Mapping class group of surfaces, free products, and trees

Question

Let $ \Sigma $ be a surface, possibly with boundary. Let $ MCG(\Sigma) $ denote the mapping class group. Is it true that $ MCG(\Sigma) $ has a quotient which is a nontrivial free product $ A \ast B $ if and only if that quotient is
$
A \ast B \cong C_2 \ast C_3 \cong PSL(2,\mathbb{Z})
$?
Correspondingly, $ MCG(\Sigma) $ has a transitive action on a regular tree if and only if that tree is $ 3 $-regular?

Moreover, is it true that that the only surfaces for which these (equivalent) properties occur are $ MCG(\Sigma)=SL(2,\mathbb{Z}),B_3,B_4 $ corresponding to the torus, the disk with 3 punctures and the disk with 4 punctures, respectively?

I think this follows from work of Culler and Vogtmann, sections "Mapping class groups of surfaces of positive genus" and "Braid groups and mapping class groups of punctured spheres." But I'm not well-versed enough in this area to feel confident.

For related answers see Are any quotients of braid groups non-trivial free products? and https://mathoverflow.net/questions/451191/action-of-braid-groups-on-regular-trees/451204#451204

Answer 1

This will be a partial answer, leaving one case unresolved. The answer is obtained by applying the results of the Culler--Vogtmann paper linked in your post. That paper focusses on the $F\mathbb R$ property of a group $G$, which says that every action of $G$ on an $\mathbb R$-tree is trivial (meaning that there is a point fixed by the whole group); this applies to actions on simplicial trees, which are a special class of $\mathbb R$-trees.

Let $M_{g,r}$ denote the mapping class group of an oriented surface of genus $g$ with $r$ punctures.

In Section 3 of the Culler Vogtmann paper you'll find an extensive list of examples of groups to which their main theorem applies, showing that each of these groups satisfies property $F\mathbb R$. Flipping to page 682 of the Culler Vogtmann paper, you'll see that this list includes:

• $M_{g,r}$ for $g \ge 2$ and $r \ge 0$
• $M_{0,r}$ for $r \ne 4$, $r \ge 0$ (for some reason that paper switches to the notation $M^r_0$ instead of $M_{0,r}$).

That settles those cases.

Regarding the three mapping class groups $M_{0,4}$ and $M_{1,0}$ and $M_{1,1}$, each has a homomorphism onto $\text{PSL}(2,\mathbb Z)$ with very small kernel. The group $\text{PSL}(2,\mathbb Z) \approx \mathbb Z / 2 \mathbb Z * \mathbb Z / 3 \mathbb Z$ has its standard Bass-Serre tree action on the 3-valent tree. It's not hard to prove that this is, essentially, the only tree action of $\mathbb Z / 2 \mathbb Z * \mathbb Z / 3 \mathbb Z$, and with a bit more thought regarding the behavior of kernels, this should settls your question for these three mapping class groups, because if a group $G$ has a finite normal subgroup $N$ then $N$ is contained in the kernel of any action of $G$ on a tree.

To be a bit more specific: $M_{1,0}$ and $M_{1,1}$ are both isomorphic to $\text{SL}(2,\mathbb Z)$ and hence both have a homomorphism onto $\text{PSL}(2,\mathbb Z)$ with kernel cyclic of order $2$. Also, $M_{0,4}$ has a split homomorphism onto $\text{PSL}(2,\mathbb Z)$ with kernel isomorphic to $\mathbb Z/2\mathbb Z \oplus \mathbb Z / 2 \mathbb Z$.

So, what's left are the groups $M_{1,r}$ for $r \ge 2$. Each of these groups has a homomorphism onto $M_{1,0} \approx \text{PSL}(2,\mathbb Z)$, by "filling in the punctures", and thus we get an action on the trivalent tree by composing this homomorphism with the standard action of $\text{PSL}(2,\mathbb Z)$.

So, your question about transitive actions remains unresolved in the case of $M_{1,r}$ for $g \ge 2$.

If I may sharpen your question just a bit, we get the following:

• Does every minimal action of $M_{1,r}$ ($r \ge 2$) on an $\mathbb R$- tree factor through the standard Bass-Serre tree action of $M_{1,0} \approx \text{SL}(2,\mathbb Z)$, via the canonical "puncture forgetting" epimorphism $M_{1,r} \mapsto M_{1,0} \approx \text{SL}(2,\mathbb Z)$ (up to automorphisms of $\text{SL}(2,\mathbb Z))$?