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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 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.

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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Mar 4, 2011 at 21:53
  • $\begingroup$ @AlexE did you find any reference for this fact? Do you have an explicit control of the upper bound on the second fundamental form in terms of the curvature and injectivity radius of the manifold? $\endgroup$
    – Raziel
    Commented Jan 4, 2019 at 16:31
  • $\begingroup$ @Raziel No, I haven't found any reference for it. $\endgroup$
    – AlexE
    Commented Jan 5, 2019 at 18:36

2 Answers 2

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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 might be sufficient. If we assume a bit better regularity (say a bound on covariant derivatives of the curvature tensor), then this could be proved along the same lines as the Nash embedding theorem.

Postscript. 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.

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  • $\begingroup$ Good point about the injectivity radius. I did not think about that carefully enough. $\endgroup$
    – Deane Yang
    Commented Mar 5, 2011 at 21:01
  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Mar 5, 2011 at 21:36
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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Mar 6, 2011 at 16:01
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    $\begingroup$ 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. $\endgroup$
    – Anton Petrunin
    Commented Mar 6, 2011 at 16:49
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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Mar 6, 2011 at 23:24
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The Gauss Equations are able to give you some coarse information immediately, see for instance the wikipedia article http://en.wikipedia.org/wiki/Gauss–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.

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  • $\begingroup$ 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? $\endgroup$
    – AlexE
    Commented Mar 4, 2011 at 21:55
  • $\begingroup$ 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. $\endgroup$
    – Ken Knox
    Commented Mar 4, 2011 at 21:59
  • $\begingroup$ 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. $\endgroup$
    – Ken Knox
    Commented Mar 4, 2011 at 22:04
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    $\begingroup$ There is "the" second fundamental form. A definition can be found, e.g., by following your own link to the wikipedia article. $\endgroup$
    – AlexE
    Commented Mar 4, 2011 at 22:06
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    $\begingroup$ I think your second comment is not right. In the sum occurring in the Gauß equation the unbounded terms may cancel with each other. Bounded curvature does not automatically imply that the second fundamental form is bounded. See also the answers to my other question (follow the link in my question here). $\endgroup$
    – AlexE
    Commented Mar 4, 2011 at 22:46

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