We know that the universal covering of a closed hyperbolic 3-manifold can be identified with the hyperbolic space $\mathbb{H}^3$. Now, what is not very clear to me is how this identification has to be intended. Of course it is a topological identification but, as we're talking about Riemannian manifold, there must be a way of talking about metrics as well. My specific case is the following: my closed hyperbolic 3-manifold $M$ is the mapping torus of a closed, hyperbolic surface $S$ (in particular hyperbolicity requires that the monodromy $\varphi$ is pseudo-Anosov). The universal covering should be, a priori, just $\mathbb{H}^3$, but if I think to $M$ as a quotient $S\times\left[0,1\right]/(x,0)\sim(\varphi(x),1)$ or, even better $S\times \mathbb{R}/(x,n)\sim(\varphi(x),n+1)$ I would be tempted to say that the universal covering has to be $\mathbb{H}^2\times \mathbb{R}$. I know that topologically they are the same space, but when I write $\mathbb{H}^2\times \mathbb{R}$ I mean that the metric is given by $ds=d\rho+dt$ where $d\rho$ is the standard metric on $\mathbb{H}^2$ and $dt$ the euclidean metric on $\mathbb{R}$. Also, I guess that $(\mathbb{H}^2\times\mathbb{R},ds)$ is quasi-isometric to $(\mathbb{H}^3,dh)$ (here $dh$ is the standard hyperbolic metric on the hyperbolic space). Could you help me understanding that? Thank you!

  • $\begingroup$ I think your first sentence is confused. For example, the universal cover of $S^3$ is certainly not $\mathbb{H}^3$! $\endgroup$ – Neal Dec 22 '11 at 15:09
  • $\begingroup$ Sorry, you are of course right! I forgot to specify 'hyperbolic'! I am going to edit the question right now! $\endgroup$ – fatoddsun Dec 22 '11 at 17:15

Okay, I don't know how helpful this answer may be, but I will give it a shot.

You seem to be asking two questions:

  • the difference between topological covers and Riemannian covers; and

  • whether or not your 3-fold is hyperbolic.

Roughly speaking, the different categories of spaces have analogous notions of cover: $X\to Y$ a "cover" means $X$ is locally "identical" to $Y$. The meaning of "identical" varies from category to category. For topological spaces, the projection is a surjective local homeomorphism. For Riemannian manifolds, the projection is a surjective local isometry.

In any dimension $n$, up to isometry and scaling the metric, there are exactly three simply connected Riemannian manifolds with constant sectional curvature: $\mathbb{H}^n$, $\mathbb{R}^n$, and $\mathbb{S}^n$. Any closed constant sectional curvature manifold is covered (in the sense of Riemannian spaces) by one of these three space forms, and the fundamental group acts by isometry.

This holds for 3-folds, but even more strongly. Mostow rigidity says that the geometry of a finite-volume hyperbolic 3-fold is uniquely determined by its fundamental group. One should therefore think of a constant-sectional-curvature 3-fold as the same as a representation $\rho:\pi_1(M)\hookrightarrow Sl(2,\mathbb{C})$, i.e., it is the quotient of $\mathbb{H}^3$ by a subgroup of $Isom(\mathbb{H}^3)$.

It is in this sense, then, that one identifies $\mathbb{H}^3$ as the universal cover of a hyperbolic 3-fold $M$, by identifying $M = \mathbb{H}^3 / \pi_1(M)$.

Now to the second point. I am a little confused here. Are you given that the mapping torus is hyperbolic, or are you looking to prove that it is hyperbolic? A mapping torus $M_f$ is hyperbolic if and only if $f$ is pseudo-Anosov. And it being hyperbolic exactly means that its universal cover is $\mathbb{H}^3$. On the other hand, if the universal cover of $M_f$ is $\mathbb{H}^2\times\mathbb{R}$, then it is not hyperbolic (and the gluing homeomorphism is not pseudo-Anosov).

  • $\begingroup$ Dear Neal, thanks for your answer! I think that you didn't get exactly what I was confusedly asking but you gave so many details that in the end I was able to understand what I had to understand! Anyway, I am given a hyperbolic 3-fold: my question was if I had to think to the universal covering as $\mathbb{H}^3$ or $\mathbb{H}^2\times\mathbb{R}$, but now I know that only the former makes sense. I was confused because on the paper I'm reading the author says that the universal cover $\tilde{M}$ is quasi-isometric to $\mathbb{H}^3$ and not just 'isometric', and I misunderstood his aim! Thanks! $\endgroup$ – fatoddsun Dec 24 '11 at 14:44

I believe here, in conjunction with the other answer, a topological covering space of a Riemannian manifold inherits a natural structure as a Riemannian manifold from the underlying structures of the covered space. This is one of those areas where a lot of 'structures' lift to the covering space in a natural way. You first construct the lifts of the TM to a TC where C covers M. Then define the metric at a pair of vectors v_x, w_x in TxC to be < v_px ,w_px > @px where p:C--->M is the covering projection.


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