TITLE: Compact surfaces of negative curvature
QUESTION [29 upvotes]: 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 $\mathbb{R}^4$?

REPLY [3 votes]: 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.