14
$\begingroup$

I originally asked this on math.stackexchange, where I asked if there could exist a closed manifold that could be given different geometric structures of constant curvature (not at the same time, of course). It was pointed out that the Chern-Gauss-Bonnet theorem shows that no such manifold exists in even dimensions. Also, it seems that by looking at the universal cover we see that no such manifold can be given both a spherical and hyperbolic, or both a spherical and euclidean structure. So, the only case that remains is showing that, in odd dimensions, no manifold can be given both a Euclidean and a hyperbolic structure.

Is there an example of a such a manifold, or a proof that no such manifold exists?

Improve this question
$\endgroup$
3
  • 13
    $\begingroup$ No such closed manifold exists. Bieberbach's theorem says that the fundamental group of a compact flat $n$-manifold contains $\mathbb{Z}^n$ as a subgroup of finite-index. But fundamental groups of compact hyperbolic manifolds cannot contain $\mathbb{Z}^2$. $\endgroup$
    – Andy Putman
    Commented May 11, 2014 at 21:22
  • 9
    $\begingroup$ Well, I suppose you could say that the circle is hyperbolic. $\endgroup$
    – Tom Goodwillie
    Commented May 11, 2014 at 21:35
  • 2
    $\begingroup$ This is essentially a duplicate of mathoverflow.net/questions/133007/… $\endgroup$
    – Misha
    Commented May 12, 2014 at 18:39

2 Answers 2

Reset to default
26
$\begingroup$

Here is another approach to impossibility. Some decades after Bieberbach, Milnor showed that the ball of radius $R$ in the universal cover of a compact manifold is basically a bunch of copies of fundamental domain, copies indexed by the ball of a similar radius in the fundamental group under the word metric. Thus a metric of negative curvature requires a fundamental group with exponential growth, while a flat metric requires a fundamental group with polynomial growth. These are not compatible, so no compact manifold can admit both metrics.

Improve this answer
$\endgroup$
1
  • 5
    $\begingroup$ It should be noted that the proof of Milnor's statement is essentially trivial (which does not mean that the statement is...), so this is a much better argument than the one based on the (rather difficult) Bieberbach theorem. $\endgroup$
    – Igor Rivin
    Commented May 12, 2014 at 0:58
19
$\begingroup$

This is impossible. By Bieberbach's Theorem a flat closed $n$-manifold $M$ has a finite cover which is a flat $n$-torus. This implies that $M$ contains a (finite-index) subgroup isomorphic to $\mathbb Z^n$. But the fundamental group of a closed hyperbolic $n$-manifold does not contain any subgroup isomorphic to $\mathbb Z^2$: this is a standard simple fact in hyperbolic geometry, see here; the crucial point is that two commuting hyperbolic isometries of hyperbolic space have the same axis.

More generally, a closed manifold whose fundamental group contains $\mathbb Z^2$ cannot have any riemannian metric of strictly negative curvature.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .