1
$\begingroup$

I'm considering the unit disk $\mathbb{D}\subset\mathbb{C}$, on which I put a hyperbolic metric such that the boundary $S^1$ is a geodesic, and $n$ equally spaced points $P$ on the x-axis are 'pushed off to infinity', or 'removed', i.e. you can never travel there. The area of the disk with this metric is finite. Now I'm looking at the Mapping Class Group of this object, i.e. at isotopy classes of diffeomorphisms: $[\phi]$, where $\phi$ is the identity on the boundary and 'permutes the holes', i.e. $\phi P=P$.

Apparently every such class can be represented by a quasi-isometry of the universal cover of that disk. This means that the unique $\tilde{\phi}$ for which $\pi\tilde{\phi}=\phi\pi$, where $\pi$ denotes the covering map, is a quasi-isometry. I would like to know why this is so.

I know how to lift a homotopy-equivalence $f:M\longrightarrow N$ between Riemannian manifolds, where $M$ is compact, and the universal covers have the pullback metric. The argument goes that since $M$ is compact, f is Lipschitz. Then one can use the fact that the covering map is a local isometry to deduce that $\tilde{f}$ is locally Lipschitz, and by 'decomposing geodesics' in the universal cover, one can show that this is in fact a global property.

Now as I see it the Lipschitz argument doesn't work in the case of the punctured disk because that disk isn't compact with the topology induced by the hyperbolic metric. Nonetheless, the statement is most likely true. I think maybe the proof has something to do with the fact that $\tilde{f}$ is equivariant with respect to the covering transformations, and those are isomorphic to the fundamental group of the punctured disk. Since the group of deck-transformation acts on the universal cover in a nice way, universal cover and fundamental group are quasi-isometric by the Svarc-Milnor Lemma. Now I'm stuck. Can somebody give a full proof or point to a reference?

$\endgroup$
2
$\begingroup$

The statement "the unique $\tilde\phi$ for which $\pi \tilde\phi = \phi \pi$ is a quasi-isometry" is false. Here's why.

Let me use $X$ to denote the universal cover of $\mathbb{D}-P$ with the lifted metric. Within any isotopy class, I will construct a diffeomorphism $\phi : \mathbb{D}-P \to \mathbb{D}-P$ such that no lift $\tilde \phi : X \to X$ is a quasi-isometry.

Pick $p \in P$, and assume that $\phi$ fixes $p$ (this assumption is not essential, but it is useful to simplify the argument). A deleted neighborhood of $p$ is called a cusp, and one can choose a parameterization of some cusp in the form $$N_p = [0,\infty) \times S^1 $$ so that for each $s \in [0,\infty)$ the subset $C_{p,s} = \{s\} \times S^1$ is a "horocircle". The key geometric feature is that for each $s,t \in [0,\infty)$ and each $x \in C_{p,s}$ the closest distance from $x$ to a point of $C_{p,t}$ equals $|s-t|$. In other words, the $[0,\infty)$ parameter can be used to measure distance between the horocircles $C_{p,s}$ and $C_{p,t}$.

If I choose any strictly increasing homeomorphism $h : [0,\infty) \to [0,\infty)$ then I can isotope $\phi$ to have the property that $$\phi(C_{p,s}) = C_{p,h(s)} $$ If I choose $h$ so that it is not a quasi-isometry of $[0,\infty)$, for instance $h(s)=s^2$ or $h(s)=e^s$, then no lift $\tilde \phi$ will be a quasi-isometry.

So what you want will not work unless you very carefully control the behavior of $\phi$ in the cusps. It is possible to do that, and so your statement that every mapping class has a representative homeomorphism of $\mathbb{D}-P$ that lifts to a quasi-isometry of $X$ is true, but it is somewhat tricky to prove and is not just a simple application of the Milnor-Svarc theorem.

$\endgroup$
  • $\begingroup$ Thank you for your anwer. I need to use this representative homeomorphism you mentioned in your last paragraph for a construction. I understand that you wrote it is tricky to prove that it exists, but could you give a short sketch or point me to a reference where this is done? $\endgroup$ – azureai Sep 6 '17 at 16:25
  • 1
    $\begingroup$ For a short sketch, first delete the interiors of the cusps (the $(0,\infty) \times S^1$ portions), then find a representative which permutes the resulting circles (the $0 \times S^1$ portions) by isometries of circles, then apply Milnor Svarc in the universal cover of the deleted object. Then extend over cusps by isometries. I think that works.... $\endgroup$ – Lee Mosher Sep 6 '17 at 18:20
  • $\begingroup$ Sure, if there's something concrete to ask. $\endgroup$ – Lee Mosher Sep 7 '17 at 20:58
  • $\begingroup$ Posted follow-up question here: mathoverflow.net/questions/280629/… $\endgroup$ – azureai Sep 7 '17 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.