Timeline for Existence of an isometric embedding into Euclidean space with bounded second fundamental form
Current License: CC BY-SA 2.5
9 events
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| Mar 5, 2011 at 0:36 | comment | added | Ken Knox | Ok I agree. Thurston gave a nice example of how the normal bundle can have unbounded curvature (but not if $N = n+1$) and and Deane mentioned that this can't happen if the second fundamental form is bounded. So my answer as stated is correct, and you have answered your own question in your first comment :-) | |
| Mar 4, 2011 at 22:46 | comment | added | AlexE | 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). | |
| Mar 4, 2011 at 22:13 | comment | added | Ken Knox | 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. | |
| Mar 4, 2011 at 22:06 | comment | added | AlexE | There is "the" second fundamental form. A definition can be found, e.g., by following your own link to the wikipedia article. | |
| Mar 4, 2011 at 22:04 | comment | added | Ken Knox | 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. | |
| Mar 4, 2011 at 21:59 | comment | added | Ken Knox | 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. | |
| Mar 4, 2011 at 21:55 | comment | added | AlexE | 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? | |
| Mar 4, 2011 at 21:31 | history | edited | Ken Knox | CC BY-SA 2.5 |
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| Mar 4, 2011 at 21:25 | history | answered | Ken Knox | CC BY-SA 2.5 |