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













4












$begingroup$


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.










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$endgroup$
















    4












    $begingroup$


    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.










    share|cite|improve this question









    New contributor



    Gari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      4












      4








      4





      $begingroup$


      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.










      share|cite|improve this question









      New contributor



      Gari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      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






      share|cite|improve this question









      New contributor



      Gari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share|cite|improve this question









      New contributor



      Gari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      YCor

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      asked 9 hours ago









      GariGari

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          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$








          • 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


















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$















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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$








            • 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















            3












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$








            • 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













            3












            3








            3





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            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

            2,0081 gold badge6 silver badges20 bronze badges




            2,0081 gold badge6 silver badges20 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
              7 hours ago












            • 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











            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Alexandre EremenkoAlexandre Eremenko

                52.9k6 gold badges148 silver badges271 bronze badges




                52.9k6 gold badges148 silver badges271 bronze badges




















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