Let $M$ be a compact, real analytic, riemannian manifold with real analytic metric $g$. Does the tangent bundle always admit an Einstein metric ?
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1$\begingroup$ Your question has answer no: en.wikipedia.org/wiki/Hitchin%E2%80%93Thorpe_inequality $\endgroup$– Ryan BudneyCommented Feb 11, 2012 at 21:58
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1$\begingroup$ This might be an interesting question, but it comes out of nowhere for me. Is there some reason why you believe that this might be the case? Is there some particular construction or approach you have in mind for obtaining the Einstein metric? $\endgroup$– Deane YangCommented Feb 11, 2012 at 22:42
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$\begingroup$ I suppose your question is ambiguous. When you refer to the tangent bundle, are you actually referring to the bundle as-stated, or are you referring to the total space of the bundle? $\endgroup$– Ryan BudneyCommented Feb 11, 2012 at 23:05
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1$\begingroup$ @marco: You aren't going to get an obstruction that way because the tangent bundle of $S^2$ does have a (complete) Einstein metric on it, in fact, a Ricci-flat one, the Eguchi-Hansen metric. $\endgroup$– Robert BryantCommented Feb 14, 2012 at 14:50
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1$\begingroup$ To my knowledge, in dimensions $>4$ no obstructions are known to the existence of Einstein metrics. Since tangent bundles are even dimensional, we are left with tangent bundle to surfaces. Tangent bundles to orientable surfaces are parallelizable. Any open parallelizable $n$-manifold immerses into $\mathbb R^n$ by Smale-Hirsch. The pullback metric under the immersion is flat (hence Einstein). Most likely the metric is incomplete but then the OP did not insist on completeness. I suspect a similar argument can work for non-orientable surfaces. $\endgroup$– Igor BelegradekCommented May 25 at 22:26
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1 Answer
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Some Particular cases are explained in Papaghiuc - On an Einstein structure on the tanent bundle of a space form.
