# Showing that Mg, the Mapping Class Group of the 1-Torus, is $SL(2,\mathbb Z)$

All:

I am trying to figure out the mapping class groupof the torus ; more accurately, I am trying to show that it is equal to $SL(2,\mathbb Z)$.

The method: every homeomorphism h: $T^2 \rightarrow T^2$ gives rise to, aka,

induces, an isomorphism g: $\pi_1(T^2) \rightarrow \pi_1(T^2)$,

and we use the fact that:

i)$\pi_1 (T^2)=\mathbb Z(+)\mathbb Z$

ii) Aut $\mathbb Z(+)\mathbb Z=SL(2,\mathbb Z)$

Now, if we can show that the homomorphism from [the group of homeomorphisms of $T^2$ to itself ] to $SL(2,\mathbb Z)$ is an isomorphism, we are done.

Now, it is not too hard (tho, I think not trivial) , to show that $SL(2,\mathbb Z)$ has a generating set with three elements ; the set of transvections (actually a set of four transvections that are generating set for the set of transvections ); the transvections are a generalization of shear maps in linear transformations $T: \mathbb R^n \rightarrow \mathbb R^m$, as maps that add a multiple of a row to another row. A (generating) shear matrix has all diagonal entries$a_{ii}$ identically equal to one, and exactly one non-diagonal entry equal to +/-1 (general shear matrices have all $a_{ii}=1$ and exactly one off-diagonal term with any non-zero value).

So to show the map is onto, I am trying to see that each of the elements of the generating set are the image of some homeomorphism from the torus to itself, i.e., to show that there are automorphisms of the torus thad induce the basis shear maps, by examing the effect of the shear maps on a standard basis {(1,0),(0,1)} of the torus, and trying to construct a self-homeo of the torus that would have the effect on homology described by the effect of the shear maps on the basis .

I was also thinking of using the fact that the mapping class group for Sg is known to have a generating set of size 3g-1 (best possible) , given by Dehn twists about the basis. I suspect these Dehn twists (about the basis elements) may induce mapsin homology describable as shear matrices. Does any know?

I will try to complete this idea, but I would appreciate some comments on whether this approach makes sense. I would appreciate your comments/suggestions.

Thanks.

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fyi: \oplus is $\oplus$ – yoyo Jul 31 '11 at 13:56
$\text{Aut}(\mathbb{Z}^2)$ is actually $\text{GL}_2(\mathbb{Z})$ (it includes matrices of determinant $-1$). – Qiaochu Yuan Aug 8 '11 at 20:53

If you use the definition of "mapping class group" as $\pi_0 Diff(S^1\times S^1)$, i.e. isotopy classes of diffeomorphisms of the torus, there's two key technical lemmas that you need.

1) If a closed smooth curve in the torus does not bound an embedded disc, then it is isotopic to a straight line. That means choose $p,q \in \mathbb Z$ coprime and look at the solution set to the equation $px+qy=0$ in $\mathbb R^2$, the image of that under the quotient map $\mathbb R^2 \to \mathbb (S^1)^2$ where you're modding out by the additive subgroup $\mathbb Z^2 \subset \mathbb R^2$.

2) Every diffeomorphism of a disc which is the identity on the boundary can be isotoped to the identity diffeomorphism on the disc, without affecting its behaviour on the boundary.

With those two technical results, plus liberal usage of things like the isotopy extension theorem, the theorem you're trying to prove falls out of the above. Also, you can prove (1) by transversality and the Schoenflies theorem (that a closed curve in the plane bounds a region diffeomorphic to a disc). (2) is more difficult, it's a theorem of Smale's. Its proof is in Thurston's (published) book, and relies on a few observations about ODE's in the plane, like the Poincare-Bendixon theorem.

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Excellent, Ryan;very helpful. – gary Jul 31 '11 at 7:35
Sorry if this comment is two unconventional, but there is another interesting post: math.stackexchange.com/questions/54519/… that I think is up your alley. I am curious about the answer too. – gary Jul 31 '11 at 8:20

Here are some remarks on your approach, and on the problem in general:

• There are lots of well-known generating sets for $SL(2,\mathbb Z)$; one useful one is $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$, $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. If you google "modular group generators" you should find more information on this.

• One way to describe a torus is as $\mathbb R^2 / \mathbb Z^2$. This gives a different perspective (more algebraic, less topological) on how $SL(2,\mathbb Z)$ might be made to act.

• It won't be the case that Homeo(Torus) $= SL(2,\mathbb Z)$. There will be a surjection, and the kernel will be homeos that are isotopic to the identity. (Actually, even to get a well-defined map, we should restrict to orientation-preserving homeos; otherwise you have to replace $SL(2,\mathbb Z)$ by $GL(2,\mathbb Z)$.) So you will have to show that a homeo which acts trivially on $\pi_1$ is isotopic to the identity. For this, it might also be helpful to regard the torus as $\mathbb R^2/\mathbb Z^2$.

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Thanks, Matt: but I wonder if showing that all the elements of a generating set are the images of homeomorphisms (i.e., that there are maps $h:T^2\rightarrow T^2$ and $g:T^2\rightarrow T^2$ that induce maps that are described by the effects of the given matrix on a basis like {(1,0),(0,1)}. Since this is low-dimensional topology, I am trying to get both the topology and geometry, and use algebra only when necessary. Still, thanks; I will consider the Torus as a quotient. – gary Jul 31 '11 at 6:16
@gary: Dear Gary, Your approach certainly makes sense, and you should be able to see the transvection topologically as a Dehn twist. I just mentioned the quotient picture because (based on your post) I thought it may not have occurred to you, and it can sometimes be helpful. I don't want to be prescriptive; ideally, you will want to understand this result from all possible perspectives! Best wishes, – Matt E Jul 31 '11 at 6:20
Matt: no problem, actually, yours is an approach I had not considered, and that itself is nice to have. I'll look into it, thanks. – gary Jul 31 '11 at 6:24
Matt: I realize how dumb my belief that the map would be a surjection: not only do any two isotopic maps induce the same map on homology, but even any two homotopic maps induce the same map on homology, so I see that we do need to consider reducing maps (at least) up to isotopy, if not homotopy. – gary Jul 31 '11 at 6:56