Timeline for Uniformisation for non simple closed curves
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Oct 10, 2017 at 2:51 | answer | added | Mohammad Ghomi | timeline score: 3 | |
Jul 7, 2017 at 19:50 | comment | added | coudy | @goette. The question is the following. Given a C1 closed curve γ:S1→R2 with finitely many self-intersections, all of them transverse, is there a continuous map (maybe even conformal) from the unit open disk to the plane, such that the number of preimages of any point in R2∖γ(S1) is equal to the absolute value of the number of turns the curve makes around the point? | |
Jul 7, 2017 at 7:54 | comment | added | Sebastian Goette | @coudy Already in the figure-8 example, you need to count multiplicities with sign. An arbitray extension will give correct multiplicities (counted with sign). Do you want an extension where no preimages cancel because of sign issues? | |
Jul 7, 2017 at 6:19 | comment | added | coudy | @Eremenko. An arbitrary extension will not necessarily give the expected multiplicity. Read again the question and look at the diagram. | |
Jul 6, 2017 at 21:12 | review | Close votes | |||
Jul 7, 2017 at 4:05 | |||||
Jul 6, 2017 at 20:49 | comment | added | Alexandre Eremenko | @coudy: The map you describe in the comment is not conformal, and not even open. Of course there is always a continuous map: a continuous map of a circle extends continuously in the disk. So what are you really asking? | |
Jul 6, 2017 at 14:11 | comment | added | coudy | @Katz Twist the disk in three space, project it on a plane. So there is a line dividing the disk that projects exactly on the double point. Perhaps asking for all the number of turns to be of the same sign is a necessary condition for a conformal map. | |
Jul 6, 2017 at 14:03 | comment | added | Mikhail Katz | What would such a map look like for the figure-8 curve? | |
Jul 6, 2017 at 13:26 | history | edited | coudy |
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Jul 6, 2017 at 8:59 | history | asked | coudy | CC BY-SA 3.0 |