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John Hubbard recently told me that he has been asking people if there are compact surfaces of negative curvature in $\mathbb{R}^4$ without getting any definite answers. I had assumed it was possible, but couldn't come up with an easy example off the top of my head.

In $\mathbb{R}^3$ it is easy to show that surfaces of negative curvature can't be compact: throw planes at your surface from very far away. At the point of first contact, your plane and the surface are tangent. But the surface is everywhere saddle-shaped, so it cannot be tangent to your plane without actually piercing it, contradicting first contact.

This easy argument fails in $\mathbb{R}^4$. Can the failure of the easy argument be used to construct an example? Is there a simple source of compact negative curvature surfaces in $\mathbb{R}^4$?

Matt Noonan
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2 Answers2

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You will find examples (topologically, spheres with seven handles) in section 5.5 of Surfaces of Negative Curvature by E. R. Rozendorn, in Geometry III: Theory of surfaces, Yu. D. Burago VI A. Zalgaller (Eds.) EMS 48.

Rozendorn tells us that «from the visual point of view, their construction seems fairly simple.» Well...

Mariano Suárez-Álvarez
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  • Mariano, is the curvature constant? Is the embedding at least $C^2$? – Victor Protsak Jul 20 '10 at 06:17
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    Victor, Rozendorn writes that «Efimov stated the supposition, at present not confirmed but also not disproved» that the surface can be deformed, preserving the sign of the curvature, to obtain in the limit a closed surface of constant negative curvature – Mariano Suárez-Álvarez Jul 20 '10 at 06:21
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    Also, the construction can be made $C^\infty$. – Mariano Suárez-Álvarez Jul 20 '10 at 06:25
  • Interesting, thank you! I even had that book in my possession within the last 3 months, but I completely missed that construction. – Matt Noonan Jul 20 '10 at 12:21
  • So... the proof that "In $\mathbb{R}^3$ [...] surfaces of negative curvature can't be compact" in the OP's question was wrong after all? – Qfwfq Jan 24 '11 at 00:54
  • @unknowngoogle: I believe The `counterexample' to the proof is in $\mathbb{R}^4$. – kangdon Apr 18 '11 at 00:53
  • The Hadamard-Cartan comparison theorem says that any complete Riemannian manifold of non-positive sectional curvature has no conjugate points.

    Does anyone know of an `easily-pictured' example of a compact (boundaryless) manifold without conjugate points?

     – kangdon Apr 18 '11 at 00:53
  • (Possibly of positive curvature) – kangdon Apr 18 '11 at 00:54
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    If anyone is not convinced by the elegant "throw tangent planes at the surface" argument, it can be translated to "for any compact surface in $\mathbb R^3$, find a point of maximum distance from the origin. At this point, the surface has curvature at least as positive as the sphere defined by that distance from the origin." – Elizabeth S. Q. Goodman Jan 27 '12 at 07:22
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In "Y. Martinez-Maure, A counter-example to a conjectured characterization of the sphere. (Contre-exemple à une caractérisation conjecturée de la sphère.) (French), C. R. Acad. Sci., Paris, Sér. I, Math. 332, 41-44 (2001), the author disproves an old characterization of the 2-sphere by giving an exemple of a "hyperbolic hedgehog" of R^3 (a sphere-homeomorphic envelope parametrized by its Gauss map whose Gaussian curvature K is everywhere negative excepted at four singular points where K is infinite).

By projective duality, this implies the existence of a 2-sphere C^2 embedded in the 3-sphere with a nonpositive extrinsic curvature but not totally geodesic.

Louis
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