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Assume there are two Riemannian metrics on a manifold ( open or closed) with the same set of all geodesics. Are they proportional by a constant? If not in general, what are the affirmative results in this direction?

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A question related in spirit: "Probing a manifold with geodesics" mathoverflow.net/questions/81622/… –  Joseph O'Rourke May 29 '13 at 17:57
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The counterexamples to 'uniqueness up to constant multiples' mentioned by alvarezpaiva and Christi Stoica, notwithstanding, it is true that for the generic metric $g$, even in dimension $2$, the only metrics that have the same geodesics (up to reparametrization) as $g$ are the constant multiples of $g$. Only very special metrics (in dimension $2$, these were classified by Dini) share their geodesics with more metrics than that. –  Robert Bryant May 30 '13 at 10:36
    
A related question: "From shortest path to manifold structure" mathoverflow.net/questions/116292/… –  Aeryk Oct 11 '13 at 1:07

5 Answers 5

up vote 22 down vote accepted

The short answer is no, the geodesics do not determine the metric. For example, in the Cayley-Klein model of hyperbolic geometry the geodesics are straight lines. It is however rather rare for two Riemannian metrics to have the same geodesics. In two dimensions these metrics were studied by Liouville (see Livre VI of Darboux's Théorie générale des surfaces). Their geodesic flow is completely integrable and, in fact, admits an additional integral of motion quadratic in the momenta.

A basic results is that if locally the geodesics are straight lines (or can be mapped to straight lines), the metric has constant curvature (Beltrami's theorem). Other rigidity results of this kind exist.

For more on this topic you can consult the works of Topalov and Matveev.

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A simple counterexample is the space $\mathbb R^n$, with a metric $g_{ab}$ independent on the point. As examples, the Euclidean space $\mathbb R^n$, but also the Minkowski spacetime (which is semi-Riemannian, but has the same geodesics as $\mathbb R^4$). The geodesics are the lines in $\mathbb R^n$, no matter how we choose the constant metric $g_{ab}$. Hence, different metrics can give the same set of geodesics.

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I hope that my ``answer'' will not be understood solely as a propaganda of my survey http://arxiv.org/abs/1101.2069 where I discussed

(1) how, given geodesics, to reconstruct a connection (in both cases: when geodesics are parameterised and not parameterised)

(2) how, given a class od projectively related connections, to reconstruct a metric, and what are advantages of additional curvature assumptions on the metric

(3) what is the freedom in reconstructing the metric by geodesics -- in particular I proved the statement wellknown to experts and mentionen by Robert Bryant in his comment that for generic metric the geodesics, even unparameterized, determine the metric.

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I attended today a talk at a conference, where Graham Hall presented some very interesting results about recovering the metric from geodesics, and mentioned some previous results too. Here are some of his papers about this:

http://www.sciencedirect.com/science/article/pii/S0393044011002269

http://link.springer.com/article/10.2478/s11533-012-0087-6

http://arxiv.org/pdf/0906.5227.pdf

http://arxiv.org/pdf/1006.5023.pdf

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The example of Cristi Stoica is isometric to Euclidean space $\mathbb R^n$ if the $g_{ab}$ defines a Riemannian metric.

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But it is still a counterexample to the question. –  HenrikRüping Jun 1 '13 at 7:32
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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 $2$. –  Gerardo Arizmendi Jun 1 '13 at 7:57
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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. –  alvarezpaiva Jun 1 '13 at 14:53
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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). –  Cristi Stoica Jun 4 '13 at 4:17
    
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/… –  Brian Rushton Feb 2 at 20:51

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