Mixing
Existence of fixed simple closed curve by polynomials
up vote
2
down vote
favorite
As the problem mentioned in the title, I wonder that if there exists a simple closed curve on the complex plane which is not circle that can be fixed by a non-linear polynomial with complex coefficients($P(C)=C$, $C$ for the curve and $P$ for the polynomial) ?
I have asked others about this problem, and some said that this is related to dynamical system.
polynomials dynamical-systems complex-dynamics fixedpoints
edited yesterday
Mark McClure
23.1k34170
asked yesterday
Steve Cheng 鄭宗弘
357
|
show 2 more comments
up vote
2
down vote
favorite
As the problem mentioned in the title, I wonder that if there exists a simple closed curve on the complex plane which is not circle that can be fixed by a non-linear polynomial with complex coefficients($P(C)=C$, $C$ for the curve and $P$ for the polynomial) ?
I have asked others about this problem, and some said that this is related to dynamical system.
polynomials dynamical-systems complex-dynamics fixedpoints
edited yesterday
Mark McClure
23.1k34170
asked yesterday
Steve Cheng 鄭宗弘
357
"that can be fixed": what does that mean ??
– Yves Daoust
yesterday
Why do you specify non-linear polynomial ?
– Yves Daoust
yesterday
For the first question, I have edited the problem to make it clear.
– Steve Cheng 鄭宗弘
yesterday
And for the second question, that is because you can draw a square and rotate it by $pi/2$ which can be represented as linear polynomial.
– Steve Cheng 鄭宗弘
yesterday
So $C$ is a simple closed curve (in a plane?), and $P(x,y)=(f(x,y),g(x,y))$ where $f$ and $g$ are polynomials in $x$ and $y$? With real coefficients, or complex, or something else?
– Servaes
yesterday
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
As the problem mentioned in the title, I wonder that if there exists a simple closed curve on the complex plane which is not circle that can be fixed by a non-linear polynomial with complex coefficients($P(C)=C$, $C$ for the curve and $P$ for the polynomial) ?
I have asked others about this problem, and some said that this is related to dynamical system.
polynomials dynamical-systems complex-dynamics fixedpoints
edited yesterday
Mark McClure
23.1k34170
asked yesterday
Steve Cheng 鄭宗弘
357
As the problem mentioned in the title, I wonder that if there exists a simple closed curve on the complex plane which is not circle that can be fixed by a non-linear polynomial with complex coefficients($P(C)=C$, $C$ for the curve and $P$ for the polynomial) ?
I have asked others about this problem, and some said that this is related to dynamical system.
polynomials dynamical-systems complex-dynamics fixedpoints
polynomials dynamical-systems complex-dynamics fixedpoints
edited yesterday
Mark McClure
23.1k34170
asked yesterday
Steve Cheng 鄭宗弘
357
edited yesterday
Mark McClure
23.1k34170
asked yesterday
Steve Cheng 鄭宗弘
357
edited yesterday
Mark McClure
23.1k34170
edited yesterday
Mark McClure
23.1k34170
edited yesterday
Mark McClure
23.1k34170
23.1k34170
asked yesterday
Steve Cheng 鄭宗弘
357
asked yesterday
Steve Cheng 鄭宗弘
357
asked yesterday
Steve Cheng 鄭宗弘
357
357
"that can be fixed": what does that mean ??
– Yves Daoust
yesterday
Why do you specify non-linear polynomial ?
– Yves Daoust
yesterday
For the first question, I have edited the problem to make it clear.
– Steve Cheng 鄭宗弘
yesterday
And for the second question, that is because you can draw a square and rotate it by $pi/2$ which can be represented as linear polynomial.
– Steve Cheng 鄭宗弘
yesterday
So $C$ is a simple closed curve (in a plane?), and $P(x,y)=(f(x,y),g(x,y))$ where $f$ and $g$ are polynomials in $x$ and $y$? With real coefficients, or complex, or something else?
– Servaes
yesterday
|
show 2 more comments
"that can be fixed": what does that mean ??
– Yves Daoust
yesterday
Why do you specify non-linear polynomial ?
– Yves Daoust
yesterday
For the first question, I have edited the problem to make it clear.
– Steve Cheng 鄭宗弘
yesterday
And for the second question, that is because you can draw a square and rotate it by $pi/2$ which can be represented as linear polynomial.
– Steve Cheng 鄭宗弘
yesterday
So $C$ is a simple closed curve (in a plane?), and $P(x,y)=(f(x,y),g(x,y))$ where $f$ and $g$ are polynomials in $x$ and $y$? With real coefficients, or complex, or something else?
– Servaes
yesterday
"that can be fixed": what does that mean ??
– Yves Daoust
yesterday
"that can be fixed": what does that mean ??
– Yves Daoust
yesterday
Why do you specify non-linear polynomial ?
– Yves Daoust
yesterday
Why do you specify non-linear polynomial ?
– Yves Daoust
yesterday
For the first question, I have edited the problem to make it clear.
– Steve Cheng 鄭宗弘
yesterday
For the first question, I have edited the problem to make it clear.
– Steve Cheng 鄭宗弘
yesterday
And for the second question, that is because you can draw a square and rotate it by $pi/2$ which can be represented as linear polynomial.
– Steve Cheng 鄭宗弘
yesterday
And for the second question, that is because you can draw a square and rotate it by $pi/2$ which can be represented as linear polynomial.
– Steve Cheng 鄭宗弘
yesterday
So $C$ is a simple closed curve (in a plane?), and $P(x,y)=(f(x,y),g(x,y))$ where $f$ and $g$ are polynomials in $x$ and $y$? With real coefficients, or complex, or something else?
– Servaes
yesterday
So $C$ is a simple closed curve (in a plane?), and $P(x,y)=(f(x,y),g(x,y))$ where $f$ and $g$ are polynomials in $x$ and $y$? With real coefficients, or complex, or something else?
– Servaes
yesterday
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Yes. The Julia set of a polynomial is always fixed in exactly the sense that you say and can often be a fractal, simple, closed curve.
The Julia set is, by definition, the closure of the set of repelling periodic points of the polynomial. For example, if $P(z)=z^2$, then the Julia set of $P$ is exactly the unit circle. If $P(z) = z^2 + c$, however, where $c$ is close to zero, then the Julia set is a somewhat distorted version of the unit circle with a fractal structure. If $c=-1/2$, for example, then the Julia set looks like so:
Thus, that simple closed curve is fixed by $P(z)=z^2-1/2$.
answered yesterday
Mark McClure
23.1k34170
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
add a comment |
up vote
1
down vote
Closed invariant curves :
- sepal ( orbit inside main chessboard box) in the parabolic case
- orbits inside Siegel disc and example image below
( external and internal rays are also invariant but they are not closed)
answered yesterday
Adam
1,1641919
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Yes. The Julia set of a polynomial is always fixed in exactly the sense that you say and can often be a fractal, simple, closed curve.
The Julia set is, by definition, the closure of the set of repelling periodic points of the polynomial. For example, if $P(z)=z^2$, then the Julia set of $P$ is exactly the unit circle. If $P(z) = z^2 + c$, however, where $c$ is close to zero, then the Julia set is a somewhat distorted version of the unit circle with a fractal structure. If $c=-1/2$, for example, then the Julia set looks like so:
Thus, that simple closed curve is fixed by $P(z)=z^2-1/2$.
answered yesterday
Mark McClure
23.1k34170
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
add a comment |
up vote
5
down vote
accepted
Yes. The Julia set of a polynomial is always fixed in exactly the sense that you say and can often be a fractal, simple, closed curve.
The Julia set is, by definition, the closure of the set of repelling periodic points of the polynomial. For example, if $P(z)=z^2$, then the Julia set of $P$ is exactly the unit circle. If $P(z) = z^2 + c$, however, where $c$ is close to zero, then the Julia set is a somewhat distorted version of the unit circle with a fractal structure. If $c=-1/2$, for example, then the Julia set looks like so:
Thus, that simple closed curve is fixed by $P(z)=z^2-1/2$.
answered yesterday
Mark McClure
23.1k34170
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Yes. The Julia set of a polynomial is always fixed in exactly the sense that you say and can often be a fractal, simple, closed curve.
The Julia set is, by definition, the closure of the set of repelling periodic points of the polynomial. For example, if $P(z)=z^2$, then the Julia set of $P$ is exactly the unit circle. If $P(z) = z^2 + c$, however, where $c$ is close to zero, then the Julia set is a somewhat distorted version of the unit circle with a fractal structure. If $c=-1/2$, for example, then the Julia set looks like so:
Thus, that simple closed curve is fixed by $P(z)=z^2-1/2$.
answered yesterday
Mark McClure
23.1k34170
Yes. The Julia set of a polynomial is always fixed in exactly the sense that you say and can often be a fractal, simple, closed curve.
The Julia set is, by definition, the closure of the set of repelling periodic points of the polynomial. For example, if $P(z)=z^2$, then the Julia set of $P$ is exactly the unit circle. If $P(z) = z^2 + c$, however, where $c$ is close to zero, then the Julia set is a somewhat distorted version of the unit circle with a fractal structure. If $c=-1/2$, for example, then the Julia set looks like so:
Thus, that simple closed curve is fixed by $P(z)=z^2-1/2$.
answered yesterday
Mark McClure
23.1k34170
answered yesterday
Mark McClure
23.1k34170
answered yesterday
Mark McClure
23.1k34170
answered yesterday
Mark McClure
23.1k34170
23.1k34170
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
add a comment |
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
I wonder if P is a bijection on that curve.
– Steve Cheng 鄭宗弘
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
@SteveCheng鄭宗弘 No. In fact, it is two-to-one, just like $z^2$ on the unit circle.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
No - $f$ acts transitively on the Julia set.
– Mark McClure
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Sorry, I deleted the question...
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
Since the Julia set may be a union of some disconnected components. I also wonder if there exists a Julia set J(f) that consisted of several components and a connected subset S of J(f) that f(S)=S?
– Steve Cheng 鄭宗弘
yesterday
add a comment |
up vote
1
down vote
Closed invariant curves :
- sepal ( orbit inside main chessboard box) in the parabolic case
- orbits inside Siegel disc and example image below
( external and internal rays are also invariant but they are not closed)
answered yesterday
Adam
1,1641919
add a comment |
up vote
1
down vote
Closed invariant curves :
- sepal ( orbit inside main chessboard box) in the parabolic case
- orbits inside Siegel disc and example image below
( external and internal rays are also invariant but they are not closed)
answered yesterday
Adam
1,1641919
add a comment |
up vote
1
down vote
up vote
1
down vote
Closed invariant curves :
- sepal ( orbit inside main chessboard box) in the parabolic case
- orbits inside Siegel disc and example image below
( external and internal rays are also invariant but they are not closed)
answered yesterday
Adam
1,1641919
Closed invariant curves :
- sepal ( orbit inside main chessboard box) in the parabolic case
- orbits inside Siegel disc and example image below
( external and internal rays are also invariant but they are not closed)
answered yesterday
Adam
1,1641919
edited 16 hours ago
answered yesterday
Adam
1,1641919
answered yesterday
Adam
1,1641919
answered yesterday
Adam
1,1641919
1,1641919
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005096%2fexistence-of-fixed-simple-closed-curve-by-polynomials%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
"that can be fixed": what does that mean ??
– Yves Daoust
yesterday
Why do you specify non-linear polynomial ?
– Yves Daoust
yesterday
For the first question, I have edited the problem to make it clear.
– Steve Cheng 鄭宗弘
yesterday
And for the second question, that is because you can draw a square and rotate it by $pi/2$ which can be represented as linear polynomial.
– Steve Cheng 鄭宗弘
yesterday
So $C$ is a simple closed curve (in a plane?), and $P(x,y)=(f(x,y),g(x,y))$ where $f$ and $g$ are polynomials in $x$ and $y$? With real coefficients, or complex, or something else?
– Servaes
yesterday