# Does a Riemannian manifold with bounded geometry admit an isometric proper embedding into Euclidean space with uniformly thick tubular neighborhood?

Suppose $$(M,g)$$ is an open Riemannian manifold with bounded geometry, i.e., the injectivity radius is $$\ge \epsilon>0$$ and each iterated covariant derivative of curvature is bounded with respect to $$g$$.

Question: Does there exist an embedding into some high dimensional $$\mathbb R^N$$ with the following properties:

• isometric
• proper (i.e., compact sets have compact inverse images)
• The normal tubular neighborhood contains a uniformly thick disk in each fiber?

The Gauss equations give a relationship between the (normal bundle valued) second fundamental form (or shape operator); but by embedding arc-length parameterized curves one can show that this is not enough to force focal points uniformly away.

There is a related question Existence of an isometric embedding into Euclidean space with bounded second fundamental form, but it does not answer this question.

By taking a product with another Euclidean space one can then have embedding where the normal bundle is trivial. Then one can use such embedding to carry over results valid on $$\mathbb R^n$$ to Riemannian manifolds with bounded geometry.

Edit: Many thanks for the comments and Anton for a very succinct answer.

I end with a wild guess: Maybe, one can characterize the Riemannian manifolds of bounded geometry admitting such embeddings as those whose ends are asymptotically flat times something compact (by which I mean an bundle)?

• If "By taking a product with another Euclidean space one can then have embedding where the normal bundle is trivial", then $M$ is parallelizable because it is an open manifold that embeds into a Euclidean space with trivial normal bundle. Did you really mean that? Mar 18, 2013 at 10:10
• On the off-chance of making a fool of myself: By "open", do you mean "open and complete"? Otherwise, gettig a proper embedding should be impossible. On the other hand, isn't proper automatically satisfied once you require completeness? Mar 18, 2013 at 10:11
• @Malte: completeness is implies by a lower injectivity radius bound (all geodesics extend by a definite amount). The graph of $sin(1/x)$ is a counterexample to your last sentence. Mar 18, 2013 at 11:07
• Peter: In view of Anton's argument, maybe you should settle for something weaker, like smooth, 1-Lipschitz, proper (but not uniformly proper), with uniform normal injectivity radius. Such map is likely to exist by a modification of the proof of Whitney embedding theorem. (Expanders will prevent existence of uniformly proper maps.) Mar 18, 2013 at 17:40

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $$\mathbb{R}^N$$ is polynomial.
Postcript. Let us say that a Riemannian manifold $$M$$ has polynomial volume growth if there is a polynomial $$p$$ such that volume of any $$r$$-ball in $$M$$ has volume at most $$p(r)$$.
Evidently, if a manifold admits an embedding with uniformly thick tubular neighborhood into $$\mathbb{R}^N$$, then it has polynomial volume growth. I suspect that polynomial volume growth is also sufficient for $$N\gg \deg p$$.