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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

Ian Gershon Teixeira
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1 Answers1

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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 settle 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 $\text{SL}(2,\mathbb Z))$?
Ian Gershon Teixeira
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Lee Mosher
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  • clarifying question, I thought $ M_{1,0} $ was the mapping class group of the torus so isn't that $ SL(2,\mathbb{Z}) $? And I might be misunderstanding something but Theorem 11 of https://web.math.ucsb.edu/~shokrian/Mapping_Class_Group_Notes.pdf seems to be saying that $ M_{0,4} $ is a semidirect product of $ PSL(2,\mathbb{Z}) $ with the the finite abelian group $ 2^2 $? I have no idea about $ M_{1,1} $, the mapping class group of the once punctured torus, but I guess I'm also a bit surprised that it is exactly $ PSL(2,\mathbb{Z}) $ – Ian Gershon Teixeira Aug 03 '23 at 18:42
  • Ah yes, I am prone to making the "SL-PSL" mistake. And yes, I forgot that $\mathbb Z_2 \oplus \mathbb Z_2$. I'll fix those. The isomorphism between $M_{1,0}$ and $M_{1,1}$ is a consequence of the fact that the torus is a Lie group. – Lee Mosher Aug 03 '23 at 22:57
  • also I notice that you didn't mention boundary in this answer, is that because the mapping class group of a surface with boundary is the same as the mapping class group of the interior? Also what about $ M_{0,5} $? That is a sphere with 5 punctures that should be same as a disk with 4 punctures which should have MCG $ B_4 $ but that should have a $ B_3 $ quotient by https://mathoverflow.net/a/451204/387190 and thus a $ PSL(2,\mathbb{Z}) $ quotient and thus an action on the 3-valent tree. How does that fit in with your 2nd bullet point? – Ian Gershon Teixeira Aug 09 '23 at 19:57
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    The mapping class group of a surface with boundary (let's say a compact surface, for simplicity) has a free abelian subgroup generated by Dehn twists around the boundary. That abelian subgroup is normal, in fact it is the kernel of the natural homomorphism to the mapping class group of the interior. If you pass to finite index subgroups fixing boundary component and punctures, that abelian subgroup is actually the center. Summary: mapping class groups of bounded surfaces have special structure, relating them closely to mapping class groups of punctured surfaces. – Lee Mosher Aug 10 '23 at 00:25