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Every metric space has a completion, and a Riemannian manifold has a structure of metric space.

Is it possible to extend a Riemannian manifold an that this extension keeps being a Riemannian manifold?

I think that the answer is no, but is there any easy counterexample? At first, I thought about the cone without vertex as submanifold of $\mathbb{R}^3$ which is not complete and if you complete it you lose the smooth structure, but possibly this is not a valid one, because I thought about the cone (with vertex) keeps being a manifold regarded as an abstract one.

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    $\begingroup$ I don't have time for a full answer, but I don't think it's true. Take the cone on $\mathbb{R}P^2$, which is not even a topological manifold because of the cone point. If we remove the cone point, what's left can be given the structure of a smooth Riemannian manifold in such a way that the inclusion map is an isometric embedding in the metric sense. Then the cone on $\mathbb{R}P^2$ is the completion, so is not a manifold. $\endgroup$ – Jason DeVito Nov 22 '17 at 13:48
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    $\begingroup$ If for you a manifold is a manifold without boundary, then the metric completion of an open interval in $\mathbb{R}$ (with the induced metric) is the closed interval which cannot be given the structure of a manifold without boundary. $\endgroup$ – levap Nov 22 '17 at 13:58
  • $\begingroup$ Possible duplicate of When is the metric completion of a Riemannian manifold a manifold with boundary? $\endgroup$ – Alex M. Jun 1 '18 at 16:13

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