Timeline for Can one recover a metric from geodesics?
Current License: CC BY-SA 3.0
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| Feb 2, 2014 at 20:51 | comment | added | Brian Rushton | You may be interested in contributing to a proposed Spanish language version of math stackexchange; it could use some input from fluent professors and students: area51.stackexchange.com/proposals/64529/… | |
| Jun 4, 2013 at 4:17 | comment | added | Cristi Stoica | That's true. But the question was whether, by assuming Riemannian metric, and by knowing the geodesics, we can recover the metric, say up to a scale factor. I just gave the simplest counterexample. Now, if you add the condition that the resulting manifolds are not isometric, we can modify the counterexample to satisfy this. A simple example is a flat torus. Flat torii have the same geodesics, but are not isometric. Moreover, in general they can't be made isometric by rescaling the metric (unless you use different scaling factor in different directions). | |
| Jun 1, 2013 at 14:53 | comment | added | alvarezpaiva | I learned from Vladimir Matveev, that even isometric examples of metrics with the same geodesics can be interesting. He considered the standard metric on the projective plane, and then its pullback by a projective transformation. Athough this seems obvious it gives rise to quadratic integrals of motion and the construction of interesting metrics such as the "Poisson spheres" (that can also be constructed using the Hopf fibration as submersive metrics obtained from metrics $SO(3)$ with a left invariant metric. | |
| Jun 1, 2013 at 7:57 | comment | added | Gerardo Arizmendi | What is the point in having an isometric counterexample? In that case why not taking $\mathbb R^2$ divide the distance in the $y$ direction by $2$. | |
| Jun 1, 2013 at 7:32 | comment | added | HenrikRüping | But it is still a counterexample to the question. | |
| May 31, 2013 at 18:53 | history | answered | Gerardo Arizmendi | CC BY-SA 3.0 |