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Is there a surface in $\mathbb R^3$ which is a closed subset and whose curvature is negative and bounded away from zero?

And the small-print...

  • By surface I mean smooth surface without boundary, and by smooth I mean at least $C^2$. If one allows a boundary the question becomes silly, as a closed disc in a catenoid will do. The smoothness requirement is subtler, but we all know about the Nash-Kuiper theorem which gives, among many things, isometric embeddings of compact sufaces of negative curvature in $\mathbb R^3$ of class $C^1$.

  • I am looking for surfaces which are closed subsets of $\mathbb R^3$. They will not be closed surfaces, though: pretty much every single textbook on the differential geometry of surfaces includes an exercise to the point that a closed surface in $\mathbb R^3$ has a point of positive curvature.

  • Ideally, the surface is embedded. At least, though, it should be inmmersed, for otherwise one can easily find examples which are even of constant negative curvature.

  • Finally, the question is only interesting if the curvature is bounded away from zero, for it is easy to produce examples of surfaces of negative curvature, like the catenoid.

Navigating between the Scylla and Charybdis of uninteresting cases is a pain :-)

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You need to say that it is geodesically complete, otherwise you can take a bounded subset of the catenoid. –  Will Jagy Nov 1 '12 at 3:37
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No, at least as long as the surface is smooth. Consider a point at maximum distance from the origin; what is its curvature there? –  Robert Bryant Nov 1 '12 at 3:51
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Notice, Robert, that I am using the term closed in he sense that the surface is a closed subset of $\mathbb R^3$, not the usual one (compact and boundaryless) –  Mariano Suárez-Alvarez Nov 1 '12 at 3:58
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Mariano, sorry, I edited in the phrase (but not compact). People (Robert, Deane, Anton) do not seem to be answering the question you want to ask. –  Will Jagy Nov 1 '12 at 4:35
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@Mariano: Sorry, for me, 'closed surface' means 'compact 2-manifold without boundary' (just as, for most topologists, 'closed manifold' means compact manifold without boundary'). By the way, you should specify that your surface is a smoothly embedded submanifold of $\mathbb{R}^3$ since, otherwise, the double pseudosphere would count as a closed surface in $\mathbb{R}^3$ that has $K=-1$. Of course, it is a cylinder (without boundary) that has a circular 'rim' along which the (topological) embedding is not an immersion, but you didn't require that it be immersed. –  Robert Bryant Nov 1 '12 at 11:57

2 Answers 2

up vote 10 down vote accepted

Efimov proved that there are no $C^2$ isometric immersions of complete surfaces with negative Guassian curvature bounded away from zero.

N.V. Efimov, "Imposibility of a complete regular surface in euclidean 3-space whose Gaussian curvature has a negative upper bound" Soviet Math. Dokl. , 4 : 3 (1963) pp. 843–846 Dokl. Akad. Nauk SSSR , 150 : 6 (1963) pp. 1206–1209

One reference I know for this is chapter 10 of the book of Han and Hong, "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces."

Edit: Tilla Klotz Milnor's paper "Efimov's theorem about complete immersed surfaces of negative curvature" is an exposition of this theorem in English. I've only looked at the introduction so far but it looks rather thorough.

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@jc: That's a good answer, but, without the $C^2$ assumption, which the OP did not make, the actual answer is 'yes', examples exist, even compact ones. As Anton points out in his answer, the Nash-Kuiper embedding theorem shows that one can isometrically embed any compact surface of negative curvature into $\mathbb{R}^3$ via a $C^1$ embedding, it's just that the embedding can't be $C^2$ everywhere. –  Robert Bryant Nov 1 '12 at 13:00
    
Thanks for the comment. In fact, I nearly added a word about the $C^1$ embeddings in my answer but then I saw Anton's answer. I believe the references I suggest mention this important point as well. –  j.c. Nov 1 '12 at 13:02
    
Thank you for the reference! This is what I was looking for. –  Mariano Suárez-Alvarez Nov 1 '12 at 13:44

You might be interested in this related result: There is a complete, bounded, negative curvature surface in $\mathbb{R}^3$, due to Nadirashvili.

Theorem. There exists a complete surface of negative Gaussian curvature minimally immersed in $\mathbb{R}^3$ which is a subset of the unit ball.

Nikolaj Nadirashvili, "Hadamard's and Calabi-Yau's conjectures on negatively curved and minimal surfaces." Invent. Math. 126(3) (1996), 457–465.

This was discussed in an earlier MO question.

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@Joseph: Yes, but the curvature of this surface is not bounded above by a negative constant. –  Robert Bryant Nov 1 '12 at 12:57
    
@Robert: Yes, I should have stated that this is not directly addressing the posed question. Thanks for clarifying. –  Joseph O'Rourke Nov 1 '12 at 14:13

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