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Nov 17, 2011 at 18:11 comment added Sam Nead If you only require a negatively curved metric on the interior then $G$ need not be hyperbolic. For example, the figure eight knot complement with the hyperbolic metric (a) has constant negative curvature and (b) is the interior of a compact three-manifold. But $G$ is not hyperbolic because the peripheral group is ${\mathbb{Z}}^2$.
Nov 17, 2011 at 16:58 answer added Ian Agol timeline score: 3
Nov 17, 2011 at 14:43 comment added Luis Jorge Actually I want to know if groups of that form (the way I difined in the first question) satisfy the k.theoretic farrell jones conjecture. I have another question. In the case of zero curvature what kind of groups may occur?
Nov 17, 2011 at 14:32 comment added Paul Siegel It is a standard result in Riemannian geometry that the fundamental group of a compact Riemannian manifold with strictly negative curvature has hyperbolic fundamental group since the universal cover is $CAT(k)$ for some $k < 0$. Does that answer your question, or are you asking about something else?
Nov 17, 2011 at 14:24 comment added Luis Jorge You are right, then change nonpositively by negatively
Nov 17, 2011 at 14:01 comment added Autumn Kent What is the last part of the first sentence supposed to say? At any rate, the fundamental group of a flat 3-torus is not hyperbolic.
Nov 17, 2011 at 13:54 history asked Luis Jorge CC BY-SA 3.0