Timeline for compatification of $\mathbb{R}^2$ under a conformal metric
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8, 2020 at 14:38 | comment | added | student | Thank you very much! This makes a lot of sense. | |
| Jul 6, 2020 at 20:42 | comment | added | Sebastian | You have $\phi^*\delta=\tfrac{1}{(\tilde x^2+\tilde y^2)^2} \delta$. Therefore you need that $e^{2u\circ\phi(\tilde x,\tilde y)} \tfrac{1}{(\tilde x^2+\tilde y^2)^2}$ extends smoothly to $0$ and that the value of this function at $0$ is bigger than 0. | |
| Jul 6, 2020 at 14:48 | comment | added | student | Sorry for the typo in the previous comment. It should be $e^{2u}=1/(1+x^2+4y^2)$. Could you give me a reference on how to use the conformal coordinates you gave to proceed computation? | |
| Jul 6, 2020 at 14:29 | comment | added | student | Thank you very much for your post. Yes, the question you posted is really what I want to understand. I have trouble in understanding why the metric completion under the metric $g=e^{2u}\delta$ implies that such completed metric is also Riemannian, or why it is also a smooth extension of the original metric defined only on $\mathbb{R}^2$? For example, given $e^{2u}=1/(x^2+4y^2)$, what are such $M$, $U$ and $\phi$? Sorry I have a very poor basis of Riemannian geometry. | |
| Jul 6, 2020 at 13:00 | history | answered | Sebastian | CC BY-SA 4.0 |