12
$\begingroup$

Good morning everyone,

I was wondering about the difference between manifolds carrying a Riemannian metric with negative sectional curvature and hyperbolic manifolds. I was told once "there are very few properties in constant negative sectional curvature that cannot be extended to the negative sectional curvature case".

Can one list the topological properties a manifold with negative (but variable) sectional curvature must have?

$\endgroup$
8
  • 3
    $\begingroup$ Haagerup property of the fundamental group is one key difference: You have it in constant curvature, but it fails for some negatively curved manifolds. $\endgroup$
    – Misha
    Jun 7, 2013 at 12:05
  • 2
    $\begingroup$ Given the form of the question, it should be community wiki. You should prbobly look at the examples constructed by Gromov of manifolds having metrics of arbitrarily pinched negative curvature, that admit no hyperbolic metric. I do not know more than their existence myself, though. $\endgroup$ Jun 7, 2013 at 13:52
  • $\begingroup$ Yes I know those examples, actually this question arised while studying them. Maybe I should precise my question, because I'm interested in the properties btoh kind of manifolds shares. I realize the way I aked my question is not clear ! $\endgroup$
    – Selim G
    Jun 7, 2013 at 14:31
  • $\begingroup$ See also mathoverflow.net/questions/107716/… $\endgroup$
    – Misha
    Jun 7, 2013 at 18:14
  • $\begingroup$ Now, as the question is rewritten, it makes less sense than before, as you are casting your net way too wide: Are you interested in common topological properties of all closed manifolds admitting metrics of negative curvature? Are you interested in topological properties which manifolds of constant curvature have and variable curvature might not have? Are you interested in geometric properties of metrics of negative curvature? All of these? If so, this is not a good question for MO as it is too unfocused; an "answer" would have to include, say, a survey of hyperbolic groups. $\endgroup$
    – Misha
    Jun 7, 2013 at 18:28

2 Answers 2

19
$\begingroup$
  1. Sullivan proved that every closed hyperbolic manifold has a stably parallelizable finite cover. This is not true for say complex hyperbolic manifolds (of real dimension $>2$). See Farrell's "Lectures on Surgical Methods in Rigidity".

  2. Real Pontryagin classes of complete hyperbolic (or more generally conformally flat) manifolds vanish.

  3. Orientable closed hyperbolic manifolds have even Euler characteristic, while there is a complex hyperbolic surface constructed by Mumford of Euler characteristic 3. (Proof: every orientable closed manifold of dimension not divisible by 4 has even Euler characteristic, and in dimensions divisible by 4 the Euler characteristic is the Betti number in the middle dimension, which vanishes for hyperbolic manifolds because their signature iz zero by the fact 2 above).

$\endgroup$
1
  • 3
    $\begingroup$ Among other examples: In locally-symmetric setting, homeomorphism implies diffeomorphism (Mostow), but not in the general variable curvature case (Farrel-Jones). $\endgroup$
    – Misha
    Jun 8, 2013 at 15:35
14
$\begingroup$

To make things interesting, I will consider two classes of manifolds (one strictly larger than the other) without compactness assumption:

  1. connected manifolds (dimension is finite but not fixed) admitting complete, negatively pinched Riemannian metrics

  2. connected manifolds (dimension is finite but not fixed) admitting metrics of constant negative curvature.

In this setting there are exactly 3 known (to me) distinctions between fundamental groups $\pi$ of manifolds which belong to these classes, the first is very simple and the other two are quite subtle:

In Class 2, we have:

a) Every nilpotent subgroup of $\pi$ is virtually abelian.

b) $\pi$ has Haagerupp property: It admits a proper isometric affine action on a Hilbert space.

c) If $K$ is a closed Kaehler manifold and $\rho: \pi_1(K)\to \pi$ is a homomorphism, then either $\rho$ factors through the fundamental group of a 1-dimensional complex orbifold or the image of $\rho$ is virtually abelian.

There are other distinctions, but they all could be reduced to these three.

All three properties are known to be violated in Class 1, even if one restricts to manifolds of finite volume. For instance, quaternionic-hyperbolic lattices (acting on ${\bf HH}^n, n\ge 2$) have property T, which is a strong negation of Haagerup. Counter-examples to (c) come from complex-hyperbolic manifolds. Thus, in all three cases, the distinction could be traced to locally-symmetric spaces of rank 1.

Conjecturally, there is one more nontrivial distinction:

d) Finitely generated groups $\pi$ in Class 2 are residually finite. (Same for the fundamental groups of other locally-symmetric spaces.)

It is expected that fundamental groups in Class 1 could violate this property, but this is a major open problem, even if one considers the class of hyperbolic groups, where things are more flexible.

There are other distinctions between classes 1 and 2, coming from topology of the manifold and not from the fundamental group, when you look, say, at manifolds which are open disk bundles with hyperbolic base or intersection pairing of open 4-manifolds. You can read this survey to find more about these.

$\endgroup$
4
  • $\begingroup$ Misha, there are torsion-free hyperbolic groups that aren't the fundamental group of a manifold in Class 1. Namely, Gromov's random groups with expanders, see Naor-Silberman's theorem 1.1 in arxiv.org/abs/1005.4084. $\endgroup$ Jun 11, 2013 at 15:52
  • $\begingroup$ In fact, these hyperbolic groups aren't fundamental groups of complete nonpositivaly curved (Riemannian) manifolds. $\endgroup$ Jun 11, 2013 at 15:54
  • $\begingroup$ @Igor: A hyperbolic group can never contain an expander. On the other hand, you are right, I somehow forgot about Gromov's examples. I will modify my answer. $\endgroup$
    – Misha
    Jun 11, 2013 at 16:13
  • $\begingroup$ By "random groups with expanders" I meant the groups constructed by adding relations to a free group randomly according to edges of a expander. Sloppy language... $\endgroup$ Jun 11, 2013 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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