3
$\begingroup$

My note says that Cartan-Hadamard theorem is an example of how negative sectional curvature "impacts" the topology of manifolds. The version of the theorem we are using is:

Let $(M,g)$ be a complete Riemannian manifold with everywhere non-positive sectional curvature. Then $\exp_p: T_pM \to M$ is a covering map.

I don`t understand what it means to "impact" the topology. How does the theorem exhibit such an effect?

$\endgroup$
  • 3
    $\begingroup$ For example, if $M$ is simply connected, then a corollary says that exp$_p$ is a diffeomorphism, hence $M \cong \mathbb{R}^m$. In particular, if you don't want $M$ to be diffeomorphic to $\mathbb{R}^m$, then $M$ cannot be simply connected. $\endgroup$ – 57Jimmy Aug 13 '16 at 13:58
  • 2
    $\begingroup$ For (another) example, the theorem implies that $M$ is covered by $\mathbb R^n$, which means that all the higher homotopy groups $\pi_n(M)$ are trivial. This exclude a lot of manifolds. $\endgroup$ – user99914 Aug 13 '16 at 17:13
  • $\begingroup$ the question is what does it mean to "impact" the topology... $\endgroup$ – user166467 Aug 15 '16 at 6:56
0
$\begingroup$

It would have helped if you were to include a complete quote and its source. As written, the quote (and, thus, your question) can be interpreted in two different ways:

  1. How does C-H theorem impacts "topology of manifolds", the latter regarded as an area of mathematics, with thousands of papers written in the last 100+ years, famous conjectures solved (like the Poincare conjecture), etc.

  2. More narrowly: If we know that a smooth manifold $M$ admits a complete Riemannian metric of negative curvature, what does C-H theorem say about the topology of the manifold $M$?

These are very different question which have very different answers. I will start with the 2nd question, since the answer is quite simple and, mostly, already mentioned in the comments:

The universal cover of $M$ is diffeomorphic to $R^n$. In particular, $\pi_i(M)=0$, $i>1$ (i.e. $M$ is aspherical), $\pi_1(M)$ has no nontrivial elements of finite order. There are other, more subtle implications, but that's already enough to get an idea, since these restrictions exclude many manifolds.

For the question in the 1st interpretation, the answer is much more ambiguous and depends on whom you ask. From my viewpoint, C-H theorem provides a source examples of manifolds $M$ whose universal covers are diffeomorphic to $R^n$ and, hence, aspherical manifolds. Study of aspherical manifolds is an important part the modern algebraic topology with the central questions revolving around conjectures of Borel and Wall. For instance, Borel conjecture states that two closed aspherical manifolds with isomorphic fundamental groups are homeomorphic. While this conjecture is currently out of reach, it was proven (in dimensions $\ne 4$) when one of the manifolds admits a metric of nonpositive curvature, see

F. T. Farrell and L. E. Jones. Topological rigidity for compact non-positively curved manifolds. In Differential geometry: Riemannian geometry (Los Angeles, CA, 1990), pages 229–274. Amer. Math. Soc., Providence, RI, 1993.

and, more recent, Survey on aspherical manifolds by W.Lueck.

Furthermore, the recent (i.e. the last 25+ years) advances towards Borel (and Wall) conjectures are along the lines of imposing nonpositive curvature-like assumptions on manifolds and then making some topological conclusions.

On the other hands, some large parts of the topology of manifolds (as an area of mathematics) were totally unaffected by the C-H theorem, for instance most of the work in 4-dimensional topology (since Donaldson and Freedman in the early 1980s) and topology of closed manifolds with finite fundamental groups.

$\endgroup$

Your Answer

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