Consider a Riemannian manifold $(M,g)$. How much regularity is required of $g$ so that for any $x\in M$ and $v\in T_xM$ with $|v|=1$ there exists a unique geodesic $\gamma\colon(-\epsilon,\epsilon)\to M$ so that $\gamma(0)=x$ and $\dot\gamma(0)=v$? All regularity is considered with respect to a fixed smooth (or somewhat less regular) atlas or (equivalently) in a fixed local coordinate chart.

If $g\in C^{1,1}$, then the Christoffel symbol is Lipschitz, whence existence and uniqueness follows from standard ODE theory. For ODEs in general Lipschitz continuity is sufficient and necessary for existence and uniqueness. It is not clear to me whether this is the case for geodesics, because the geodesic equation has a very specific structure and that may provide uniqueness in even lower regularity.

Existence requires less regularity than uniqueness, and this old question is about existence for continuous metrics. Since existence is known in very low regularity, my question concerns uniqueness (although I want to have both existence and uniqueness in as low regularity as possible). If a metric can have a jump discontinuity, then both existence and uniqueness can be made fail, but I am not aware of any smoother counterexamples.

A refined (but equivalent) version my question is: What are the best sufficient and necessary regularity conditions we know (whether or not they coincide in our current knowledge) on a Riemannian metric for uniqueness of geodesics? I am not aware of any positive or negative uniqueness results for $C^{0,\alpha}$ or $C^{1,\alpha}$ below $C^{1,1}$. For example, are geodesics unique if $g\in C^{1,\alpha}$ for some $\alpha>0$, or are there perhaps counterexamples for all $\alpha<1$? If you find the question unclear, please ask for details.

At least for $g\in C^1$ the geodesic equation is a well-defined classical ODE. I guess that satisfying the geodesic equation and minimizing arc length locally are equivalent in this regularity, but I may be mistaken.

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    $\begingroup$ In some sense, regularity of the metric is not the right condition to use, since it is coordinate-dependent. You want to distinguish between metrics that are smooth with respect to some coordinate system and metrics that are not smooth in every coordinate system. Perhaps one should consider generalized curvature bounds, defined in terms of geodesic triangles (i.e., the distances between 3 points). $\endgroup$
    – Deane Yang
    Jun 8, 2018 at 19:33
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    $\begingroup$ @DeaneYang That seems to depend on the regularity of the atlas. If the transition functions are smooth, a $C^{1,\alpha}$ metric in some coordinates is also $C^{1,\alpha}$ in any other coordinates, so regularity of the metric is invariant. (Or did I miss something?) If the atlas has less regularity than the metric, then regularity is indeed coordinate-dependent. $\endgroup$ Jun 8, 2018 at 20:02
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    $\begingroup$ @DeaneYang Yes, regularity may vary between different atlases, but given a fixed atlas (which I considered here to come with a Riemannian manifold), regularity of the metric is invariant. Pretty much all regularity can change when you change an atlas, but not when you change coordinates within a smooth atlas. Smoothness depends on the smooth structure, and meant to work within a fixed atlas. $\endgroup$ Jun 8, 2018 at 20:51
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    $\begingroup$ I think the issue about harmonic coordinates is that if your metric is $C^{k,\alpha}$ in some local coordinates, not necessarily with respect to a smooth atlas, then in harmonic coordinates it is $C^{k+1,\alpha}$. However, if the metric is $C^{k,\alpha}$ in a chart, then it is $C^{k,\alpha}$ in harmonic coordinates. As compared for example to normal coordinates where the metric is only $C^{k-2,\alpha}$. $\endgroup$ Jun 9, 2018 at 9:35
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    $\begingroup$ @DeaneYang As an afterthought: I think we are not talking about the same thing. For me a (Riemannian) manifold comes with its smooth structure. Charts and regularity are defined with respect to this. In this setting you cannot do better than by using harmonic coordinates. But this does not mean that there is another smooth structure where the regularity is better. And this seems to be what you mean. $\endgroup$ Jun 9, 2018 at 20:38

1 Answer 1


There is a classical example by Hartman that shows failure of uniqueness for $C^{1,\alpha}$ metrics. (P. Hartman, On the local uniqueness of geodesics, Amer. J. Math. 1950).

You could lower the regularity if the metric is smooth off some hypersurface but globally only $C^{0,1}$, see for example our recent review: On geodesics in low regularity

Moreover, being minimzing and solving the geodesic equations is not the same below $C^{1,1}$; see also Hartman or our review.

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    $\begingroup$ Thanks! It seems that Hartman's paper gives an example of non-uniqueness for $g\in C^{1,1/3}$ in dimension two. It can probably be made work for any $\alpha<1$, but I didn't see it in the paper. That review looks like an excellent resource on the topic. $\endgroup$ Jun 8, 2018 at 19:21
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    $\begingroup$ Yes, it works for any $\alpha<1$. See also P. Hartman and A. Wintner. On the problems of geodesics in the small. Amer. J. Math. , 73:132–148, 1951 or Ex. 2.1 in our review. $\endgroup$ Jun 9, 2018 at 20:45

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