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…
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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 $C^1$.
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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.
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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.
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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 🙂
Best Answer
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.