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Is there a nontrivial Riemannian metric for which a geodesic equation doesn't have any solutions? What would such a metric mean physically?

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    $\begingroup$ No solutions anywhere, or no solutions at a particular point? $\endgroup$
    – TimRias
    Feb 9, 2021 at 20:29
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    $\begingroup$ Differential equations have solutions even if they are not expressible in terms of functions that we have given names to. $\endgroup$
    – G. Smith
    Feb 9, 2021 at 20:43
  • $\begingroup$ @mmeent No solutions anywhere. $\endgroup$ Feb 9, 2021 at 21:42

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No, it does not exist. Given a Riemannian manifold with metric of class $C^2$, for every initial point and every initial vector, there is a unique maximal geodesic satisfying those initial conditions. That is a straightforward consequence of the existence and uniqueness theorem for integral curves on manifolds (here the tangent space manifold).

For the existence $C^1$ is enough, but it does not guarantee uniqueness. With less regularity it is not possible to write down the geodesic equation. (Actually one could require that the metric is differentiable but the derivatives are not continuous, in this case nor the existence is guaranteed. On the other hand such metric has continuous derivatives on a dense set.)

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  • $\begingroup$ To what extent, if at all, do the set of all geodesics on a manifold determine the metric (assume any and all smoothness, etc.)? I mean we assign in every direction to every point a curve and declare it to be a geodesic (of course, same geodesic for any point on it), would this determine the metric? $\endgroup$
    – hyportnex
    Feb 10, 2021 at 17:29
  • $\begingroup$ The squared length of the geodesics completely determines the metric. $\endgroup$ Feb 10, 2021 at 20:07
  • $\begingroup$ The point is that a geodesic admits a class of affine parameters. When you speak about the class of geodesics which affine parameter you consider? $\endgroup$ Feb 10, 2021 at 20:10
  • $\begingroup$ I was not considering that at all for I thought the geodesic equation being independent of linear reparametrization $\lambda ' = a\lambda + b$ they should not matter. Is there a "natural" metric that might be independent of any "reasonable" nonlinear reparametrization of the geodesic curves even before we assign length to them? In other words, to what extent the points of the geodesic without assigning length to the curve determine the metric that is compatible with all the geodesics. $\endgroup$
    – hyportnex
    Feb 10, 2021 at 21:14
  • $\begingroup$ They cannot completely, think of $R^n$. Knowing only the geodesics up to reparametrization you cannot distinguish between the Euclidean metric and that metric rescaled with a constant factor. The same fact is true for every manifold. $\endgroup$ Feb 10, 2021 at 21:26

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