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Swenson proved in "A cut point theorem for ${\rm CAT}(0)$ groups" that a locally compact Hadamard space with a geometric action by a group $G$ admits a hyperbolic isometry (that lie in $G$). Is it still true if we only assume the action to be cocompact? I tried to modify his proof but I didn't succeed...

Thanks

flag	asked Mar 24 2011 at 13:21 Aurelien 19●3

Anon · Answer 1 · 2011-03-24 15:00:52Z UTC

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There is an answer, under perhaps some conditions, in Section 6.C of

Caprace, Pierre-Emmanuel; Monod, Nicolas Isometry groups of non-positively curved spaces: structure theory. J. Topol. 2 (2009), no. 4, 661-700.

Regarding the notations in that reference: notice that if X is any proper CAT(0) space, then the group G=Isom(X) will automaticall act properly on X.

answered Mar 24 2011 at 15:00

Anon
261●2●2

	Thanks! Caprace and Monod's article just jumped on the top of my stack of "to be read" articles that take the dust on my desktop. However, the question still holds... And sorry, I don't have enough reputation points to give you a thumbs up... – Aurelien Mar 24 2011 at 17:40
	Are you sure of your lat remark? They do not define proper action in their paper, but Bridson and Haefliger define an action to be proper in for every compact $K$ the number of $g$ in the group that satisfy $g(K)\cap K\neq\void$ is finite. With this definition, the isometry group of a homogeneous manifold is not proper! – Aurelien Mar 29 2011 at 13:44
	This is because they give the definition in the particular case of discrete groups. The "true" definition" replaces "finite" by "compact". But in general, the topology of the group of isometries is defined in such a way that the action is proper almost by definition. So this is definitely not a problem. – Anon Apr 5 2011 at 14:02

Hyperbolic isometries in cocompact Hadamard (i.e. cat(0) proper simply connected) spaces

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