Extended mapping class group of $S^p \times S^q$

Question

If I understand correctly,

(1) the extended mapping class group of $S^2 \times S^1$ is $$ \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2, $$

• how do I understand the third generator that is not within the mapping class group of MCG($S^2 \times S^1)=\mathbb{Z}_2 \times \mathbb{Z}_2$?

(2) What is the extended mapping class group of $S^p \times S^q$ in general?

Say,

$$ \text{ extended MCG$(S^p \times S^q)$ = $\mathbb{Z}_2 \times $ MCG$(S^p \times S^q)$ }? $$ Is differed by just an additional $\mathbb{Z}_2$ factor more than that of Mapping class group of $S^p \times S^q$?

Here $S^d$ is a d-sphere.

Thank you in advance!

Answer 1

I guess "extended mapping class group" means $\pi_0 \text{Diff}(M)$, the set of isotopy classes of all diffeomorphisms. (I have never heard this notation in my life.) What you call the mapping class group is what I would call the oriented mapping class group $\pi_0 \text{Diff}^+(M)$, the set of orientation-preserving isotopy classes of diffeomorphisms. I will continue to use this standard notation instead of yours, writing $MCG(M)$ and $MCG^+(M)$, respectively. In my previous post, I wrote $MCG(M)$ everywhere, but what I meant was $MCG^+(M)$ - I made that restriction at the start of the post. It seems that this is what has caused the notation confusion; I am sorry about that.

I hope that you can use what follows to answer questions like this yourself from now on (as they are not too difficult).

If $M$ is an orientable manifold, the set $\text{Diff}(M)$ carries an homomorphism to $\Bbb Z/2$ (by whether or not a map preserves orientation), where the kernel is $\text{Diff}^+(M)$. So long as the map to $\Bbb Z/2$ is surjective, there is a short exact sequence $$0 \to \text{Diff}^+(M) \to \text{Diff}(M) \to \Bbb Z/2 \to 0.$$ For the map to $\Bbb Z/2$ to be surjective means precisely that $M$ carries some orientation-reversing diffeomorphism. If not, then $\text{Diff}^+(M) = \text{Diff}(M).$

Whenever you have a short exact sequence of topological groups, it is a fibration, and you may apply the long exact sequence of homotopy groups of a fibration:

$$\cdots \to \pi_1(\Bbb Z/2) \to \pi_0 \text{Diff}^+(M) \to \pi_0 \text{Diff}(M) \to \Bbb Z/2 \to 0.$$ Of course, $\pi_1 (\Bbb Z/2) = 0$, so we see we have a short exact sequence $$0 \to MCG^+(M) \to MCG(M) \to \Bbb Z/2 \to 0.$$

That is, the full mapping class group is always an extension of $\Bbb Z/2$ by the oriented mapping class group. Deciding what the extension is tends to be quite difficult.

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Because every manifold $S^p \times S^q$ has an orientation-reversing diffeomorphism, we see that their mapping class groups an oriented mapping class groups fit into an extension as above. It is not always trivial: there is a surjection $MCG(S^{2n+1} \times S^{2n+1}) \to GL_2 \Bbb Z$, given by taking the induced map on $H_{2n+1}$. The oriented mapping class group to $SL_2 \Bbb Z$, and if the extension $$0 \to MCG^+(S^{2n+1} \times S^{2n+1}) \to MCG(S^{2n+1} \times S^{2n+1}) \to \Bbb Z/2 \to 0$$ was trivial, then so too would be the extension $$1 \to SL_2 \Bbb Z \to GL_2 \Bbb Z \to \Bbb Z/2 \to 0.$$ But this extension is nontrivial.