Timeline for $L^{n/2}$ norm of scalar curvature
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2017 at 17:38 | vote | accept | Matthias Ludewig | ||
| Jan 31, 2017 at 14:56 | answer | added | Deane Yang | timeline score: 5 | |
| Jan 31, 2017 at 14:35 | comment | added | Matthias Ludewig | Yes, that is the way I am used to look at the Yamabe functional, formulating it in terms of the conformal factor. That was why I find the statement in the Wikipedia article very curious, and thy I wondered whether there is some obvious reason why this norm should be bounded. But apparently there isn't and the Wikipedia article should be updated. | |
| Jan 31, 2017 at 3:44 | comment | added | Deane Yang | You can find the formula for the scalar curvature of the metric $e^{2\phi}g$ in terms of the scalar curvature of $g$ in terms of $\phi$ and its first and second covariant derivatives here: en.wikipedia.org/wiki/… | |
| Jan 31, 2017 at 0:27 | comment | added | Deane Yang | Just calculate the Yamabe functional for a metric conformal to $g$ in terms of the Yamabe functional of $g$ and the first and second derivatives of the conformal factor. Observe that there is an inequality relating the two Yamabe functionals. Now apply the Holder's inequality. The $L^{n/2}$ norm of scalar curvature is useful, because it is scale invariant, which is why there is no volume factor on the right side of the inequality. | |
| Jan 30, 2017 at 23:19 | answer | added | Sam | timeline score: -1 | |
| Jan 30, 2017 at 11:57 | comment | added | Matthias Ludewig | The Wikipedia defines $Y(g)$ as the infimum of the Yamabe energy $\mathcal{E}(g)$, taken over the set of metrics conformal to $g$. It the proceeds to argue why the infimum is not $-\infty$. The argument is definitely not there to say why $\mathcal{E}(g)$ is finite; in particular, on would not need to use Hölder inequality for this. In conclusion, I have no idea how you come to your interpretation of the relevant passage of the Wikipedia article. | |
| Jan 27, 2017 at 8:34 | comment | added | Gunnar Þór Magnússon | As far as I can tell, the Wikipedia article uses that inequality to say that the Yamabe functional is not $-\infty$. In this case, this should be because we're on a compact manifold and the scalar curvature is a smooth function on it, so the right-hand side of the inequality is $> -\infty$. | |
| Jan 27, 2017 at 8:16 | history | asked | Matthias Ludewig | CC BY-SA 3.0 |