# 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:
$\bullet$ isometric
$\bullet$ proper (i.e., compact sets have compact inverse images)
$\bullet$ 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 (here), 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)?

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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? – Igor Belegradek Mar 18 '13 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? – Malte Mar 18 '13 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. – Igor Belegradek Mar 18 '13 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.) – Misha Mar 18 '13 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.