MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


suppose we are given a complete, non-compact Riemannian manifold $(M,g)$. Is it possible to embed (or just immerse) it isometrically into some $R^N$, such that the second fundamental form is bounded? Maybe under some additional assumptions on our manifold and/or metric on it?

This question is a follow-up question to this one: Riemannian manifold of bounded geometry has a normal bundle of bounded geometry.

Thanks, Alex

edit: With "second fundamental form" I mean the quadratic form on the tangent space defined via taking the covariant derivative in $R^N$ and then orthogonally project it onto the normal bundle. So it is not defined not only for hypersurfaces.

Anton Petrunin claimed in his answer that bounded curvature of M and bounded injectivity radius are sufficient for the existence of such an embedding. I this is true, I would be grateful for a reference.

share|cite|improve this question
I'm pretty sure bounded sectional curvature implies existence of an isometric immersion with bounded second fundamental form but I don't know where to find a proof. Maybe in Gromov's book Partial Differential Relations. – Deane Yang Mar 4 '11 at 21:53
up vote 6 down vote accepted

The curvature tensor can be expressed in terms of second fundamental form. Therefore bounded curvature is a necessary condition.

Yet injectivity radius has to be bounded below.

These two conditions are sufficient. It seems that this could be proved along the same lines as the Nash embedding theorem.

P.S. In the formulation, you had to say what you mean by "second fundamental form". Most people think it is only defined for hypersurfaces, but you mean a quadratic form on tangent space with values in the normal space.

share|cite|improve this answer
Good point about the injectivity radius. I did not think about that carefully enough. – Deane Yang Mar 5 '11 at 21:01
I also agree that the proof is basically just Nash's original proof (or one of the simplified versions) but keeping more careful track of what happens to the second fundamental form and, if needed, doing some of the local-to-global constructions a little more carefully. – Deane Yang Mar 5 '11 at 21:36
I do not know about any explicit verification of this condition. The book by Han and Hong, "Isometric Embedding of Riemannian Manifolds in Euclidesn Spaces" appears to have a nice self-contained not-too-long proof of Nash's theorem for a complete Riemannian manifold. I suggest working through that. You just need to verify the following: 1) The construction of an approximate isometric embedding gives bounded second fundamental form. 2) The deformation of the approximate isometric embedding into an exact one changes the second fundamental form by a bounded amount. – Deane Yang Mar 6 '11 at 16:01
I agree with Deane. here few more comments: The proof of Nash theorem is quite involved. The deformation (2 in Deane's comment) is the hardest part. BUT I am quite sure one can build a proof and use deformation part as a black box. BUT in any case one has to learn the rest of proof. – Anton Petrunin Mar 6 '11 at 16:49
And I agree with Anton. You do want to focus on the construction of the approximate isometric embedding and verify that it can be done with bounded second fundamental form under the assumptions stated by Anton. Since the construction is essentially pointwise linear algebra, this seems extremely likely to be true. As for the second step, you don't need to go through the proof. You just need to verify that the topology used in the implicit function theorem or the contraction mapping argument (depending on the presentation) preserves the second fundamental form property. – Deane Yang Mar 6 '11 at 23:24

The Gauss Equations are able to give you some coarse information immediately, see for instance the wikipedia article–Codazzi_equations here. For instance, if $M$ is $n$ dimensional, and you have such an isometric embedding, then about some point $p \in M\cap \mathbb{R}^N$ there is a basis of $N-n$ vectorfields, say $\{e_i\}$ normal to $M$ in $R^N$. Then for each $e_i$ one gets an operator $X \to \nabla_X e_i$ so in this sense there are $N- n$ second fundamental forms. Now the gauss equation writes the curvature of $M$ in terms of the sum of these operators, so if $M$ has unbounded curvature, so must this sum. In particular if $N = n + 1$ then a necessary condition for an embedding (with bounded second fundamental form) is that the curvature of $M$ is bounded.

share|cite|improve this answer
Why the restriction to the case N=n+1? Using the Gauß equation, if the second fundamental form is bounded, so is the curvature of M. Or am I missing something? – AlexE Mar 4 '11 at 21:55
If $N > n+1$ then there isn't "the" second fundamental form. You can think about the "bending" of $M$ in $R^N$ in any direction orthogonal to $M$. In this case, the bounded curvature of $M$ is still a necessary condition if you require that all of the "second fundamental forms" are bounded. If you let some of them go to infinity, then you can let some of the curvatures of $M$ go to infinity. – Ken Knox Mar 4 '11 at 21:59
As another side note, if you don't care how large N is, then, if M has bounded curvature, you can use the nash embedding theorem to embed M in to R^N, and then the gauss equations will tell you that each second fundamental form is bounded. The really hard part about isometric embeddings is bounding N. – Ken Knox Mar 4 '11 at 22:04
There is "the" second fundamental form. A definition can be found, e.g., by following your own link to the wikipedia article. – AlexE Mar 4 '11 at 22:06
You are right, in that sense you can talk about "the" second fundamental form, but then a bound on that operator is the same as individually bounding each of the operators associated to the $e_i$ that I mentioned above. – Ken Knox Mar 4 '11 at 22:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.