Let $\mathbb{D}\subset\mathbb{C}$ be the unit disk, and remove $n\geq 2$ of its points $P$. The resulting object will be called the punctured disk $\mathbb{D}_n$ in the following. I am interested in its mapping class group, that is, the isotopy classes of orientation-preserving self-homeomorphisms/self-diffeomorphisms $\phi:\mathbb{D}_n\rightarrow\mathbb{D}_n$ which fix the boundary, and permute the points in $P$, where isotopy is defined as homotopy 'through' such homeomorphisms. I can equip $\mathbb{D}_n$ with a hyperbolic metric so that the boundary is a geodesic in that metric, and the punctures are cusps. The metric can be lifted to the universal cover $\widetilde{\mathbb{D}_n}$, which can then be embedded into the hyperbolic plane $\mathbb{H}^2$.

Specifically, I am interested in how the mapping class group acts on geodesic rays in the universal cover. It is apparently 'well-known' that the mapping class group acts on part of the boundary at infinity of $\mathbb{H}^2$, which is given by equivalence classes $\Gamma$ of geodesic rays, where two geodesic rays are considered equivalent iff they stay a bounded distance apart/are asymptotic. This is used in proving that the braid-groups (which are just the above mapping class groups) admit uncountably many left-orderings by an approach due to Thurston, but the action is never given explicitly. See for example page 113 in this paper. (The paper also has a nice picture of the embedding $\widetilde{\mathbb{D}_2}\subset\mathbb{H}_2$ on page 114.)

I am trying to construct this action explicitly, rather than working with high-level arguments. What I tried so far: Fix a point $x\in\mathbb{D}_n$, and a point $y\in\widetilde{\mathbb{D}}_n$ with $p(y)=x$, where $p$ is the covering map. Now I want to create the action: Take an element $[\phi]$ of the mapping class group, and a geodesic ray $\Gamma:[0,\infty)\rightarrow\widetilde{\mathbb{D}}_n$ starting at $y$. The first guess is $[\phi].[\Gamma]:=[\widetilde{\phi}\circ\Gamma]$, where $\widetilde{\phi}$ is the unique lift of $\phi\circ p$ to the universal cover such that $\widetilde{\phi}(y)=y$. This doesn't work though, because $\widetilde{\phi}\circ\Gamma$ need not be a geodesic.

Next I thought that maybe the lift $\widetilde{\phi}$ as above is a quasi-isometry, which would allow me to conclude that $\widetilde{\phi}\circ\Gamma$ is a quasi-geodesic ray, which I could then associate to a unique geodesic ray since I am operating in hyperbolic space. According to the answer to this question $\widetilde{\phi}$ need not be a quasi-isometry though. But the answer says there is a fix, where I can construct a representative homeomorphism $\psi$ of each isotopy class which has a quasi-isometry $\widetilde{\psi}:\mathbb{D}_n\rightarrow\mathbb{D}_n$ as lift, and this would give me what I need to construct my action. Sadly, the construction is only sketched and I don't have the knowledge to do it myself, so my first question would be if someone could please elaborate on the sketch given there in the comments. It goes like this: 1. Delete the interior $(0,\infty)\times S_1$ of the cusps. 2. Find a representative which permutes the resulting circles by isometries of circles. 3. Apply Milnor Svarc to the universal cover of the deleted object. 4. Extend over cusps by isometries. I have no idea how to make this precise, but assume for the moment that I accept this result.

I assume that I can then explicitly state my action as $[\phi].[\Gamma]=[\widetilde{\phi}\circ\Gamma]$ where I assume w.l.o.g. that $\phi$ lifts to a quasi-isometry. I then still have to show well-definedness of $[\widetilde{\phi}\circ\Gamma]$ for different representatives of the same class which lift to quasi-isometries, which brings me to my second question: Is $\pi_1(\mathbb{D}_n)$ quasi-isometric to $\widetilde{\mathbb{D}}_n$? If I knew that this was true, then I came up with a proof (won't include it here because the post is already so long). But I still need to know whether the statement is true for my proof to hold. I know that at least for the deleted object the analogous statement holds, because the deleted object is compact and thus the fundamental group acts geometrically on the universal cover by deck-transformations. This means that I can apply the Milnor-Svarc-Lemma which states that if a group $G$ acts cocompactly, properly discontinuously, and by isometries on a nice-enough metric space, then group and space are quasi-isometric. But $\mathbb{D}_n$ isn't compact (the points are removed), so I can't apply Milnor-Svarc like I'd like to.

Now if somebody could explain to me

  1. How to construct the representative as in the sketch in a little more detail and/or
  2. Tell me if/why $\pi_1(\mathbb{D}_n)$ is quasi-isometric to $\widetilde{\mathbb{D}}_n$. This is apparently wrong.

Most importantly:

  1. Since 2) isn't the case, how can I assure well-definedness of $[\phi].[\Gamma]=[\widetilde{\phi}\circ\Gamma]$ for different representatives $\phi_1,\phi_2$ from the same isotopy class and which both lift to quasi-isometries? Or if this doesn't work, how can I construct the action instead? I would be very grateful for answers.

Edit 2: An idea regarding 3). Let $D_n$ denote the object I obtain by deleting very small circles around the punctures of $\mathbb{D}_n$. Is it true that the mapping class group of $D_n$ is the same as that of $\mathbb{D}_n$? If this were so, maybe then for any homeomorphism $\phi:D_n\rightarrow D_n$ its lift $\widetilde{\phi}:\widetilde{D}_n\rightarrow\widetilde{D}_n$ would be a quasi-isometry (why?), and because $D_n$ is compact I could use my earlier proof with Svarc-Milnor that relies on $\pi_1(D_n)\overset{quasi-isometric}{\simeq}\widetilde{D_n}$ to show well-definedness of $[\phi].[\Gamma]=[\widetilde{\phi}\circ\Gamma]$.

Edit 1: I meant to write $\widetilde{\mathbb{D}}_n$ instead of $\mathbb{D}_n$ in 2).

  • 2
    $\begingroup$ $\pi_1(\mathbb{D}_n)$ is free on $n$ generators, so has infinitely many ends for $n\ge 2$, while $\mathbb{D}_n$ has finitely many ends ($n+1$). So they are not quasi-isometric. For $n=1$, $\mathbb{D}_1$ (punctured disk) can't be endowed with a hyperbolic metric for which both ends are cusps. Anyway, for any complete hyperbolic metric, it will have exponential growth, while the fundamental group $\mathbf{Z}$ has linear growth, so again they are not QI for $n=1$ as well (a similar argument works for $n=0$). $\endgroup$
    – YCor
    Sep 7, 2017 at 22:22
  • $\begingroup$ 2) is mis-copied from earlier in the question where it asks whether $\pi_1(\mathbb{D}_n)$ is quasi-isometric to $\widetilde{\mathbb{D}}_n$. However, that's not really a sensible question either. It is certainly sensible to ask whether the orbit map $\pi_1(\mathbb{D}_n) \to \widetilde{\mathbb{D}}_n$ of the action is a quasi-isometry, but the answer is "no" because the action is not cobounded. $\endgroup$
    – Lee Mosher
    Sep 8, 2017 at 1:36
  • $\begingroup$ $\pi_1(D_n)$ is not QI to $\tilde{D}_n$ because the former is a free group (so has 0, 2 or $\infty$ ends), while the second one is the hyperbolic space and hence has 1 end. $\endgroup$
    – YCor
    Sep 8, 2017 at 8:14
  • $\begingroup$ How about what I proposed in my second edit. Do the deleted object and $\mathbb{D}_n$ have the same mapping class group? $\endgroup$
    – azureai
    Sep 9, 2017 at 0:46

1 Answer 1

  1. This is an exercise. Your second step is the part that will require some thought. For a related discussion, see Rolfsen's book where he classifies simple closed curves in the annulus.

  2. They are not quasi-isometric. The fundamental group of a hyperbolic surface with punctures (and perhaps with boundary) is never isometric to its universal cover. This is because the cusps introduce distortion: a group element that winds about the cusp can be geometrically shortened (by entering the cusp) but cannot be algebraically shortened.

  3. If $\phi_1$ and $\phi_2$ are a pair of representatives, then there is an isotopy $F$ between them. We have to choose $F$ to fix $x$, to respect the system of peripheral circles, and near the punctures be an isometry. That is, every $F_t$ is a homeomorphism of the same quality as the $\phi_i$ are. Building such an $F$ is an exercise similar to, but harder than (1) above. Now, consider $\tilde{F}$, the lift of $F$ that fixes $y$.

  4. You ask in the comments if the punctured disk and the disk with disks removed have the same mapping class group. This depends on how you define the mapping class group. Must isotopies fix boundaries pointwise? Or is each boundary equipped with a marked point, and the set of points is fixed setwise? Or perhaps there is no condition? See the book by Farb-Margalit for a discussion.

  • $\begingroup$ Thank you. I was reading Farb-Margalit just now, and have been doing this for a few weeks already. Since I'm not familiar with all the terminology there it's a slow read for me. My aim is to understand how the braid groups act on the real line, and those are apparently isomorphic to the mapping class of a punctured disk as I have described in the OP. Regarding 4) then, I am of course interested in mapping class groups isomorphic to the braid groups. Still, I need their universal cover to embed into the hyperbolic plane. I will have a look at the rest of your answer later after I got some sleep $\endgroup$
    – azureai
    Sep 9, 2017 at 8:24
  • $\begingroup$ Which book of Rolfsen are you referring to in 1)? $\endgroup$
    – azureai
    Sep 9, 2017 at 18:53
  • 1
    $\begingroup$ The title is "Knots and links". Here is a link: ams.org/publications/authors/books/postpub/chel-346 $\endgroup$
    – Sam Nead
    Sep 9, 2017 at 19:47
  • $\begingroup$ I think I came up with something: Let $\phi$ and $\psi$ be isotopic in $\mathbb{D}_n$. Let $D_n$ be the object with very little circles around the punctures deleted, and $X:=\mathbb{D}_n - D_n$. Then for any deleted circle-neighborhood, I isotope $\phi$ to be the identity there using Alexander's trick for the once-punctured disk. Gluing together $\phi|_{D_n}$ and the homotopies on the little disks, I get another map in the same mapping class. Same for $\psi$. I consider then the lifts separately on $p^{-1}(X)$ and $p^{-1}(D_n)$. $\endgroup$
    – azureai
    Sep 11, 2017 at 1:20
  • $\begingroup$ They agree on the former, and on the latter I have a lifted (continuous) isotopy on the pre-image of the compact set $D_n$ which is itself compact. This means in particular that the homotopic $\psi$ and $\phi$ are bounded in the sup-norm there, and I'm done with 3). (In the above, $p:\widetilde{\mathbb{D}_n}\rightarrow\mathbb{D}_n$ is the covering map, and I meant to write 'gluing together $\phi|_{D_n}$ and the identities on the little disks'. Now I still need to understand why $\phi$ and $\psi$ (after I changed them) lift to quasi-isometries. $\endgroup$
    – azureai
    Sep 11, 2017 at 1:28

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