13
$\begingroup$

If the fundamental group of a smooth closed aspherical manifold is a hyperbolic group, does that manifold admit a metric with non-positive sectional curvature?

If not, what's the obstruction to admitting a metric of non-positive curvature?

Here we are all talking about smooth closed manifold and the curvature are sectional curvature. Aspherical manifold are the maniflod which's universal covering is contractable.

We know that any manifold with a metric of non-positive curvature is aspherical by the Cartan-Hadamard theorem, and when in addition the manifold is negatively curved, then the fundamental group is a hyperbolic group in the context of Gromov. I am curious about the converse.

By the way, does there exist an aspherical manifold which admits a metric with non-negative sectional curvature?

$\endgroup$
13
  • $\begingroup$ It is not true that closed nonpositively curved manifolds have hyperbolic fundamental groups - one needs negative curvature for that. There are many examples of the type you want, see section 2 of arxiv.org/abs/1002.4235 for a survey. $\endgroup$ Oct 26, 2016 at 15:32
  • $\begingroup$ @Igor the question whether hyperbolic $\pi_1$ and aspherical implies the existence of a non-positively curved Riemannian metric is a reasonable question. It's a big question whether hyperbolic groups are all CAT(0). $\endgroup$
    – YCor
    Oct 26, 2016 at 16:06
  • $\begingroup$ @YCor: the question is reasonable and the answer is known, e.g., there are locally $CAT(-1)$ closed manifolds that are not homotopy equivalent to closed smooth manifolds (see section 2.1 in the paper I linked above). One can ask a number of questions of this type and some of them are still open, but again the paper linked above surveys what is known. $\endgroup$ Oct 26, 2016 at 16:24
  • $\begingroup$ @IanAgol: actually I was answering to Yves, who I thought did not assume smoothness. I think the smooth example can be also constructed. Davis-Januszkiewicz in [Hyperbolization of polyhedra, J. Diff. Geom. 34 (1991), 347--388], people.math.osu.edu/davis.12/old_papers/djJDG.pdf, in theorem 5b.1 constructed a closed $CAT(0)$ manifold whose universal cover is not simply-connected at infinity and hence it admits no Riemannian nonpositively curved metric. Apply strict hyperbolization to their example. I think their proof that the manifold is not simply-connected at infinity still holds. $\endgroup$ Oct 26, 2016 at 17:57
  • 3
    $\begingroup$ I've deleted my answer referring to Davis-Jaunszkiewicz, since Misha Kapovich pointed out that there was an error in the paper. This may be fixed by a construction in Charney-Davis, but this needs to be checked. I might check with one of them who are both here at MSRI now. $\endgroup$
    – Ian Agol
    Oct 27, 2016 at 21:02

0

Your Answer

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