Smooth Julia set for quadratic polynomialsWhat are the shapes of rational functions?How is the Julia set of $fg$ related to the Julia set of $gf$?Is this a Julia set (and if so, for which function family is it the Julia set)?Invariant curves of rational functions“Explicit” examples of Irrational numbers very well approximated by rationnal numbersdegree of a rational map on infinitely connected fatou componentA question about Julia set for quadratic familyInfinitely renormalizable parameters for quadratic polynomialsLimit cycles of quadratic systems and closed geodesics(Finitness of $H(2)$)Julia set containing smooth curve
Smooth Julia set for quadratic polynomials
What are the shapes of rational functions?How is the Julia set of $fg$ related to the Julia set of $gf$?Is this a Julia set (and if so, for which function family is it the Julia set)?Invariant curves of rational functions“Explicit” examples of Irrational numbers very well approximated by rationnal numbersdegree of a rational map on infinitely connected fatou componentA question about Julia set for quadratic familyInfinitely renormalizable parameters for quadratic polynomialsLimit cycles of quadratic systems and closed geodesics(Finitness of $H(2)$)Julia set containing smooth curve
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This question is related to a classification of rational maps in terms of properties of their Julia set.
Let $f= z^2 + c$, with $cin mathbbC$ be a quadratic polynomial such that its Julia set $J(f)$ is connected.
- Q1: If there exists a relatively open set of $J(f)$ that is (the support of) a smooth curve. Is $f$ conjugate (resp equal) to a Tchebychev polynomial or power map $z^2$?
- Q2: Is the answer to Q1 yes under the additional assumption that $J(f)$ is also locally connected?
- Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $c$?
Thanks a lot.
ds.dynamical-systems complex-dynamics
New contributor
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add a comment |
$begingroup$
This question is related to a classification of rational maps in terms of properties of their Julia set.
Let $f= z^2 + c$, with $cin mathbbC$ be a quadratic polynomial such that its Julia set $J(f)$ is connected.
- Q1: If there exists a relatively open set of $J(f)$ that is (the support of) a smooth curve. Is $f$ conjugate (resp equal) to a Tchebychev polynomial or power map $z^2$?
- Q2: Is the answer to Q1 yes under the additional assumption that $J(f)$ is also locally connected?
- Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $c$?
Thanks a lot.
ds.dynamical-systems complex-dynamics
New contributor
$endgroup$
add a comment |
$begingroup$
This question is related to a classification of rational maps in terms of properties of their Julia set.
Let $f= z^2 + c$, with $cin mathbbC$ be a quadratic polynomial such that its Julia set $J(f)$ is connected.
- Q1: If there exists a relatively open set of $J(f)$ that is (the support of) a smooth curve. Is $f$ conjugate (resp equal) to a Tchebychev polynomial or power map $z^2$?
- Q2: Is the answer to Q1 yes under the additional assumption that $J(f)$ is also locally connected?
- Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $c$?
Thanks a lot.
ds.dynamical-systems complex-dynamics
New contributor
$endgroup$
This question is related to a classification of rational maps in terms of properties of their Julia set.
Let $f= z^2 + c$, with $cin mathbbC$ be a quadratic polynomial such that its Julia set $J(f)$ is connected.
- Q1: If there exists a relatively open set of $J(f)$ that is (the support of) a smooth curve. Is $f$ conjugate (resp equal) to a Tchebychev polynomial or power map $z^2$?
- Q2: Is the answer to Q1 yes under the additional assumption that $J(f)$ is also locally connected?
- Q3: If the answer to Q1 is no, can one describe the set of such counterexamples in terms of the parameter $c$?
Thanks a lot.
ds.dynamical-systems complex-dynamics
ds.dynamical-systems complex-dynamics
New contributor
New contributor
edited 7 hours ago
YCor
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29.9k4 gold badges89 silver badges144 bronze badges
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asked 9 hours ago
GariGari
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2 Answers
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From this paper of Bedford and Kim (arxiv link):
Fatou showed that if the
Julia set $J$ is a smooth curve, then either $J$ is the unit circle, or $J$ is a real interval. If $J$ is the circle, then $f$ is equivalent to $z → z^d$
, where $d$ is an integer with $|d| ≥ 2$; if $J$ is the interval, then $f$ is equivalent to a Chebyshev polynomial.
$endgroup$
2
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
add a comment |
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919. There are many generalizations of this fact. For one generalization, and further references you may look to
Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.
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2 Answers
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2 Answers
2
active
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active
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$begingroup$
From this paper of Bedford and Kim (arxiv link):
Fatou showed that if the
Julia set $J$ is a smooth curve, then either $J$ is the unit circle, or $J$ is a real interval. If $J$ is the circle, then $f$ is equivalent to $z → z^d$
, where $d$ is an integer with $|d| ≥ 2$; if $J$ is the interval, then $f$ is equivalent to a Chebyshev polynomial.
$endgroup$
2
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
add a comment |
$begingroup$
From this paper of Bedford and Kim (arxiv link):
Fatou showed that if the
Julia set $J$ is a smooth curve, then either $J$ is the unit circle, or $J$ is a real interval. If $J$ is the circle, then $f$ is equivalent to $z → z^d$
, where $d$ is an integer with $|d| ≥ 2$; if $J$ is the interval, then $f$ is equivalent to a Chebyshev polynomial.
$endgroup$
2
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
add a comment |
$begingroup$
From this paper of Bedford and Kim (arxiv link):
Fatou showed that if the
Julia set $J$ is a smooth curve, then either $J$ is the unit circle, or $J$ is a real interval. If $J$ is the circle, then $f$ is equivalent to $z → z^d$
, where $d$ is an integer with $|d| ≥ 2$; if $J$ is the interval, then $f$ is equivalent to a Chebyshev polynomial.
$endgroup$
From this paper of Bedford and Kim (arxiv link):
Fatou showed that if the
Julia set $J$ is a smooth curve, then either $J$ is the unit circle, or $J$ is a real interval. If $J$ is the circle, then $f$ is equivalent to $z → z^d$
, where $d$ is an integer with $|d| ≥ 2$; if $J$ is the interval, then $f$ is equivalent to a Chebyshev polynomial.
edited 7 hours ago
YCor
29.9k4 gold badges89 silver badges144 bronze badges
29.9k4 gold badges89 silver badges144 bronze badges
answered 8 hours ago
Bullet51Bullet51
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2
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
add a comment |
2
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
2
2
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
$begingroup$
They take $f$ arbitrary holomorphic. What's meant by "is the unit circle"? If one conjugates $zmapsto z^2$ by an affine map, one can get any circle.
$endgroup$
– YCor
7 hours ago
add a comment |
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919. There are many generalizations of this fact. For one generalization, and further references you may look to
Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.
$endgroup$
add a comment |
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919. There are many generalizations of this fact. For one generalization, and further references you may look to
Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.
$endgroup$
add a comment |
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919. There are many generalizations of this fact. For one generalization, and further references you may look to
Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.
$endgroup$
The answer to a) is yes, and this was proved by Fatou in 1919. There are many generalizations of this fact. For one generalization, and further references you may look to
Meromorphic functions with linearly distributed values and Julia sets of rational functions, Proc. AMS. 137 (2009), 2329-2333.
answered 4 hours ago
Alexandre EremenkoAlexandre Eremenko
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Gari is a new contributor. Be nice, and check out our Code of Conduct.
Gari is a new contributor. Be nice, and check out our Code of Conduct.
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