# Splitting lemma for non abelian groups and mapping class group

Hi everybody! I'm in trouble with a version of the splitting lemma in the non abelian case. To be more specific, I'm working with mapping class groups of surfaces, i.e. the groups of isotopy classes of automorphisms of the surface. In the abelian case, the splitting lemma implies that if the short exact sequence of abelian groups $$1 \rightarrow A \rightarrow B \rightarrow C \rightarrow 1$$ splits (i.e. there is a section $C \rightarrow B$ of the projection $B \rightarrow C$) then we have $B \cong A \oplus C$. If the groups are not abelian, we can't conclude this, but we have $B \cong A \rtimes C$. The problem is that the semidirect product $B \cong A \rtimes C$ is not always uniquely defined. I'm looking for results that enable me to conclude that, under certain assumptions, the semidirect product is in fact a direct product, or at least that it is uniquely determined. One such condition is that the above sequence is a left split, i.e. we have a map $B \rightarrow A$ which composes with $A \rightarrow B$ to the identity of A. But this is not enough for the groups I'm working with.

Let $\mathcal{MCG}(S)$ be the group of isotopy classes of orientation preserving automorphisms of the surface S and let $\mathcal{MCG}^{\pm}(S)$ be the group of isotopy classes of all the automorphisms of S. These groups are highly non-abelian and we obviously have:

$$1 \rightarrow \mathcal{MCG}(S) \rightarrow \mathcal{MCG}^{\pm}(S) \rightarrow \mathbb{Z}_{2} \rightarrow 1$$

which is exact and right split, because every surface has an orientation reversing automorphism of order 2. My problem is how to determine $\mathcal{MCG}^{\pm}(S)$ knowing just $\mathcal{MCG}(S)$. I mean, in general we do know it is a semidirect product with $\mathbb{Z}_{2}$...but when is this uniquely determined? when is this a real direct product? In the general case I've not been able to prove that the sequence is also a left split.

Moreover I have this example putting me in trouble: let $D$ be the closed unit ball of the plane. By Alexander's Trick we have that $\mathcal{MCG}^{\pm}(D)$ is trivial; so must be $\mathcal{MCG}(D)$, being one of its subgroups. But then, on the other hand I would expect that

$$\mathcal{MCG}^{\pm}(D) \cong \mathbb{Z}_{2}$$ because of the semidirect product relation. Where am I doing wrong?

So, my questions are:

1) when is a semidirect product of non abelian groups uniquely determined?

2) when is it a direct product? (without using the left split conditions because I'm not able to prove it in my case...or using it and proving it works in my situation)

3) how do you handle the example of the unit closed ball?

Thank you, Lor

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I don't think any of these questions is 'MO-level' (see the FAQ), so I'm voting to close. (By the way, I think it's very unusual for the exact sequence you are considering to be a direct product.) –  HJRW Mar 26 '11 at 18:04
I think you are confusing two different definitions of the mapping class group. There is the group of homeomorphisms, up to isotopy, and then there is the group of homeomorphisms that fix the boundary pointwise, again up to isotopy. As an additional suggestion - suppose that $f$ is an orientation reversing involution of the surface $S$. Then you want to know if $fgf^{-1} = g$ for all classes $g$. But suppose that $g$ is a left Dehn twist... –  Sam Nead Mar 26 '11 at 18:23
For closed surfaces of positive genus this is never a direct product. On way to see this is because the action of $\mathbb{Z}/2$ on the Mumford--Morita--Miller class $\kappa_1 \in H^2(\mathcal{MCG}(\Sigma_g);\mathbb{Z})$ is by a sign, and this class is non-trivial. –  Oscar Randal-Williams Mar 26 '11 at 18:25