# Can every Riemmanian Manifold be completed?

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?
• Geologically complete? You mean geodesically complete, right? – user_of_math May 19 '16 at 16:24
• Yes, unfortunately it was auto-pseudo-corrected – AIM_BLB May 19 '16 at 16:25
• 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. – Anubhav Mukherjee May 19 '16 at 16:30
• But is the complete metric space a manifold itself afterwards or only a metri space? – AIM_BLB May 19 '16 at 16:32
• Possible duplicate of When is the metric completion of a Riemannian manifold a manifold with boundary? – 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.
• 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. – Lee Mosher May 19 '16 at 19:37