When is a Riemannian manifold an open subset of a complete one? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:04:37Z http://mathoverflow.net/feeds/question/8622 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8622/when-is-a-riemannian-manifold-an-open-subset-of-a-complete-one When is a Riemannian manifold an open subset of a complete one? Akhil Mathew 2009-12-11T23:06:32Z 2010-04-05T03:23:45Z <p>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? </p> <p>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).</p> <p>Also, is there any kind of uniqueness (there is for $C^{\omega}$ manifolds--implied by the Myers-Rinow theorem)?</p> http://mathoverflow.net/questions/8622/when-is-a-riemannian-manifold-an-open-subset-of-a-complete-one/8628#8628 Answer by some guy on the street for When is a Riemannian manifold an open subset of a complete one? some guy on the street 2009-12-11T23:45:27Z 2009-12-13T00:05:14Z <p>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 <em>not</em> 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.</p> <p>Worse examples could be constructed, but that's the general idea.</p> <p><hr /></p> <p>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.)</p> http://mathoverflow.net/questions/8622/when-is-a-riemannian-manifold-an-open-subset-of-a-complete-one/20320#20320 Answer by Benoît Kloeckner for When is a Riemannian manifold an open subset of a complete one? Benoît Kloeckner 2010-04-04T17:03:56Z 2010-04-04T17:03:56Z <p>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 <em>conformally</em>, non-trivially embed it into another?</p> <p>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.</p> <p>Details can be found at the following adress: <a href="http://www.math.u-psud.fr/~frances/boundary-frances.pdf" rel="nofollow">http://www.math.u-psud.fr/~frances/boundary-frances.pdf</a></p> http://mathoverflow.net/questions/8622/when-is-a-riemannian-manifold-an-open-subset-of-a-complete-one/20324#20324 Answer by Charlie Frohman for When is a Riemannian manifold an open subset of a complete one? Charlie Frohman 2010-04-04T18:06:50Z 2010-04-04T21:27:42Z <p>This isn't an answer its a conjecture. Nice question.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/8622/when-is-a-riemannian-manifold-an-open-subset-of-a-complete-one/20360#20360 Answer by Deane Yang for When is a Riemannian manifold an open subset of a complete one? Deane Yang 2010-04-05T03:23:45Z 2010-04-05T03:23:45Z <p>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.</p> <p>You need to have some control over the topology and metric near the metric boundary.</p> <p>Some possible sufficient conditions that come to mind:</p> <p>a) The metric completion of the incomplete Riemannian manifold is a smooth Riemannian manifold with boundary.</p> <p>b) The metric completion of the incomplete Riemannian manifold is smooth Riemannian manifold without boundary.</p> <p>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 <em>not</em> 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.</p>