Timeline for Can the topologist's sine curve be realized as a Julia set?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 24 at 14:55 | comment | added | KhashF | @SophieM Thanks for your comments. I think the easiest way to argue that $f$ cannot admit a Siegel disk or Herman ring is to notice that $f$ should be injective on such a component. But there is only one Fatou component and a map of degree $d\geq 2$ restricts to a $d$-to-$1$ self-map on its Fatou set. | |
| May 24 at 14:24 | vote | accept | KhashF | ||
| May 24 at 13:33 | history | became hot network question | |||
| May 24 at 11:52 | answer | added | Alexandre Eremenko | timeline score: 18 | |
| May 24 at 7:41 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Furter tagged+minor formatting
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| May 24 at 6:48 | comment | added | Sophie M | Slightly subtle aspect of the simple connectedness issue (because this caused me to doubt myself after I posted that comment): $U$ isn't a Hermann ring because it's simply connected in the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, and it's not a Siegel disk because it's not simply connected in the plane $\mathbb{C}$. | |
| May 24 at 0:07 | comment | added | Sophie M | The complement of $T$ is simply connected in the Riemann sphere, so the Fatou set of any map $f$ with $J(f) = T$ has a single component $U$, and the classification of Fatou components tells us what $U$ could be like. $U$ is neither a Hermann ring (because it's simply connected) nor a Siegel disk (because it's clearly not the image of the unit disk under an analytic map). So either $U$ is parabolic or it contains a single attracting fixed point. Have you investigated what either of these possibilities would imply about $f$? | |
| May 23 at 22:44 | history | edited | KhashF | CC BY-SA 4.0 |
Expanded the discussion.
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| May 23 at 22:09 | history | asked | KhashF | CC BY-SA 4.0 |