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)?

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    $\begingroup$ 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? $\endgroup$ Mar 18, 2013 at 10:10
  • $\begingroup$ 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? $\endgroup$
    – Malte
    Mar 18, 2013 at 10:11
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    $\begingroup$ @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. $\endgroup$ Mar 18, 2013 at 11:07
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    $\begingroup$ 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.) $\endgroup$
    – Misha
    Mar 18, 2013 at 17:40

1 Answer 1


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 $N\gg \deg p$.

If we assume a bit better regularity (say a bound on covariant derivatives of curvature tensor), then it can be proved. It is done by applying the result in "The intrinsic dimensionality of graphs" by Krauthgamer and Lee (thaks to aorq, see my question) together with the Nash embedding theorem.

It seems to be unknown if any complete Riemannian manifold with bounded geometry admits an isometric immersion with bounded normal curvatures. But once this problem is solved, the same argument could be used.


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