In the complex analytic setting, it is easy to see that the moduli space of a sphere with four punctures is $\mathcal{M}=\mathbb{CP}^1 / { 0,1,\infty }$, since I can use a Moebius transformation to send three of the punctures at those location whilst the fourth puncture is free to move.

On the other hand, the moduli space can be obtained as the quotient of the Teichmüller space by the mapping class group. Teichmüller space can be described using e.g. Fenchel-Nielsen coordinates, for each simple closed curve on the sphere we have two coordinates (a length and an angle). The mapping class group can be described using Dehn twists.

I always assumed that the three boundary points ${0,1,\infty}$ could be interpreted as sending to zero the length of three distinct curves separating the punctures into one of the three sets $(12)(34)$, $(23)(41)$ or $(13)(24)$, but I have recently realized that these curves are related to each other by suitable elements of the mapping class group. So I would naively say that the corresponding limits should be identified as well.

How do I reconcile the above picture of the moduli space with the one coming from Teichmüller modulo mapping class group? How do I see three boundary points in this picture?

  • $\begingroup$ The very recent issue of BAMS (57, N3, 2020) contains a nice big paper of Milnor addressing this subj. $\endgroup$ Mar 26, 2020 at 21:10
  • $\begingroup$ Could you tell me what BAMS stands for? $\endgroup$ Mar 27, 2020 at 3:26
  • $\begingroup$ Ok, I could find it but it seems to me that there is very little material relevant for my question. It does not seems to me that this particular moduli space is mentioned in any example. Could you point me at the pag. You think is relevant? $\endgroup$ Mar 27, 2020 at 7:11

1 Answer 1


In your description of moduli space you say:

I can use a Moebius transformation to send three of the punctures at those location whilst the fourth puncture is free to move.

That assumes that the punctures have names. Let's call them $a, b, c, d$ and we will agree to send them to $0, 1, \infty, z$. (I prefer to use letters for the names, as then we can't confuse the punctures with numbers.)

In your discussion of Teichmuller space, you identify three curves. The first curve, call it $\beta$, separates $a$ and $b$ from $c$ and $d$, while the second curve, call it $\gamma$, separates $a$ and $c$ from $b$ and $d$. Now, since these punctures have names, there is no mapping class that sends $b$ to $c$. Thus there is no mapping class that sends $\beta$ to $\gamma$.

Except there obviously is...

The solution lies in understanding which mapping class group you are using. In your definition of moduli space you are using the pure mapping class group (fixing all punctures). In your definition of Teichmuller space you are using the full mapping class group (of orientation preserving homeomorphisms, up to isotopy).

  • $\begingroup$ In order to send the two curves into each other I only need Dehn twists, which are in the pure MCG. More precisely a Dehn twist along Beta followed by one along alpha sends alpha into beta. $\endgroup$ Mar 27, 2020 at 3:29
  • $\begingroup$ Perhaps you are using half-twists instead of full twists? (But half-twists are not pure.) Or perhaps you think that $\alpha$ and $\beta$ meet once, and that the product of the twists behaves as it does on a torus? (But $\alpha$ and $\beta$ meet twice...) Hmmm. I suggest you draw some pictures. The easiest way is to draw a round disk $D$ in the plane, add three dots inside (labelled $a$, $b$, and $c$) and draw the curves $\beta$ and $\gamma$ as simply as is possible. They intersect an even number of times (aka twice) - after all, they are curves in the plane! $\endgroup$
    – Sam Nead
    Mar 27, 2020 at 7:34
  • $\begingroup$ yes, I am trying to draw some. I was using this software to experiment flipper.readthedocs.io/en/latest Just to make it clear, you would disagree with me that a sequence of pure mapping class group elements (Dehn full twists) would send those three lines into each other? $\endgroup$ Mar 27, 2020 at 8:46
  • $\begingroup$ I think that using that software I was applying half twists, if i do the drawings by hand I get the same as the the software applying twice its MCG elements... $\endgroup$ Mar 27, 2020 at 9:38
  • $\begingroup$ How can you prove formally that up to pure mappings class group any simple closed curve is equivalent to one of the three described in my question? $\endgroup$ Mar 27, 2020 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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