# When is a Riemannian manifold an open subset of a complete one?

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)?

• 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. Dec 11 '09 at 23:49
• 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. Dec 12 '09 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? Dec 12 '09 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? Dec 12 '09 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). Dec 12 '09 at 4:12

I take the opportunity to advertise the work of a colleague Charles Frances, which is somehow related. There are counter-examples to a more flexible question: given a (pseudo-)riemannian manifold, is it always possible to conformally, non-trivially embed it into another?

A counter-example to this question gives a counter-example to yours since a conformal class on a (non-compact !) manifold contains a non-complete riemannian metric.

Details can be found at the following adress: http://www.math.u-psud.fr/~frances/boundary-frances.pdf

• Cool, thanks! So I guess at least the answer to "can any Riemannian manifold be embedded in a compact one" is no then? Apr 4 '10 at 22:24
• Yes, but that already follows from previous answers (a revolution cone without the apex is the simplest counter-example, proposed by Mariano Suárez-Alvarez). Apr 5 '10 at 8:51

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.)

• 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. Dec 12 '09 at 0:32
• 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? Dec 12 '09 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. Dec 12 '09 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. Dec 12 '09 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. Dec 13 '09 at 0:06

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.

• I was about to post a version of conjecture a). "If the new points in the metric completion of an n-dimensional riemannian manifold form a metric space isometric to a (possibly empty) n-1 dimensional riemannian manifold B, then M can be completed by adding Bx[0,+\infty)" Apr 5 '10 at 3:41
• If the metric completion of a 2-dimensional manifold adds a single point, which is not a 1-dimensional manifold, I thinking of a cone without tip or a punctured otherwise-complete surface. To distinguish between those I would parallel transport a vector along a small loop around the new point. In the case of the cone, when the length of the loop goes to zero, the angle between the original and the transported vector would converge to zero in the punctured surface, and non-zero in the tipless cone. Apr 5 '10 at 4:18
• In the same line of conjecturing: "An n-dimensional riemannian manifold M is embeddable in a complete one iff its metric completion is a metric subspace of an n-dimensional riemannian manifold." Apr 5 '10 at 4:23
• @Marcos: In the case of the 2-d manifold, how do you know that a neighborhood of the missing point has the right topological type (that of a punctured Euclidean ball)? Apr 5 '10 at 11:00
• Deane, I respectfully disagree. I think there are going to be conditions that allow you to bootstrap. First make sure a point at "infinity" has a tangent space, then make sure it has a metric, then the metric has a curvature tensor, then construct normal coordinates on a neighborhood, then get the normal coordinates to fit together to be a manifold. The hypotheses will start out looking hokey, but after fiddling, there should be a decent theorem here. Apr 5 '10 at 15:02

This isn't an answer its a conjecture. Nice question.

Suppose that $M, N$ are a Riemannian manifolds and $M\subset N$ is an open subset and $N$ is complete. Lets assume that $M$ is path connected, so that there is no funny business in defining the distance between $p, q \in M$ to be the infimum of the length of a path joining $p$ to $q$. Also lets assume that that path metric is bounded, so you don't have infinite ends.

There is a map from the metric space completion of $M$ into $N$ and its image will be the closure of $M$ in $N$. There is now a plethora of obstructions to the embedding, derived from this map.

For instance: Let $CI(\overline{M})$ be those continuous functions on the metric space completion of $M$ whose restriction to $M$ is smooth. Let $I$ be the ideal of all functions in $CI(\overline{M})$ that vanish at a point $p$ of the completion. It should be the case that $T=(I/I^2)^*$ is isomorphic to $\mathbb{R}^n$ where $n$ is the dimension of the manifold. Next, the metric tensor should extend to the completion, where you interpret it a point at infinity as a tensor on $T$, and its coefficients should be elements of $CI(\overline{M})$. Next you should be able to extend the Riemann curvature tensor appropriately as a map from the tensor square of T to itself, and the coefficients of the extension should also be in $CI(\overline{M})$ and they should satisfy all the restrictions on the tensor that the Riemann curvature tensor of a smooth manifold satisfies.

Here is my conjecture : The condition above is necessary and sufficient. The reason is you should be able to build a candidate piece of the manifold $N$ with normal coordinates, and those normal coordinate patches should glue together coherently.

• Its probably wrong, because I didn't pick CI(overline{M}) carefully enough. There is probably something like what I said that is true though. Apr 4 '10 at 22:09