Timeline for What is known about analogous results of Kazdan and Warner in higher dimensions?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2013 at 12:13 | comment | added | Ritwik | Thank you. I see why the scalar curvature can not equal the $*e(TM)$. | |
| Jan 5, 2013 at 19:11 | comment | added | Otis Chodosh | *by my first sentence, I mean an identity between scalar curvature and the Gauss-Bonnet integrand (as @unknown comments) | |
| Jan 5, 2013 at 19:10 | comment | added | Otis Chodosh | ...giving a contradiction. You may be interested in the following MO post: mathoverflow.net/questions/30035/…. | |
| Jan 5, 2013 at 19:09 | comment | added | Otis Chodosh | @Ritwik, there is no possibility of such an identity holding. If $*e(TM)$ was some multiple of the scalar curvature $R$, this would imply that $\int R = \int *e(TM) dV = <e(TM),[M]>$, and this is a topological invariant of the (smooth structure) topology of $M$. (see en.wikipedia.org/wiki/…). However, one can show that in dimensions $\geq 3$, all manifolds admit a metric of negative scalar curvature, e.g. $S^3$. So, if the above identity held, we could evaluate $<e(TM),[M]>$ on the standard metric and this other one... | |
| Jan 5, 2013 at 19:00 | comment | added | John Pardon | I assume that by $\ast e(TM)$ you mean the Gauss--Bonnet integrand. In dimensions at least three, it is different from the scalar curvature. So prescribing scalar curvature is a different question in dimensions at least three. | |
| Jan 5, 2013 at 15:54 | comment | added | Ritwik | I am sorry if this is a trivial question or a well known fact, but is the Scalar curvature the same as $*e(TM)$ (maybe upto a constant factor)? | |
| Jan 5, 2013 at 15:43 | comment | added | Liviu Nicolaescu | Check this Wikipedia page en.wikipedia.org/wiki/Prescribed_scalar_curvature_problem | |
| Jan 5, 2013 at 15:17 | history | asked | Ritwik | CC BY-SA 3.0 |