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

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.

• What is an extended mapping class group?
– user98602
Jan 31, 2019 at 4:52
• (1) Isn't that just the orientation-reversing map $(x, y) \to (-x, y)$? Jan 31, 2019 at 5:03

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.

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.

• thank you +1! this is so nice as your previous post Jan 31, 2019 at 19:11
• It is defined here: "If we modify the definition to include all homeomorphisms we obtain the extended mapping class group" Jan 31, 2019 at 19:13
• en.wikipedia.org/wiki/… Jan 31, 2019 at 19:13
• @MikeMiller It is pretty standard for people who only think about mapping class groups of orientable surfaces(for example this is how MCG is defined in the primer). If I had to guess it is because Dehn twists generate the full orientation preserving mapping class group and not that much changes when restricting to this setting. Although one theorem that comes to mind where the (extended)mapping class group is preferable is Dehn-Nielsen-Baer theorem that the mapping class group is isomorphic to $Out(\pi_1 S)$ for closed surfaces $S$.
– user29123
Feb 8, 2019 at 19:15
• I am not sure how hard it would be to check. Geometrically I would think about it at the induced map on the boundary of the cayley graph/hyperbolic plane. The orientation preserving/reversing shouldn't be that hard to detect by looking at how two words intersect and how the "axis" change orientation in the lift. There is probably a pretty easy/algorithmic way to do this with just looking where the words are mapped to, but I would have to think about it.
– user29123
Feb 8, 2019 at 19:58