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?



Failure of Mostow rigidity in dimension 2
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.


