Timeline for Metric on tangent bundle
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2017 at 10:33 | answer | added | Olga | timeline score: 2 | |
| Feb 19, 2013 at 3:08 | vote | accept | Cecilia | ||
| Feb 18, 2013 at 13:12 | answer | added | Robert Bryant | timeline score: 15 | |
| Feb 18, 2013 at 3:05 | vote | accept | Cecilia | ||
| Feb 19, 2013 at 3:08 | |||||
| Feb 18, 2013 at 2:54 | comment | added | Cecilia | Yes, that's exactly what I'm asking. Thank you again. | |
| Feb 18, 2013 at 2:54 | answer | added | Peter Michor | timeline score: 17 | |
| Feb 18, 2013 at 2:31 | comment | added | Robert Bryant | @Cecilia: Are you asking when there is a metric $\hat g$ on $TM$ that takes the above form in some local tangential coordinates around a given point? Note that the form you give is not invariant under change of tangential coordinates $(x,v)$, since a local coordinate change $y=F(x)$ would give tangential coordinates of the form $(y,w) = (F(x), F'(x)v)$, and you won't, in general maintain the $\hat g$-orthogonality of $dy$ and $dw$ even if $dx$ and $dv$ are $\hat g$-orthogonal. | |
| Feb 18, 2013 at 2:10 | comment | added | Cecilia | Ok, I see I wasn't expressing myself correctly, so I mislead you. Suppose $(x_1,...,x_n)$ are coordinates on $M$ and $(x_1,...,x_n,v_1,...,v_n)$ are the coordinates on $TM$. I would like to know when does $TM$ admit a metric of the form $$g_{ij}(x) dx_i \otimes dx_j + a_{ij}(x,v) dv_i \otimes dv_j.$$ And, in particular, I would like to know wether one can choose $a_{ij}(x,v)=\delta_i^j$. | |
| Feb 18, 2013 at 2:03 | comment | added | Ryan Budney | One thing I don't understand about your question Cecilia, you mention $M$ is a Riemann manifold, but the Riemann metric on $TM$ is presumably unrelated to the Riemann metric on $M$. So your question does not seem to depend on $g$. For example, you can't put a complete flat metric on $TS^2$, regardless of what metric you put on $S^2$. A Euclidean manifold is covered by the regular Euclidean space, so all of its higher homotopy groups must be trivial. | |
| Feb 18, 2013 at 2:00 | comment | added | Robert Bryant | @Anton: Well $M$ could be a product of circles $S^1$. That would work. Of course, any flat, compact manifold $M$ would have the property that its tangent bundle carries a flat complete metric. I'm not sure that anything else would, though. | |
| Feb 18, 2013 at 1:59 | comment | added | Ryan Budney | @Anton: I think the question is about the curvature vanishing, she's not asking for a global isometry with Euclidean space. | |
| Feb 18, 2013 at 1:54 | comment | added | Anton Petrunin | @Cecilia: Then $\dim =0$, otherwise the tangent bundle is not even homeomorphic to the Euclidean space. | |
| Feb 18, 2013 at 1:14 | comment | added | Cecilia | Yes, I mean a flat metric. Yes, I do require the metric to be complete. Thank you for your time. | |
| Feb 18, 2013 at 1:02 | comment | added | Robert Bryant | What do you mean by 'eucldean Riemannian metric'? Do you mean a Riemannian metric whose curvature vanishes? Do you require the metric to be complete? | |
| Feb 18, 2013 at 0:56 | history | asked | Cecilia | CC BY-SA 3.0 |