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Under what conditions can a Riemannian manifold be embedded isometrically as a submanifold of a complete one of the same dimension? There should some kinds of necessary conditions. For instance, any ball in $M$ (considered as a metric space) must be totally bounded. Is this sufficient?

I am curious because it seems that many theorems are stated and proved only for the complete case, and I was wondering how to what extent they could be generalized using a completion tool (if it existed).

Also, is there any kind of uniqueness (there is for $C^{\omega}$ manifolds--implied by the Myers-Rinow theorem)?

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Your question is more or less the same as the one in this thread: mathoverflow.net/questions/8513/… As "some guy on the street" (username) says, Riemann manifolds are metric spaces in a natural way, so you form the completion and your question is whether or not that completion is a Riemann manifold. – Ryan Budney Dec 11 at 23:49
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As Akhil says, ""completion" may not be the right word"; the question is about isometric imbeddability. Density would be a problem, for example thinking of (0,1) with standard structure where the completion isn't a manifold. Regarding "espcecially of the same dimension": perhaps "especially" should be removed or it should be made more specific about how much dimension growth is allowed, because otherwise the answer is "always" by Nash's imbedding theorem. – Jonas Meyer Dec 12 at 0:25
Sure the completion of $(0,1)$ is a manifold -- a manifold with boundary. If you want embeddability in boundaryless manifold I agree the question would be more subtle. Is that what Akhil wants? – Ryan Budney Dec 12 at 0:48
Good question, I shouldn't have assumed that is what is desired. Perhaps my trivial counterexample is to the wrong question. Clarification please, Akhil Mathew? – Jonas Meyer Dec 12 at 0:50
I should not have used the word completion. I have edited it out. In the case of (0,1) with the standard structure, I was fine with its being embedded as an open submanifold of $\mathbb{R}$. @Jonas: Good point about Nash's theorem, I've restricted to the case of a submanifold of the same dimension (basically, I'm curious when a Riemannian manifold is an open submanifold of a complete one, but when I tried to make it more general I botched things). – Akhil Mathew Dec 12 at 4:12
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If I understand you right, you're assuming that there is already a Riemann metric chosen for you. This of course integrates to a distance function ("metric" in the sense of metric spaces) whether or not the manifold is complete. Then one can do the usual thing of freely forming limits of Cauchy sequences which must be equal for any two sequences that are mutually tethered. Then one has to ask whether the completed space is a manifold. Of course it might not be. For example, take any orbifold you like that isn't a manifold, put a sensible Riemannian metric on it by unfolding singularities and averaging in an equivariant way --- and then the usual partition of unity lets you glue the bits together just as you'd like. Then the complement of the singular subspace is a manifold, and of course it doesn't sit in a complete manifold, because its completion really is what you started with --- so it sits an a complete orbifold.

Worse examples could be constructed, but that's the general idea.


Edits to the original question suggest that what's really sought is is something like a smooth extrapolation of a Riemannian metric, and sufficient conditions on the manifold to get such a thing. I believe the orbifold examples are still "bad" for this purpose; one can also build spaces such that the intrinsic diameter of the boundary diverges, but still have finite diameter on the whole. This can come in a variety of shapes --- for instance, the Koch snowflake is a flat example and clearly makes a nice open subset of the plane. One can also build variants with unbounded curvature but finite (gross) diameter --- here the unbounded curvature will be the obstruction to smooth extension... (more to come. I'm cw/ing this answer now.)

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There are easy examples of Riemannian manifolds whose completions as metric spaces are not manifolds, but I don't think that is what is asked for. – Jonas Meyer Dec 12 at 0:32
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Is a humble connected component of $\{(x,y,z):x^2+y^2=z^2\}\setminus0$ with the induced metric from $\mathbb R^3$ also an example? – Mariano Suárez-Alvarez Dec 12 at 0:33
@sgots, But I wonder if this can be modified to give examples where there is no isometric imbedding into a complete Riemannian manifold of the same dimension. – Jonas Meyer Dec 12 at 0:38
As Ryan Budney pointed out above I should have found out whether or not only boundaryless manifolds are allowed instead of assuming that such "easy example" would be allowed. I therefore retract the first part of my first comment with apologies. – Jonas Meyer Dec 12 at 0:57
@Mariano, yes, the half-cone is a classic of an orbifold; the nonsingular part is also an affine manifold. That is, you can build it out of paper. – some guy on the street Dec 13 at 0:06

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