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I have recently been studying a proof of Mostow rigidity (along the lines of Mostow's original argument), and I'm left a little confused about something. We start with an isomorphism $\alpha: \Gamma \to \Gamma'$ between cocompact lattices in $\mathbb{H}^n$, $n \geq 3$, and observe that such a map lifts to a quasi-isometry $\phi: \mathbb{H}^n \to \mathbb{H}^n$. We declare that two quasi-isometries are close if the pointwise distance between them is uniformly bounded, and we prove that the group $QI$ of quasi-isometries on $\mathbb{H}^n$ modulo closeness is isomorphic to the group $QC$ of quasi-conformal homoemorphisms of the boundary sphere. With the aid of some detailed analysis of quasi-conformal maps and an ergodic theorem, we complete the proof by showing that the map on the boundary sphere induced by $\phi$ is actually conformal.

My question is about the part of the argument where we prove that $QI$ is isomorphic to $QC$. To prove that every quasi-isometry on $\mathbb{H}^n$ extends to a homeomorphism of the boundary sphere, that this extension map is injective, and that its image lies in $QC$ involves only standard ideas in large scale geometry and hyperbolic geometry (mainly the fact that hyperbolic space is Gromov hyperbolic). Surjectivity, on the other hand, is much harder; it basically involves a tricky compactness theorem in analysis. Most of the references that I have used make a point to discuss this argument, but it is not clear to me why we even need surjectivity since in the proof of Mostow rigidity we show that an a-priori quasi-conformal map is secretly conformal and conformal maps are much more easily seen to be induced by honest isometries. Could someone familiar with this argument explain why we need surjectivity (assuming we do)?

Thanks!

flag	asked Apr 20 2010 at 16:40 Paul Siegel 9,600●2●33●74

Agol · Accepted Answer · 2011-02-24 18:21:56Z UTC

As you say, $QI\cong QC$ is not necessary for the proof of Mostow rigidity. I'm not sure which reference you are referring to, but Gromov's proof (which is popular among topologists) does not lead to this result.

I suspect it is mentioned because there is a general program to try to classify quasi-isometry groups of $\delta$-hyperbolic spaces, and this is the motivating example. There is a nice discussion of how to extend quasiconformal isotopies of the sphere to quasi-isometries of hyperbolic space by "visual extension" in Appendix B of Curt McMullen's book "Renormalization and 3-manifolds which fiber over the circle".

bb · Answer 2 · 2010-04-21 18:46:00Z UTC

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I don't think you really need to use surjectivity explicitly in the remainder of the proof, but you also don't need analysis to prove that the extension is surjective. You can use basic hyperbolic geometry to show that the extension is injective and continuous. Any continuous injective map from a sphere to itself is necessarily surjective.

See "Lectures on hyperbolic geometry" by Benedetti and Petronio or Thurston's notes for easy proofs that the extension is injective and continuous.

answered Apr 21 2010 at 18:46

bb
1,191●5●9

Sorry, I was not very clear about what I meant. To be explicit, there is a map f: QI -> QC obtained by extending a quasi-isometry from H^n to H^n to a map from the boundary sphere S^{n-1} to itself, and I am referring to the fact that f is itself surjective rather than the fact that the image of f consists of surjective maps (as you say, this is standard hyperbolic geometry). In other words, I am asking if to prove Mostow rigidity we need to know that every quasi-conformal homeomorphism from the boundary sphere to itself is the extension of a quasi-isometry from hyperbolic space to itself. – Paul Siegel Apr 21 2010 at 19:02

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Sorry to answer the wrong question. I don't think you need to use that the map f is surjective. As you say, once you know that the map on the boundary sphere induced by phi is a quasi-conformal homeomorphism, then some theorems from analysis tell you that the induced map on the boundary sphere is actually conformal hence extends to an isometry of H^n. I don't see where you would need surjectivity of the map f. – bb Apr 21 2010 at 21:49

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