Timeline for Einstein metrics on the tangent bundle
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| May 25 at 22:26 | comment | added | Igor Belegradek | 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. | |
| May 25 at 18:16 | history | edited | LSpice | CC BY-SA 4.0 |
Further proofreading, while this is on the front page
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| S May 25 at 18:13 | history | suggested | Ali Taghavi | CC BY-SA 4.0 |
einstein to Einstein
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| May 25 at 16:39 | review | Suggested edits | |||
| S May 25 at 18:13 | |||||
| May 25 at 16:30 | answer | added | Ali Taghavi | timeline score: 1 | |
| Feb 15, 2012 at 6:53 | comment | added | william | and what about the torus, by making the same constructions as in the above post ? | |
| Feb 14, 2012 at 15:36 | comment | added | william | so what would be then a counter example ? | |
| Feb 14, 2012 at 14:50 | comment | added | Robert Bryant | @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. | |
| Feb 14, 2012 at 7:51 | comment | added | william | What if you choose $X \subset TS^{2}$ as $X := \{(x,v) \in T_{x}S^{2} | g_{x}(v,v) <= 1\}$, where $g$ is some metric in $S^{2}$. How can one compute $\tau(X)$ and $\chi(X)$ (as in the hitchin-thorpe formula)? | |
| Feb 12, 2012 at 14:35 | comment | added | william | do you have some counterexample using hitchin - thorpe formula? just some counterexample. or is there any, where one is considering the tangentbundle of some given manifold ? | |
| Feb 12, 2012 at 12:49 | comment | added | Deane Yang | It's not known in general which manifolds admit Einstein metrics. Why should the question be easier for the total space of a tangent bundle? And you still have not said whether you want the metric to be complete or not. | |
| Feb 12, 2012 at 7:47 | comment | added | william | can you give an example, please ? | |
| Feb 12, 2012 at 6:51 | comment | added | william | what do you mean by that ? | |
| Feb 12, 2012 at 0:22 | comment | added | user39719 | What about restricting to the unit tangent bundle? | |
| Feb 11, 2012 at 23:53 | comment | added | william | because $TM$ is not compact and the hitchin-thorpe formula does not necessary apply ?! | |
| Feb 11, 2012 at 23:51 | comment | added | william | actually i am refering to the following: if $M$ is a manifold as stated above (compact, riemannian , real analytic ...). Now consider the tangent bundle as a manifold $TM$ (as a new manifold). Does this manifold always carry a einstein metric? If no why exactly? | |
| Feb 11, 2012 at 23:09 | comment | added | Deane Yang | Ryan, I assumed he was asking for an Einstein metric on the total space. | |
| Feb 11, 2012 at 23:05 | comment | added | Ryan Budney | 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? | |
| Feb 11, 2012 at 22:42 | comment | added | Deane Yang | 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? | |
| Feb 11, 2012 at 21:58 | comment | added | Ryan Budney | Your question has answer no: en.wikipedia.org/wiki/Hitchin%E2%80%93Thorpe_inequality | |
| Feb 11, 2012 at 21:48 | history | asked | william | CC BY-SA 3.0 |