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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 surfaces 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 immersed, 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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    $\begingroup$ No, at least as long as the surface is smooth. Consider a point at maximum distance from the origin; what is its curvature there? $\endgroup$ Nov 1, 2012 at 3:51
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    $\begingroup$ 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) $\endgroup$ Nov 1, 2012 at 3:58
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    $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Nov 1, 2012 at 4:35
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    $\begingroup$ @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. $\endgroup$ Nov 1, 2012 at 11:57
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    $\begingroup$ @Mariano: I'm still trying to understand why you are asking this question in the form that you are. For example, are you requiring that the surface have the induced topology and be connected? If not, then the induced metric might not be complete, and Efimov's result wouldn't apply. (For example, the union of the open half-plane $y=0,\ z > 0$ with the plane $z=0$ is a closed subset that is smoothly parametrized by a 1-1 immersion of a surface, but the surface's topology is not the induced topology. It's not obvious to me why something like this couldn't happen locally in negative curvature.) $\endgroup$ Nov 2, 2012 at 21:25

2 Answers 2

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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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    $\begingroup$ @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. $\endgroup$ Nov 1, 2012 at 13:00
  • $\begingroup$ 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. $\endgroup$
    – j.c.
    Nov 1, 2012 at 13:02
  • $\begingroup$ Thank you for the reference! This is what I was looking for. $\endgroup$ Nov 1, 2012 at 13:44
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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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    $\begingroup$ @Joseph: Yes, but the curvature of this surface is not bounded above by a negative constant. $\endgroup$ Nov 1, 2012 at 12:57
  • $\begingroup$ @Robert: Yes, I should have stated that this is not directly addressing the posed question. Thanks for clarifying. $\endgroup$ Nov 1, 2012 at 14:13

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