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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 29, 2011 at 21:46 comment added Joseph O'Rourke @Ryan: Great! Terminological variations are converging...
Jul 29, 2011 at 21:40 comment added Ryan Budney For 3-manifolds the Riemann curvature tensor is recoverable from the Ricci tensor, so "Ricci flat" implies its a Euclidean manifold -- locally isometric to open subsets of $\mathbb R^3$.
Jul 29, 2011 at 21:09 history edited Will Jagy CC BY-SA 3.0
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Jul 29, 2011 at 20:52 comment added Joseph O'Rourke Thanks, Will, I just ordered the Thorpe book. For the record, Hick's Cor.6: "A hypersurface is Ricci flat if and only if it is Einstein with total curvature zero. If $n=3$ and $M$ is Ricci flat, then the second mean curvature is also zero so at points $m$ on $M$ that are not flat points ($L_m \neq 0$), the multiplicity of the nonzero principle curvature is unity."
Jul 29, 2011 at 20:13 history edited Will Jagy CC BY-SA 3.0
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Jul 29, 2011 at 19:38 history answered Will Jagy CC BY-SA 3.0