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Mar 4, 2010 at 0:04 vote accept José Figueroa-O'Farrill
Mar 3, 2010 at 2:32 comment added Igor Belegradek Jose, do not worry about these things; my contribution was tiny.
Mar 2, 2010 at 23:53 comment added José Figueroa-O'Farrill I'm minded to accept this answer, but I will wait a little while to see if anyone else has anything to say whether some further progress has been made on this question.
Mar 2, 2010 at 17:15 comment added José Figueroa-O'Farrill @Igor: I've just taken a look at Petersen. Thanks. So indeed this is the "euclideanisation" of the lorentzian Schwarzschild solution. I will be updating the question to incorporate your answer. Many thanks.
Mar 2, 2010 at 17:01 comment added Igor Belegradek Jose, the Schwarzschild metric I know is Riemannian. :) Look e.g. in Petersen's "Riemannian Geometry" text for detailed study of Riemannian Schwarzschild metric
Mar 2, 2010 at 16:56 comment added Igor Belegradek Jose, yes this is the book. Another standard source is "Surveys in Differential Geometry: Essays on Einstein manifolds" by International Press which was also published a decade ago. I do not think there have been any major breakthroughs since then in regard to your question.
Mar 2, 2010 at 16:55 comment added José Figueroa-O'Farrill Another question: the Schwarzschild metric I know is lorentzian, so it has holonomy $\mathrm{SO}(3,1)$. Hence this answers the bonus question! This prompts another question: are there noncompact Ricci-flat riemannian manifolds with generic holonomy?
Mar 2, 2010 at 16:48 comment added José Figueroa-O'Farrill Is the book by Berger that you mention the same as A panoramic view of Riemannian geometry?
Mar 2, 2010 at 15:28 comment added Deane Yang Berger's book is pretty old by now, but I think the situation remains as you describe. As far as I know, every construction of a closed Ricci-flat manifold proceeds by building a metric with a holonomy group that implies Ricci-flat. The holonomy assumption implies algebraic conditions on the curvature tensor that in turn play crucial roles in the elliptic estimates used to prove existence. In some sense, you're replacing a second order elliptic PDE by a stronger first order system that implies the second order PDE.
Mar 2, 2010 at 15:22 history answered Igor Belegradek CC BY-SA 2.5