Timeline for Conformally flat manifold with zero scalar
Current License: CC BY-SA 3.0
11 events
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| Jan 18, 2016 at 15:44 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Jan 13, 2016 at 12:33 | history | undeleted | Mikhail Katz | ||
| Jan 13, 2016 at 12:33 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Jan 13, 2016 at 3:08 | history | deleted | Mikhail Katz | via Vote | |
| Jan 13, 2016 at 3:08 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Jan 13, 2016 at 3:07 | comment | added | Mikhail Katz | For sure, sorry. | |
| Jan 13, 2016 at 0:12 | comment | added | Deane Yang | In dimensions 4 or higher, conformally flat means the Weyl tensor vanishes. If the scalar curvature also vanishes, there is still the possibility that the trace-free part of the Ricci curvature is non-vanishing. As Robert points out, there are indeed local metrics satisfying this. | |
| Jan 12, 2016 at 20:45 | comment | added | Robert Bryant | @katz: Indeed, your local argument cannot be correct. In dimensions $n>2$, there are many conformally flat metrics with vanishing scalar curvature that are not flat. Essentially, they depend on $2$ arbitrary functions of $n{-}1$ variables, and the generic one does not have constant sectional curvature. | |
| Jan 12, 2016 at 19:05 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Jan 12, 2016 at 18:57 | comment | added | Holonomia | @ katz: Just a doubt: your product $S^2 \times X$ is conformally flat as the OP asked ? it is well-known that $S^2 \times H^2$ is conformally flat where $H^2$ unit disk with the hyperbolic metric i.e. the universal covering of X. So $S^2 \times X$ is locally conformally flat. It is obvious that a multiple of the product metric is flat?. | |
| Jan 12, 2016 at 16:01 | history | answered | Mikhail Katz | CC BY-SA 3.0 |