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
$begingroup$
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
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
$endgroup$
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
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
$endgroup$
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
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
$endgroup$
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
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
edited Jun 21 at 17:36
YCor
30k4 gold badges89 silver badges144 bronze badges
30k4 gold badges89 silver badges144 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
asked Jun 21 at 15:21
GariGari
284 bronze badges
284 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919.
Sur les équations fonctionnelles
Bulletin de la S. M. F., tome 48 (1920), p. 208-314.
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 Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
$endgroup$
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
2
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
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.
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
$endgroup$
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
Jun 21 at 17:41
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919.
Sur les équations fonctionnelles
Bulletin de la S. M. F., tome 48 (1920), p. 208-314.
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 Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
$endgroup$
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
2
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
add a comment |
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919.
Sur les équations fonctionnelles
Bulletin de la S. M. F., tome 48 (1920), p. 208-314.
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 Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
$endgroup$
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
2
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
add a comment |
$begingroup$
The answer to a) is yes, and this was proved by Fatou in 1919.
Sur les équations fonctionnelles
Bulletin de la S. M. F., tome 48 (1920), p. 208-314.
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 Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
$endgroup$
The answer to a) is yes, and this was proved by Fatou in 1919.
Sur les équations fonctionnelles
Bulletin de la S. M. F., tome 48 (1920), p. 208-314.
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 Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
edited Jun 22 at 16:23
answered Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
answered Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
answered Jun 21 at 19:59
Alexandre EremenkoAlexandre Eremenko
53k6 gold badges149 silver badges271 bronze badges
53k6 gold badges149 silver badges271 bronze badges
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
2
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
add a comment |
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
2
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
Thanks a lot! I haven't gone through Fatou's argument yet but I am more convinced now that it is in fact true.
$endgroup$
– Gari
Jun 22 at 11:05
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
$begingroup$
I don't see yet why the Julia set being equal to a circle or an arc of a circle implies that the answer to Q1 is yes but I will try.
$endgroup$
– Gari
Jun 22 at 11:06
2
2
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
$begingroup$
@Gari: This is explained in Fatou, archive.numdam.org/article/BSMF_1920__48__208_1.pdf
$endgroup$
– Alexandre Eremenko
Jun 22 at 16:22
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.
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
$endgroup$
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
Jun 21 at 17:41
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
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.
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
$endgroup$
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
Jun 21 at 17:41
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
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.
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
$endgroup$
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 Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
edited Jun 21 at 17:37
YCor
30k4 gold badges89 silver badges144 bronze badges
30k4 gold badges89 silver badges144 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
answered Jun 21 at 15:49
Bullet51Bullet51
2,0551 gold badge6 silver badges21 bronze badges
2,0551 gold badge6 silver badges21 bronze badges
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
Jun 21 at 17:41
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
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
Jun 21 at 17:41
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
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
Jun 21 at 17:41
$begingroup$
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
Jun 21 at 17:41
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
$begingroup$
Thanks for the answer. However I wasn't able to track down the statement using their reference. Maybe it is somewhere in Milnor and I have overlooked it.
$endgroup$
– Gari
Jun 22 at 10:53
add a comment |
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