TITLE: When is a Riemannian manifold an open subset of a complete one?
QUESTION [18 upvotes]: 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)?

REPLY [3 votes]: I doubt it's possible to give necessary and sufficient conditions for an incomplete Riemannian manifold to be embeddable in a complete Riemannian manifold of the same dimension. It's too easy to construct incomplete Riemannian manifolds that do terrible things at its metric boundary.
You need to have some control over the topology and metric near the metric boundary.
Some possible sufficient conditions that come to mind:
a) The metric completion of the incomplete Riemannian manifold is a smooth Riemannian manifold with boundary.
b) The metric completion of the incomplete Riemannian manifold is smooth Riemannian manifold without boundary.
I don't even know how to resolve the following simple case (a point singularity): an incomplete Riemannian manifold with bounded sectional curvature whose metric completion is the manifold plus one additional point. It's easy enough to give examples where completion is not a complete Riemannian manifold. But what I don't know how to do, even in this example, is how to give necessary and sufficient conditions for the completion to be a complete Riemannian manifold.