13
$\begingroup$

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?

This a folow up to question that turned up to be open Existence of finite index torsion free subgroups of hyperbolic groups

$\endgroup$
18
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ For the examples mentioned by Igor, the groups are known to admit an infinite image linear representation, hence have finite index subgroups. $\endgroup$ – Pierre Oct 5 '18 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.