Timeline for Simply-connected domain around a curve
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2010 at 14:49 | comment | added | Lasse Rempe | 0 and $\infty$ are interchanged. In the restatement, the Jordan domain is supposed to be contained in $\mathbb{C}$ and have 0 on the boundary. If you transform this picture via 1/z, you get a Jordan domain that is contained in $\mathbb{C}\setminus\{0\}$ and contains $\infty$ on the boundary. | |
| Jul 29, 2010 at 14:08 | comment | added | BS. | Then, what takes the role of $0$ in the restatement ? | |
| Jul 29, 2010 at 5:14 | answer | added | fedja | timeline score: 3 | |
| Jul 28, 2010 at 22:24 | comment | added | Lasse Rempe | No, but it doesn't matter as you can apply the transformation $z\mapsto \frac{1}{z}$; hence my restatement. | |
| Jul 27, 2010 at 13:10 | comment | added | BS. | In view of your comment, I think that in the main question, you meant $\gamma(t)\to 0$ when $t\to\infty$. Am I right ? | |
| Jul 27, 2010 at 13:00 | comment | added | Lasse Rempe | Note that the question can be restated, perhaps more concisely, as follows: Let be a Jordan arc in the complex plane, connecting 0 and 1. Is there a Jordan domain V, containing $\gamma\setminus\{0\}$, such that the convergence of gamma to 0 is horocyclic in V? | |
| Jul 27, 2010 at 12:59 | history | edited | Lasse Rempe | CC BY-SA 2.5 |
added 6 characters in body
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| Jul 26, 2010 at 16:52 | history | edited | Greg Kuperberg |
edited tags
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| Jul 26, 2010 at 15:38 | history | asked | Lasse Rempe | CC BY-SA 2.5 |