Timeline for Possible sign of scalar curvature for Einstein warped product manifold with Ricci-flat
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2022 at 21:04 | vote | accept | MathDG | ||
| Feb 20, 2022 at 20:27 | history | edited | Vitali Kapovitch | CC BY-SA 4.0 |
added 16 characters in body
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| Feb 20, 2022 at 20:26 | comment | added | Vitali Kapovitch | @exxxit8 Yes, By Bonnet-Myer's theorem. | |
| Feb 20, 2022 at 20:24 | comment | added | MathDG | Thank you so much again dear Professor Kapovitch! So in conclusion the question, regarding my initial question, is independent of the warped product, an Einstein manifold with Riemannian metric to have positive Ricci curvature must be compact, am I right? | |
| Feb 20, 2022 at 20:15 | comment | added | Vitali Kapovitch | If by semi-Riemannian you mean a metric of arbitrary signature then I don't know if this is possible. Positivity of Ricci does not imply compactness in this case so my argument does not apply. | |
| Feb 20, 2022 at 20:10 | comment | added | MathDG | Thank you! So if $Ric_F = 0$ then $M$ cannot be compact. But this is also true in the case of semi-Riemannian metric and conformal base-manifold $\bar{g_B} = \frac{1} {\phi^2} g_B $? | |
| Feb 20, 2022 at 20:02 | comment | added | Vitali Kapovitch | Bonnet-Myers theorem says that if $Ric_{M^n}\ge (n-1)$ then $diam(M)\le \pi$. This is completely independent of warped products. | |
| Feb 20, 2022 at 19:59 | comment | added | MathDG | Thank you for your answer. You say "Constant positive Ricci curvature implies that $M$ is compact", but you mean that constant positive Ricci curvature implies that $M$ is compact if $Ric_F = 0$, or in general independent of $Ric_F$, to have constant positive Ricci curvature, does $M$ have to be compact? | |
| Feb 20, 2022 at 19:42 | history | answered | Vitali Kapovitch | CC BY-SA 4.0 |