A Cute Application of Mapping Class Groups

Let’s start with a fairly innocuous question:

Question: What elements of \pi_1(\mathbb T^2) can be represented by a  (homotopy class of a) simple closed curve?

A naive guess might be that with some effort, every element can be obtained by some simply closed curve, but further inspection shows that this cannot be the case. For example, first drawings of the element (2,0) will show that the curve crosses itself. On the other hand, there are elements that surely can be represented: (1,0), (2,1), (3,2) and so on. In fact, given any (a,b) with \mathrm{gcd}(a,b)=1, we have the following procedure:

which fails for when (a,b) have a common factor, since it produces multiple curves!

However, we claim that indeed these are the only elements of \pi_1(\mathbb T) that can be represented by a simple closed curve.

 

To do this, we need to make the following definition:

Definition: A separating curve (in the sense of the Jordan curve theorem) is one that bounds a disk  on the Torus. Otherwise, a curve is said to be nonseperating.

Since we will only consider curves up to homotopy, it is clear that we care mostly about nonseparating curves. Here is the crux of the argument:

Cutting along a non-separating curve produces a cylinder.  To see this, note that \chi(\mathbb T)=0, and that cutting along a nonseperating curve also has euler characteristic 0, almost by construction (in particular, consider a CW-structure,and notice that it doubles the 0-cell, but also adds a 1-cell, so it leaves \chi invariant. Using the classification of surfaces with boundary, we see that this must be a cylinder.

 

But then, consider the cylinder given by cutting along (1,0) and call this C_1. Similarly, take any other nonseparating curve, and consider its defining cylinder, C_2. There is obviously a homeomorphism h:C_1  \to C_2, and by passing to the quotient that identifies the two connected components of the boundary (along the curve we cut), the homeomorphism descends to \tilde{h}:T \to T, and it maps (1,0) to the other simple closed curve. In particular, we have reduced the question to determining where (1,0) can go under a homeomorphism. This essentially rephrases the question: how does \mathrm{Homeo}^{+}(\mathbb T) act on \pi_1(\mathbb T)? Clearly, any homeomorphism isotopic to identity acts trivially so this gives us the reduction

Question 1.1: How does \mathcal{M}(\mathbb T) act on \pi_1(\mathbb T)=\mathbb Z \oplus \mathbb Z?

Well, clearly every homeomorphism h:\mathbb T \to \mathbb T gives a  automorphism h_*: \mathbb  Z^2 \to \mathbb Z^2 by functoriality. Hence, we have a homomorphism \phi: \mathcal{M}(\mathbb T) \to \mathrm{Aut}(\mathbb Z^2)=SL_2(\mathbb Z). This is injective, since we are considering homotopy classes of homeomorphisms. Indeed, we can check that it is surjective, by observing that any element T \in \mathbb SL_2(\mathbb Z) has the property that  for any real numbers x,y \in \mathbb R and m,n \in \mathbb Z, we have T(x+m,y+n)=T(x,y)+T(m^{\prime},n^{\prime} for some integers m^{\prime},n^{\prime}). However, \mathbb R^2 is the universal cover for the torus, and this condition means exactly that in the covering map p:\mathbb R^2 \to \mathbb T, this map also descends to some homeomorphism \tilde{T}: \mathbb T \to \mathbb T, since it is well defined in the quotient. Hence, we see that every element of SL_2(\mathbb Z) can be realized has a self-homeomorphism of \mathbb T^2, showing that our map \phi was surjective.

 

Hence, \mathcal{M}(\mathbb T)=SL_2(\mathbb Z), and by studying the image of (1,0), we need only look at an element of T \in SL_2(\mathbb Z), and note that there exists a map T(1,0)=(a,b) if and only if gcd(a,b)=1, which completes the proof.