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My question is the following:

Let M be a simply connected Riemannian manifold whose sectional curvatures are all nonpositive and let G be a group. Suppose that G acts in M properly discontinuous and cocompactly by isometries. Is G a CAT(0) group?

flag	asked Dec 10 2011 at 16:13 Luis Jorge 205●1●3

Igor Rivin · Answer 1 · 2011-12-10 16:16:32Z UTC

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The answer is yes, since $M$ is a $CAT(0)$ space, and the group is quasi-isometric to it. See (for example) Jim Cannon's article in Bedford Keane Series:

The theory of negatively curved spaces and groups J Cannon - t. Bedford, M. Keane, and C. Series. Oxford University …, 1991

answered Dec 10 2011 at 16:16

Igor Rivin
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By definition, a group is $CAT(0)$ if it acts isometrically, properly and co-compactly on a $CAT(0)$ space... So it is enough to say that $M$ is $CAT(0)$. See aimath.org/pggt/… – Alain Valette Dec 10 2011 at 18:33

Igor, you should say "yes by definition", but not "yes, since". – ε-δ Dec 10 2011 at 21:22

@ε-δ: I believe that is the essence of Alain's commentary... – Igor Rivin Dec 11 2011 at 10:08

To be fair, it is not a trivial fact that a simply connected Riemannian manifold whose sectional curvatures are all nonpositive is CAT(0). – Anon Dec 15 2011 at 10:01

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@Anon: I believe that this is why the T is in CAT(0)... – Igor Rivin Dec 15 2011 at 10:37

Examples of CAT(0)-groups

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