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I am trying to understand why Mostow rigidity fails in dimension 2. More concretely, I have the following question:

(1) What is an example of a quasiisometry $f$ of the hyperbolic plane $\mathbb H^2$ to itself such that for the continuation $\partial f$ of $f$ to the boundary $\partial \mathbb H^2$, there does not exist an isometry $\varphi$ of $\mathbb H$ such that $\bar{\varphi} |_{\partial \mathbb H^2} = \partial f$?

The reason I am asking this question is that one way to prove classical Mostow rigidity (two closed hyperbolic manifolds of the same dimension $n \geq 3$ with isomorphic fundamental groups are isomorphic) can be deduced from the statement that for $n \geq 3$, every quasiisometry $f$ of hyperbolic space $\mathbb H^n$ extends to a homeomorphism $\partial f: \partial \mathbb H^n \rightarrow \partial \mathbb H^n$ which is the restriction of an isometry of $\mathbb H^n$.

The proof of this statement decomposes into two pieces: First, one shows that $\partial f$ is a homeomorphism, secondly, one proves that it is conformal and hence has to come from an isometry of $\mathbb H^n$. The first step also works in dimension 2, but the second does not. So more general than the above, we could even ask the following (by identifying $\partial \mathbb H^2$ with $S^1$):

(2) Which self-homeomorphisms of $S^1$ arise as extensions of quasiisometries $\mathbb H^2 \rightarrow \mathbb H^2$ to the boundary?

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  • $\begingroup$ I believe the answer to Question 2 is precisely the 'quasi-symmetric' ones, but I'm not expert enough to tell you what this means. $\endgroup$ – HJRW Nov 2 '14 at 9:43
  • $\begingroup$ an answer to (1) is: take any $C^1$-diffeomorphism of the circle which is not in $PGL_2(\mathbf{R})$; then it's the boundary map of some QI, which can't be at bounded distance to an isometry. I guess that it's possible to define some such maps piecewise explicitly on $H^2$. $\endgroup$ – YCor Nov 2 '14 at 10:07
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ad (i):

First consider a Dehn twist at some simple, closed curve in a closed hyperbolic surface. It is obviously a quasi-isometry (as any smooth map between closed surfaces) but not an isometry.

Then lift this Dehn twist to a self-map of the hyperbolic plane, equivariant with respect to the cocompact group action from the surface group. The lifted map is again a quasi-isometry.

You can check that the extension to the ideal boundary has infinitely many fixed points. (This uses the fact that the Dehn twist fixed all geodesics not intersecting the simple closed curve.)

However if your boundary map were the extension of some isometry, then it could have at most 2 fixed points. So the Dehn twist can not be homotopic to some isometry.

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    $\begingroup$ You need to choose the lift of the Dehn twist rather carefully to ensure that it fixes infinitely many points on the ideal boundary. Not all lifts have this property, in fact "most" lifts have exactly two fixed points just like an isometry. $\endgroup$ – Lee Mosher Nov 2 '14 at 16:24
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If $f$ is a quasisymmetric homeomorphism of the unit circle to itself, then it extends to a quasiconformal homeomorphism of the disc. Such a qc homeomorphism will be a quasi-isometry in the Poincaré disc model of the hyperbolic plane.

Conversely, every quasi-isometry extends to a quasisymmetric homeomorphism on the boundary.

Hence the answer to (ii) is provided precisely by the quasisymmetric homeomorphisms of the unit circle to itself, as HJRW mentions.

The answer to (i) is that the boundary extensions of quasi-isometries form an infinite-dimensional space (note that, as YCor mentions, this includes all $C^1$-diffeomorphisms). On the other hand, the isometries are of course all given by Möbius transformations.

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The actual questions you asked have already been answered, and so I am hesitant to chime in with anything.

Anyway, here is an different response to your opening line:

"I am trying to understand why Mostow rigidity fails in dimension 2."

Here is my first thought:

The structure of the fundamental group of surfaces allows for local deformations in the moduli space of hyperbolic structures.

For every hyperbolic structure on a closed $n$-manifold there is a discrete faithful representation (hence irreducible) from $\pi_1(M)\to \mathrm{Isom}(M)$, unique up to conjugacy.

When $M$ is a surface, $\mathrm{Isom}(M)$ is $\mathrm{PSL}(2,\mathbb{R})$ and $\pi_1(M)$ has a presentation with $2g$ generators and 1 non-trivial relation in those generators. One can use this to conclude that: $\mathrm{Hom}(\pi_1(M),\mathrm{Isom}(M))/\mathrm{Isom}(M)$ has dimension $3(2g-1)-3=6g-6$; and locally so at any irreducible representation.

So there are non-trivial infinitesimal deformations in that case. However, the infinitesimal deformations can be "integrated", since the obstructions are in second cohomology which vanishes since $n=2$. Hence there is no rigidity (local and hence not global).

For $n>2$, there are no such deformations; which follows from Weil's Remarks on the cohomology of groups (see here) also using information about a presentation of $\pi_1(M)$. One concludes that there are no infinitesimal deformations and thus no local deformations.

Mostow's Rigidity says there is global rigidity (for $n>2$); admittedly a stronger statement than local rigidity (although infinitesimal rigidity is stronger than local rigidity). But the difference between $n=2$ and $n>2$ is seen locally as well and mostly is a consequence of nature of $\pi_1(M)$ in those cases.

So one can say:

Mostow Rigidity Fails in dimension 2 since there is no local rigidity in dimension 2, which follows mostly from the structure of $\pi_1(M)$.

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    $\begingroup$ What you are saying is that there are infinitesimal deformations - it is not a priori clear that these can be "integrated". $\endgroup$ – Igor Rivin Nov 2 '14 at 16:57
  • $\begingroup$ You are right Igor (and I knew that). I did not mean to suggest that I was giving complete arguments (just trying to suggest a different point of view). Weil's result also requires more work too. Let me add an edit. $\endgroup$ – Sean Lawton Nov 2 '14 at 17:03
  • $\begingroup$ And thank you Igor for pointing it out. I should have mentioned that. $\endgroup$ – Sean Lawton Nov 2 '14 at 17:16

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