most recent 30 from http://mathoverflow.net 2013-06-19T23:56:11Z http://mathoverflow.net/feeds/question/40170 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40170/does-a-compact-negatively-curved-manfiold-of-dimension-4-admit-a-cover-of-finite Dmitri 2010-09-27T15:57:50Z 2010-09-27T16:55:13Z

<p>A $3$-dimensional compact manifold of negative sectional curvature admits (by geometrisation?) a metric of curvature $-1$, and so its fundamental group has subgroups of finite index. I wonder if an analagous question is open already in dimension $4$ -- i.e. it is not known that $M^4$ with negative sectional curvature always have a cover of finite degree? Are there any positive results in this direction?</p> <p>This a folow up to question that turned up to be open <a href="http://mathoverflow.net/questions/40120/existence-of-finite-index-torsion-free-subgroups-of-hyperbolic-groups" rel="nofollow">http://mathoverflow.net/questions/40120/existence-of-finite-index-torsion-free-subgroups-of-hyperbolic-groups</a></p>

http://mathoverflow.net/questions/40170/does-a-compact-negatively-curved-manfiold-of-dimension-4-admit-a-cover-of-finite/40181#40181 Igor Belegradek 2010-09-27T16:55:13Z 2010-09-27T16:55:13Z

<p>This is a well-known open problem. In fact, there are very few tools for studing general negatively curved manifolds. Even in dimension 3 it is unknown (I think) how to prove existence of proper finite index subgroups without using the geometrization. Geometrization implies residual finiteness of f.g. 3-manifold groups, and hence existence of proper finite index subgroups. As for positive results, lattices in semisimple Lie groups are residually finite. There is only one known method of constructing compact negatively curved manifolds that are not homotopy equivalent to locally symmetric ones, namely, branched covers (with examples given by Mostow-Siu, Gromov-Thurston, and Deraux). I do not know the answer to your question for these examples.</p>