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Let G be the fundamental group of a compact 3-manifold which supports on its interior a complete non positively curved Riemannian metric and is a cilinder near de metric. Is G hyperbolic?

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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. –  Richard Kent Nov 17 '11 at 14:01
    
You are right, then change nonpositively by negatively –  Luis Jorge Nov 17 '11 at 14:24
    
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? –  Paul Siegel Nov 17 '11 at 14:32
    
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? –  Luis Jorge Nov 17 '11 at 14:43
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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 ${\mathbb{Z}}^2$. –  Sam Nead Nov 17 '11 at 18:11

1 Answer 1

Yes, this follows from the geometrization theorem. A negatively curved complete manifold is atoroidal, and the manifold is irreducible, so there is a complete hyperbolic metric on the interior by geometrization.

In your comments, you ask about zero curvature. Then it is just a Euclidean manifold, with fundamental group a Bieberbach group.

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I think he means word-hyperbolic (I didn't say the question made any sense...) –  Igor Rivin Nov 17 '11 at 17:56
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I see Igor - I always confuse 3-manifolds and their fundamental groups. –  Ian Agol Nov 17 '11 at 19:34

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