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Space forms are complete (connected) Riemannian manifolds of constant sectional curvature. These fall into three classes: Euclidean, with universal covering isometric to $\mathbb{R}^n$, spherical, with universal covering isometric to $S^n$, and hyperbolic, with universal covering isometric to $\mathbb{H}^n$.

Does there exist compact space forms of the same dimension from two different classes having isomorphic fundamental groups?

This cannot happen for $n = 2$, since the Gauss-Bonnet theorem

$$ \int_M K{ \ }\mathrm{vol}_M = 2{\pi}\chi(M) $$

shows that the Euler characteristic $\chi(M)$ is positive, zero, or negative when the space form $M$ is Euclidean, spherical or hyperbolic respectively. But for (closed) surfaces the fundamental group determines the Euler characteristic.

It is essential for the question to require compactness, otherwise there are trivial examples. Dividing out $\mathbb{R}^2$ by the group of isometries generated by $(x,y) \mapsto (x+1,y)$ yields a Euclidean space form with fundamental group isomorphic to $\mathbb{Z}$, while dividing out $\mathbb{H}^2$ by the group of isometries generated by the hyperbolic isometry $(x,y) \mapsto (x+1,y)$ yields a hyperbolic space form also with fundamental group isomorphic to $\mathbb{Z}$.

The standard reference on space forms is Spaces of Constant Curvature by Joseph A. Wolf. The classification of Euclidean space forms is given in Chapter 3, and of spherical ones in Chapter 7. Wolf does not treat hyperbolic space forms, possibly because not much was known about them in 1967. Unfortunately, the fundamental groups are infinite for the compact Euclidean space forms, and finite for the spherical space forms (which are necessarily compact, being quotients of $S^n$). So a hypothetical example has to involve a hyperbolic space form.

An example might drop out of the theory of three-manifolds. In dimension three the space forms belong to three of the eight Thurston model geometries. A pair yielding an example would have to be one Euclidean and one hyperbolic, since it follows from Perelman's geometrization theorem that the spherical ones are precisely those with finite fundamental group.

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+1 for well-written question. –  Theo Johnson-Freyd Mar 25 '10 at 23:32
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4 Answers

up vote 10 down vote accepted

The fundamental group of a compact hyperbolic space form has exponential growth, according to a well-known theorem of Milnor [Milnor, J. A note on curvature and fundamental group. J. Differential Geometry 2 1968 1--7. MR0232311]. Bieberbach groups are, on the other hand, polynomial: indeed, their translations subgroups have finite index, so by Gromov's theorem they have polynomial growth.

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Thanks for the quick answer! –  engelbrekt Mar 25 '10 at 20:31
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Actually Gromov's theorem is about what follows from polynomial growth. The converse implication (used here) is trivial. –  Sergei Ivanov Mar 25 '10 at 21:02
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The converse to Gromov's theorem (that nilpotent groups have polynomial growth) may not be hard, but it's not completely trivial -- the formula for the degree of growth is due to Bass. But in this case the groups are virtually abelian, and that case really is trivial. –  Tom Church Mar 26 '10 at 1:26
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No, they are always different. Mostow Rigidity tells you that complete, finite-volume hyperbolic manifolds are determined by their fundamental groups. These groups never contain a $\mathbb Z \oplus \mathbb Z$, and any group acting on $\mathbb E^2$ contains this as a subgroup of finite index. Groups acting on the three-sphere are finite. This generalizes to all $n$: Mostow Rigidity remains true (and these fundamental groups never contain $\mathbb Z^n$; they are Gromov hyperbolic groups), Bieberbach Theorems say that the groups acting on $\mathbb E^n$ are virtually abelian, and the groups acting on the $n$-sphere are finite.

EDIT: I forgot to say compact above. Peripheral subgroups of a nonuniform lattice acting on $\mathbb H^n$ are abelian of rank $n-1$, so they aren't Gromov hyperbolic. However, they contain nonabelian free subgroups, so they still don't act nicely on $\mathbb E^n$ or the sphere.

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Thanks for the quick answer! I am only allowed to check one answer as accepted, but morally I accept yours too, of course. –  engelbrekt Mar 25 '10 at 20:31
    
How do you know that hyerbolic groups are not virtually abelian? I mean, I certainly believe it to be true -- I have studied Fuchsian groups, where this follows from the explicit presentation of the group in terms of its signature -- but what technique or fact are you using to see this? (You could use Milnor's theorem, as above, but maybe you have something else in mind?) –  Pete L. Clark Mar 26 '10 at 5:57
    
For hyperbolic groups, a $\mathbb{Z}\oplus\mathbb{Z}$ subgroup has triangles which are as fat as triangles in the Euclidean plane, which violates the thin triangles condition which is the definition of hyperbolic group. –  Tom Church Mar 26 '10 at 9:57
    
For fundamental groups of negatively curved manifolds, the answer above should say "compact hyperbolic manifolds never contain a $\mathbb{Z}\oplus\mathbb{Z}$". You can see this directly without appealing to the fact that they are Gromov-hyperbolic from the fact that centralizers are virtually cyclic. To see this, note that a hyperbolic isometry has two fixed points on the boundary at infinity. The centralizer of a given hyperbolic isometry must preserve these fixed points, and so sits inside $\mathbb{R}\times SO(n-1)$. A discrete subgroup of this group is virtually $\mathbb{Z}$. –  Tom Church Mar 26 '10 at 10:02
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These groups are not even quasi-isometric. (Two metric spaces are said to be quasi-isometric if they contain bi-Lipschitz equivalent nets. In the case of finitely generated groups, word metrics are assumed.)

Indeed, it is well-known (and easy to prove) that the fundamental group of a compact Riemannian manifold $M$ is quasi-isometric to the universal cover $\tilde M$ (with the lifted metric). And $S^n$, $R^n$ and $H^n$ are not quasi-isometric by a simple volume growth argument.

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Thanks for your answer! –  engelbrekt Mar 25 '10 at 20:49
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For the record, no compact 3-manifold can ever admit two different geometries; this isn't just for space forms. You should be able to show this using quasi-isometric techniques, as in Sergei Ivanov's answer above (although I haven't thought about this in a while). One nice corollary is that you can determine the geometry just from the fundamental group without a lot of technical conditions (this is copied from Wikipedia with minimal editing, since it is laid out so nicely there).

Assume $M$ is a compact 3-manifold which admits one of the 8 geometries. Then:

  • If $\pi_1(M)$ is finite then the geometry on $M$ is spherical.
  • If $\pi_1(M)$ is virtually cyclic but not finite then the geometry on $M$ is $S^2×\mathbb{R}$.
  • If $\pi_1(M)$ is virtually abelian but not virtually cyclic then the geometry on $M$ is Euclidean.
  • If $\pi_1(M)$ is virtually nilpotent but not virtually abelian then the geometry on $M$ is Nil geometry.
  • If $\pi_1(M)$ is virtually solvable but not virtually nilpotent then the geometry on $M$ is Sol geometry.
  • If $\pi_1(M)$ virtually splits as a semidirect product with $\mathbb{Z}$ but is not virtually solvable then the geometry on $M$ is the universal cover of $SL_2(\mathbb{R})$.
  • If $\pi_1(M)$ has an infinite normal cyclic subgroup but not of the above form and is not virtually solvable then the geometry on $M$ is $\mathbb{H}^2\times\mathbb{R}$.
  • If $\pi_1(M)$ has no infinite normal cyclic subgroup and is not virtually solvable then the geometry on $M$ is hyperbolic.

This is false for finite-volume non-compact manifolds, for example the complement of a trefoil knot, which admits both $\widetilde{SL_2(\mathbb{R})}$ and $\mathbb{H}^2\times\mathbb{R}$ geometries.

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There is a nice flowchart of this in Thurston's Three-Dimensional Geometry and Topology (the published one). –  Matthew Stover Mar 26 '10 at 1:50
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