User anon - MathOverflow

Answer by Anon for Is $SL(n,\mathbb{Z})$ a CAT(0) group?

2011-11-28T09:35:06Z

If n=2, yes: it acts on its Bass-Serre tree.

If n>2, no: your group contains distorted elements, i.e. elements conjugated to a proper power of themselves (look at unipotent matrices).

Such an element will have zero displacement length, which is impossible for an infinite order element in a group acting discretely cocompactly.

For this and much more, see the monograph of Bridson and Haefliger.

For even more restrictions on SL_n actions, see Theorem 1.14 in:

Caprace-Monod, Isometry groups of non-positively curved spaces: structure theory Journal of Topology 2 No. 4 (2009), 661–700

Answer by Anon for bounded cohomology of subgroups of groups

2011-10-18T10:52:46Z

In general, no. There is not even a natural map (in general) in the direction you want.

There is a natural map in the opposite direction, namely restriction, and this is sometimes an embedding, but not always. See chapter 8.6 in "Continuous bounded cohomology of locally compact groups", Lecture Notes in Mathematics 1758

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

2011-03-24T15:00:52Z

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.

Comment by Anon

2011-12-15T10:01:46Z

To be fair, it is not a trivial fact that a simply connected Riemannian manifold whose sectional curvatures are all nonpositive is CAT(0).

Comment by Anon

2011-04-05T14:02:52Z

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.