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Let $(M,g)$ be a connected Riemannian manifold of dimension $n>1$. Then the Hopf-Rinow theorem states that $(M,g)$ is geodesically complete if and only if $(M,d_g)$ is complete as a metric space ($d_g$ is the induced intrinsic metric).

I need to know if a similar result is true under weaker hypothesis:

1) Suppose $(M,d)$ is a connected metric space, then is it true that if $(X,d)$ is geodetically complete then it is also complete?

2) If the answer to the previous question is no, would it change something if we consider $(M,d_g)$, where $d_g$ is the intrinsic induced metric of the Riemannian metric $g=\sum_{i,j=1}^ng_{ij}dx_idx_j$ where the $g_{ij}$ are just continuous, not $C^\infty$ (in this case I already know from this other question Geodesics for non differentiable riemannian metric that $(M,d_g)$ is locally geodesic).

Thank you!

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2 Answers 2

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It seems that the answer to 1 is no: The Hopf-Rinow Theorem is false in infinite Dimensions. If you add the assumption that $M$ is a locally compact length space then the answer is "yes" by Theorem 2.5.28 in this book. I think this is strong enough to answer "yes" to 2.

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  • $\begingroup$ Thank you! I'm sorry to ask you the same question I asked to Ben McKay, but you both gave the same answer: is there a particular reason why we extend the geodesic $\gamma:[0,a)\rightarrow X$ on $[0,a]$ instead of the whole $\mathbb{R}$? $\endgroup$
    – user99087
    Commented Oct 5, 2016 at 18:21
  • $\begingroup$ If you can extend to $[0, a]$ then extending to all of $\mathbb{R}$ is trivial: the path which takes the value $\gamma(0)$ on $(-\infty, 0)$ and $\gamma(a)$ on $(a, \infty)$ does the job. Note that you can't do much better than this in general; consider the unit speed geodesic $[0,a)$ in the metric space $[0, a]$, for example. $\endgroup$
    – Paul Siegel
    Commented Oct 5, 2016 at 19:11
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The answer to (2) is yes. I wrote up a proof of Hopf-Rinow for metric spaces in

http://euclid.ucc.ie/Mckay/analysis/analysis.pdf p. 293

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  • $\begingroup$ Thank you! so I get that this theorem is true only if the metric space is locally complete. Is there a particular reason why we extend the geodesic $\gamma:[0,a)\rightarrow X$ on $[0,a]$ instead of the whole $\mathbb{R}$? $\endgroup$
    – user99087
    Commented Oct 5, 2016 at 18:09

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