I had two trails of though.. is either of them fruitful?

  1. I know every metric space can be completed, my question is: can a Riemmanian manifold $M$ be embedded smoothly and isometrically into it's metric completion $\hat{M}$? If so I could conclude using the Hopf–Rinow theorem that $M$ can be embedded into a geodesically complete manifold $\hat{M}$...
  2. Can we adjoin the boundary to M and then it becomes compact since Riemmanian manifolds satisfy the HB property?
  • $\begingroup$ Geologically complete? You mean geodesically complete, right? $\endgroup$ – user_of_math May 19 '16 at 16:24
  • $\begingroup$ Yes, unfortunately it was auto-pseudo-corrected $\endgroup$ – AIM_BLB May 19 '16 at 16:25
  • $\begingroup$ It is true that any on a manifold $M$ compatible with the topology of $M$ turned into a complere metric giving the same topology of $M$...If you want I can give you a sketch of the proof of this fact. $\endgroup$ – Anubhav Mukherjee May 19 '16 at 16:30
  • $\begingroup$ But is the complete metric space a manifold itself afterwards or only a metri space? $\endgroup$ – AIM_BLB May 19 '16 at 16:32
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    $\begingroup$ 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

The metric completion of $M$ might not be a manifold. For an example, take the Alexander horned sphere $A \subset S^3$. There are two complementary components of $A$; let $M$ be one of them. Then the metric completion of $M$ is $M \cup A$ which is not a manifold-with-boundary.

This example has a coincidental side-effect that $M$ can be isometrically embedded into a geodesically complete manifold, namely under its inclusion into $S^3$. But I think it won't be too hard to construct an $M$ which does not have this coincidental property either.

  • $\begingroup$ Thank you Lee, that was very illustrative. $\endgroup$ – AIM_BLB May 19 '16 at 16:36
  • $\begingroup$ but your original question was"can every Riemannian manifold be completed"? $\endgroup$ – Anubhav Mukherjee May 19 '16 at 16:36
  • $\begingroup$ This is a morally correct answer, but you have to be more careful with the choice of the wild sphere. Namely, it may happen that the "wild points" of the wild sphere do not belong to the metric completion of M (every path connecting them to a base point in M will have infinite length). $\endgroup$ – Moishe Kohan May 19 '16 at 17:22
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    $\begingroup$ Good point, one has to choose $M$ to be the 3-ball side of $A$, and one has to make sure that the construction of the arms shrinks at an appropriate rate so that each point in the limiting Cantor set is the limit of a finite length path in the interior. $\endgroup$ – Lee Mosher May 19 '16 at 19:37

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