle.

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

3. If the parameter lies

at a buried point in the

“Cantor necklaces” in

the parameter plane,

J is again a Sierpinski

curve.

parameter plane

n = 4

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

3. If the parameter lies

at a buried point in the

“Cantor necklaces” in

the parameter plane,

J is again a Sierpinski

curve.

parameter plane

n = 4

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

3. If the parameter lies

at a buried point in the

“Cantor necklaces” in

the parameter plane,

J is again a Sierpinski

curve.

parameter plane

n = 4

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

4. There is a Cantor set

of circles in the parameter

plane on which each

parameter corresponds

to a Sierpinski curve.

n = 3

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

4. There is a Cantor set

of circles in the parameter

plane on which each

parameter corresponds

to a Sierpinski curve.

n = 3

Sierpinski curves arise in lots of different half plane, each with an endpoint

ways in these families:

4. There is a Cantor set

of circles in the parameter

plane on which each

parameter corresponds

to a Sierpinski curve.

n = 3

Theorem: half plane, each with an endpoint All these Julia sets are the same topologically, but they are all (except for symmetrically located parameters) VERY different from a dynamics point of view (i.e., the maps are not conjugate).

Problem: Classify the dynamics on these

Sierpinski curve Julia sets.

Corollary: half plane, each with an endpoint

Corollary: half plane, each with an endpoint

Yes, those planar topologists

are crazy, but I sure wish I were one of them!

Corollary: half plane, each with an endpoint

Yes, those planar topologists

are crazy, but I sure wish I were one of them!

The End!

website: half plane, each with an endpoint

math.bu.edu/DYSYS