Seeking an intuitive explanation of the Mapping Class Group

Question

For a surface $S$ the mapping class group $MCG(S)$ of $S$ is defined as the group of isotopy classes of orientation preserving diffeomorphisms of $S$: $$MCG(S)=Diff^+(S)/Diff_0(S).$$

I understand this definition as well as all of its component pieces. What I don't understand is why this quotient is a natural thing to study.

Specifically, I can see why the full diffeomorphism group $Diff(S)$ would be natural to study, and if $S$ happens to be orientable, I can see why it would be reasonable to restrict ones attention to $Diff^+(S)$. However, I don't see why the quotient is a natural or intuitive next step. Is there a good explanation why diffeomorphisms that are isotopic to the identity are 'uninteresting'? Thanks!

Answer 1

One reason is that elements of $\operatorname{Diff}_0(X)$ are diffeomorphisms $X \to X$ homotopic to the identity (through diffeomorphisms), so they act triviailly on homology, cohomology, and homotopy groups. From the point of view of algebraic topology, these maps aren't any different from $\operatorname{id} : X \to X$, so we identify them with $\operatorname{id}$ by forming the quotient $\operatorname{Diff}^+(X)/\operatorname{Diff}_0(X)$.

Another reason is that in some geometric constructions, only the homotopy class of the diffeomorphism matters, i.e. two isotopic diffeomorphisms give rise to isomorphic objects. One example where the mapping class group naturally arises is in the construction of mapping tori.

Let $X$ be a smooth manifold of dimension $n$, and $f : X \to X$ a diffeomorphism, then the mapping torus of $f$ is the quotient $M_f = X\times I/\sim$ where $(x, 0) \sim (f(x), 1)$. Intuitively, we construct the 'cylinder' $X\times [0, 1]$ and glue the ends together via the diffeomorphism $f$. The mapping torus is a smooth manifold of dimension $n+1$, and there is a map $p : M_f \to S^1$ given by $[(x, t)] \mapsto t$ which turns $M_f$ into a fiber bundle over $S^1$ with fiber $X$. Two diffeomorphisms which are isotopic give isomorphic fiber bundles, so there is a well-defined map between the mapping class group of $X$ and isomorphism classes of fiber bundles over $S^1$ with fiber $X$. A word of warning, this map is not injective; two elements of the mapping class group give the same isomorphism class of bundle if and only if they are conjugate.

Mapping tori associated to diffeomorphisms of surfaces are particularly important in the study of three-manifolds.

Answer 2

If you view an isotopy as a path in the space of diffeomorphisms, each element of the mapping class group corresponds to a path component of the orientation-preserving diffeomorphism group.

Answer 3

Our mental pictures of diffeomorphisms are already vague - it's not like our mind has graphing softwhere in it that plots specific points with a high level of precision, it just has a general idea of some stretching and bending and so forth - which means we already implicitly interpret two diffeomorphisms as essentially the same if they are sufficiently close.

One way to formalize this a bit is to pick a sufficiently small neighborhood $U$ of the identity, then declare $a\sim b$ (meaning "$a$ and $b$ are essentially the same") whenever $a,b\in U$, then translate this relation by declaring $a\sim b$ whenever $a,b$ are in some translate of $U$. If we want to extend this to an equivalence relation, that forces us to declare $a$ and $b$ to be related if one can be obtained from the other by a sequence of more or less negligible perturbations, even if the sum total of the perturbations may not be something we'd intuitively accept as negligible. That may be paradoxical, but this paradox is hardly unique to this situation: it's called the sorites paradox.

I think we do lose interesting information in the process. Certainly the mapping class group is coarser than the diffeomorphism group. For instance, rotating a 2-sphere around an axis by a really small angle is essentially the same as the identity, but doing it a large number of times exhibits a nontrivial loop in the diffeomorphism group which the mapping class can't see. In other words $\pi_0$ doesn't know anything about $\pi_1$. Just because it's a lesser amount of information doesn't change it's being an interesting feature, though.

If we forget the fact we're talking about diffeomorphism groups for the moment, hypothetically someone could apply the same sort of skepticism to any topological invariant which blurs together things that are sufficiently close and then accepts the formal consequences of the blurring. One could ask, "why should I consider two points that are close together in a space approximately the same, why is that an interesting thing to do?" as a reaction to the idea of taking the $\pi_0$ of a space, but ultimately knowing how many connected components a space has seems like a pretty fundamental feature of the space doesn't it?

Going back to the situation of topological groups of transformations on a space, consider how we think about smaller groups of transformations, namely Lie groups. Like, $\mathrm{O}(n)$ has two components - one for rotations, one for reflections, which are fundamentally different transformations. Indeed, an indefinite orthogonal group $\mathrm{O}(p,q)$ has four components, and it means something that $\pi_0$ looks like Klein four instead of cyclic order four.

Answer 4

To explain why this is interesting, let's look at an analysis of a particular case, namely $S = T^2$.

From algebraic topology we learn about the homology functor, from which we deduce that there is a homomorphism $$\text{Diff}^+(T^2) \to \text{Aut}(H_1(T^2;\mathbb{Z})) \approx \text{SL}(2,\mathbb{Z}) $$ Thinking about the construction of $T^2$ as the quotient $\mathbb{R}^2 / \mathbb{Z}^2$, and thinking about the fact that elements of the group $\text{SL}(2,\mathbb{Z})$ act on $\mathbb{R}^2$ preserving the orbits of the additive action of $\mathbb{Z}^2$, one quickly concludes that the above homomorphism is surjective.

Once you've gone that far, nothing is more natural from an algebraic perspective to ask: what is the kernel of that homomorphism? Because, once you figure out that the kernel is $\text{Diff}_0(T^2)$, you get an amazing theorem: $$\text{MCG}(T^2) \approx \text{Diff}^+(T^2) / \text{Diff}_0(T^2) \approx \text{SL}(2,\mathbb{Z}) $$ And once you have that theorem in front of you, what is more natural than to generalize, and to ask "What can I say about $\text{Diff}^+(S)/\text{Diff}_0(S)$ in the general case?"