Skip to main content
10 events
when toggle format what by license comment
Mar 26, 2010 at 12:23 history edited Matthew Stover CC BY-SA 2.5
added 2 characters in body
Mar 26, 2010 at 12:14 history edited Matthew Stover CC BY-SA 2.5
Added a condition
Mar 26, 2010 at 10:02 comment added Tom Church For fundamental groups of negatively curved manifolds, the answer above should say "compact hyperbolic manifolds never contain a $\mathbb{Z}\oplus\mathbb{Z}$". You can see this directly without appealing to the fact that they are Gromov-hyperbolic from the fact that centralizers are virtually cyclic. To see this, note that a hyperbolic isometry has two fixed points on the boundary at infinity. The centralizer of a given hyperbolic isometry must preserve these fixed points, and so sits inside $\mathbb{R}\times SO(n-1)$. A discrete subgroup of this group is virtually $\mathbb{Z}$.
Mar 26, 2010 at 9:57 comment added Tom Church For hyperbolic groups, a $\mathbb{Z}\oplus\mathbb{Z}$ subgroup has triangles which are as fat as triangles in the Euclidean plane, which violates the thin triangles condition which is the definition of hyperbolic group.
Mar 26, 2010 at 5:57 comment added Pete L. Clark How do you know that hyerbolic groups are not virtually abelian? I mean, I certainly believe it to be true -- I have studied Fuchsian groups, where this follows from the explicit presentation of the group in terms of its signature -- but what technique or fact are you using to see this? (You could use Milnor's theorem, as above, but maybe you have something else in mind?)
Mar 25, 2010 at 20:31 comment added engelbrekt Thanks for the quick answer! I am only allowed to check one answer as accepted, but morally I accept yours too, of course.
Mar 25, 2010 at 20:27 vote accept engelbrekt
Mar 25, 2010 at 20:27
Mar 25, 2010 at 20:26 vote accept engelbrekt
Mar 25, 2010 at 20:26
Mar 25, 2010 at 20:25 history edited Matthew Stover CC BY-SA 2.5
Gave better justification
Mar 25, 2010 at 20:19 history answered Matthew Stover CC BY-SA 2.5