CSP for MCGs

 In this post I record what I can remember from Henry Wilton's explanation of his conditional result that mapping class groups have CSP. I am very grateful to Henry for taking the time to explain this to me. All errors are naturally my own.

This post aims to discuss a theorem of Wilton that, assuming all hyperbolic groups are residually finite, mapping class groups of hyperbolic surfaces of finite type have the congruence subgroup property. Let me begin by explaining what CSP means.

Definition: Consider a subgroup $\Delta\leq\mathrm{Out}(G)$. Let $Q$ be a finite, characteristic quotient of $G$. The kernel of the natural map
\[
\Delta \to \mathrm{Out}(G)\to\mathrm{Out}(Q)
\]
is called a principal congruence subgroup of $\Delta$. A subgroup of $\Delta$ that contains a principal congruence subgroup is called a congruence subgroup, and a quotient of $\Delta$ by a normal congruence subgroup is called a congruence quotient. If every subgroup of finite index in $\Delta$ is congruence, then $\Delta$ is said to have the congruence subgroup property.

Note the slight abuse of terminology: whether or not $\Delta$ enjoys the congruence subgroup property depends on $G$.

Recall that
\[
\mathrm{Out}(\mathbb Z^2)\cong GL_2(\mathbb Z)\cong\mathrm{Out}(F_2)
\]
where $F_2$ is the free group of rank 2. This is the starting point of a well-known heuristic that there should be many similarities between mapping class groups, $\mathrm{Out}(F_n)$, and $\mathrm{GL}_n(\mathbb{Z})$.

Asada proved that $\mathrm{Out}(F_2)$ has the congruence subgroup property. On the other hand, it is a classical fact, known to Fricke and Klein, that $\mathrm{Out}(\mathbb Z^2)$ does not have the congruence subgroup property. 

This stands in contrast to the higher-rank case: Bass--Milnor--Serre proved that $GL_k(\mathbb{Z})\cong\mathrm{Out}(\mathbb Z^n)$, as well as many other higher rank arithmetic groups, does have the congruence subgroup property when $k\geq3$.

Motivated by the analogy, we may now ask whether mapping class groups have CSP. From now on, $\Sigma$ will be an orientable, hyperbolic surface of finite type and $\Sigma_*$ be the result of adding a single puncture. For a mapping class $\phi\in\mathrm{PMod}(\Sigma)$, a choice of preimage will be denoted by $\phi_*$. Note that $\phi_*$ is naturally an automorphism of $\pi_1(\Sigma)$, while $\phi$ is the corresponding outer automorphism.  

Mapping class groups act on a wide variety of rich geometric structures, so it is only natural that we try to consider some sort of action. Here is one choice that will be useful to us. A filling curve $\alpha \subset \Sigma$ is said to be asymmetric if its stabiliser in the mapping class group is the identity. The point is that for two mapping classes $\phi, \psi \in \mathrm{Mod}(\Sigma)$, they are equal if and only if $\phi(\alpha)=\psi(\alpha)$, so one can understand a mapping class just by its action on this specific curve.

I will now attempt to talk through Wilton's proof.

We are given a finite index subgroup $\Gamma$ of the mapping class group. Fix an asymmetric filling curve $\alpha \subset \Sigma $. Let $\phi_1=1,  \dots, \phi_n$ be a choice of coset representatives of $\Gamma$ in $\mathrm{Mod}(\Sigma)$ and define $\alpha_i \colon= \phi_i(\alpha)$, which are other filling curves that form orbit representatives for the action of $\Gamma$ on $\Sigma$.

Now fix a basepoint $*$, which we will also view as a puncture, and WLOG assume that it lies on $\alpha$. 
The Birman exact sequence is
\[
1\to\pi_1(\Sigma)\stackrel{\iota}{\to} \mathrm{PMod}(\Sigma_*)\to\mathrm{PMod}(\Sigma)\to 1
\] 
where $\iota$ is the point-pushing map and the forgetful map $\mathrm{PMod}(\Sigma_*)\to\mathrm{PMod}(\Sigma)$ forgets the puncture. Algebraically, these maps can be understood via the assertion that the natural diagram

commutes. When $\Sigma$ is hyperbolic, the vertical arrows are injective.

We may thus regard $\alpha$ as an element of the fundamental group, which is a subgroup of $\mathrm{Mod}(\Sigma_*)$ by the point-pushing map. Kra proved that for any filling curve $\alpha$ passing through the basepoint which isn't a proper power in $\pi_1(\Sigma)$, the image of $\alpha$ in $\mathrm{Mod}(\Sigma_*)$ is a pseudo-Anosov.

Let $\Gamma_*$ denote the preimage of $\Gamma$ in $\mathrm{Mod}(\Sigma_*)$. It turns out that the $\langle \alpha_i \rangle$ form a malnormal family in $\Gamma_*$ because $\alpha$ is an asymmetric curve.

At this point we need to produce a congruence subgroup contained in $\Gamma$, and this requires a bit of insight/creativity. 

Definition: If $H$ is a normal subgroup of finite index in $G$ then
\[
\Delta_H:=\{\phi\in\Delta \mid \phi(H)=H\}
\]
is called a Veech subgroup. (Note that, because $H$ is normal in $G$, $H$ is preserved by all inner automorphisms, so the assertion that $\phi(H)=H$ is independent of the choice of automorphism representing $\phi$.)

Every Veech subgroup is a congruence subgroup, and hence one can prove the congruence subgroup property by showing that every subgroup of finite index contains a Veech subgroup. It is also useful to notice that elements of Veech subgroups descend to outer automorphism groups of quotients. Specifically, let $H$ be a normal subgroup of finite index in $G$. Then $\phi\in\mathrm{Out}(G)$ descends to an outer automorphism $\bar{\phi}$ of $G/H$ if and only if $\phi$ is contained in the Veech subgroup $\Delta_H$.

A standard generalisation of the Dehn--Nielsen--Baer theorem asserts that the natural map $\mathrm{Mod}(\Sigma)\to\mathrm{Out}(\pi_1(\Sigma))$ is injective, so the congruence subgroup property for $\mathrm{Mod}(\Sigma)$ makes sense.

We will write $\mathrm{Mod}^\Lambda(\Sigma)$ for the Veech subgroup corresponding to a finite-sheeted normal covering space $\Lambda\to\Sigma$; it consists of the mapping classes in $\Sigma$ that lift to $\Lambda$. Moreover, it will be technically convenient to replace $\mathrm{Mod}(\Sigma)$ by $\mathrm{Mod}_0(\Sigma)$, a torsion-free subgroup of finite index in the pure mapping class group of $\Sigma$, and we will set
\[
\mathrm{Mod}_0^\Lambda(\Sigma):=\mathrm{Mod}_0(\Lambda)\cap\mathrm{Mod}^\Lambda(\Sigma)\,,
\]
the Veech subgroup of $\mathrm{Mod}_0(\Sigma)$ corresponding to $\Lambda$.

Veech subgroups have been introduced because we would like to show that our finite index subgroup $\Gamma$ contains a Veech subgroup. So which one?

Suppose we have some finite Galois cover $\Lambda \to \Sigma$ such that $\mathrm{Mod}^\Lambda(\Sigma)$ is our desired Veech subgroup of $\Gamma$. This is equivalent to saying that nothing in $\mathrm{Mod}^\Lambda(\Sigma)$ descends to an action on the finite group $\mathrm{Out}(\pi_1(\Sigma)/\pi_1(\Lambda))$. One way to ensure this would be to ask that the images of the $\alpha_i$ have distinct orders in $\pi_1(\Sigma)/\pi_1(\Lambda)$ (perhaps this is a good moment to reflect on this). It is also here that we use the fact that the set of cyclic subgroups generated by the $\alpha_i$ form a malnormal family, since otherwise the orders of the $\alpha_i$ could in principle be equal in every finite quotient.

So how can we ensure this? We only need some Veech subgroup, so it makes sense to start by fixing the hardest thing which needs to be fixed: finding this $\Lambda$ such that the orders in the quotient are distinct. This is where the residual finiteness of hyperbolic groups is needed, and we will return to what exactly goes on here afterwards, but for the moment let's assume the following: there is a finite group $Q$ and a morphism $f: \mathrm{Mod}(\Sigma_*) \to Q$ such that $f(\alpha_i)$ have distinct orders. 

 Let $\Lambda$ be the finite-sheeted normal covering space of $\Sigma$ corresponding to the kernel of $f\circ\iota$. It remains to prove that $\mathrm{Mod}^\Lambda_0(\Sigma)\subseteq\Gamma$.

To this end, consider $\psi\in\mathrm{Mod}^\Lambda_0(\Sigma)$ with a choice of preimage $\psi_*$ under the forgetful map. At the level of fundamental groups, this means that $\psi_*$ preserves $\pi_1(\Lambda)$, and so descends to an automorphism $\bar{\psi}_*$ of the finite group $\pi_1(\Sigma)/\pi_1(\Lambda)\cong f\circ\iota(\pi_1(\Sigma))\leq Q$. Therefore,

\[o(f\circ \iota(\psi_*(a)))=o(\bar{\psi}_*(f\circ\iota(a)))=o(f(\alpha_1))=K\, \tag{1}.\]

On the other hand, because the $\phi_i$ are coset representatives for $\Gamma_*$, $\psi_*=\gamma_*\phi_i$ for some $i$ and some $\gamma_*\in\Gamma_*$. The commutativity of the diagram from the Birman exact sequence enables us to compute the action of elements of $\phi_*\in\mathrm{PMod}(\Sigma_*)$ on the point-pushing subgroup $\iota(\pi_1(\Sigma))$. Specifically,\[\iota(\phi_*(g))=\phi_*\iota(g)\phi_*^{-1}\]for any $g\in \pi_1(\Sigma)$ and any $\phi_*\in\mathrm{PMod}(\Sigma_*)\leq\mathrm{Aut}(\pi_1(\Sigma))$. We obtain
\[
\iota(\psi_*(a))=\psi_*\alpha_*\psi_*^{-1}=\gamma_*\phi_i\alpha_*\phi_i^{-1}\gamma_*^{-1}=\gamma_*\alpha_i\gamma_*^{-1}
\]
, so

\[o(f\circ\iota(\psi_*(a)))=o(f(\gamma_*\alpha_i\gamma_*^{-1}))=o(f(\alpha_i))\, \tag{2}.\]

Putting equations 1 and 2 together, we get $o(f(\alpha_i))=K$ which implies, by the choice of $f$, that $i=1$. Since $\phi_1=1$ it follows that $\psi_*=\gamma_*\in\Gamma_*$, so $\psi\in\Gamma$. This proves that $\mathrm{Mod}^\Lambda_0(\Sigma)\subseteq\Gamma$.

In summary, $\Gamma$ contains a Veech subgroup, so $\Gamma$ is a congruence subgroup as required.

HHG stuff

It remains to explain how to ensure there is such a morphism to a finite group to begin with. Now the mapping class group isn't hyperbolic, but indulge me the following: how would you find such a map for a hyperbolic group, assuming all hyperbolic groups are residually finite?

The answer, as it often is for non-constructive questions in hyperbolic groups, is small cancellation. Add on large (distinct) powers of these words to the relations, so that one gets a small cancellation thing and the quotient is hyperbolic. The torsion theorem for small cancellation says that the images of the words we started with have the orders we want them to have, and then residual finiteness means we can find a finite quotient where the orders are distinct.

Although the mapping class group isn't hyperbolic, it is what is called hierarchically hyperbolic. I will not attempt to give a definition, but maybe the reader will get some sort of feeling about hierarchical hyperbolicity from the properties we use below. Instead, here's a picture of the mapping class group as a hierarchically hyperbolic space:

To find the finite quotient, one iterates the following procedure. Start at the lowest level, which is shaded gray in the picture. This is a hyperbolic group, so one finds a finite quotient as above. The long definition of hierarchical hyperbolicity means that one can do this procedure while maintaining the geometry of the levels above it, so that one obtains a hierarchically hyperbolic group/space in the quotient, except the quotient has one level less, somewhat akin to how small cancellation on a hyperbolic group produces another hyperbolic group. We iterate this procedure until there is only one level, consisting of a finite group.

The formal proof that this cartoon sketch actually works is due to Sisto. Incidentally, this shows that one only needs residual finiteness of the hyperbolic groups arising in this construction, and not all of them, but I certainly have no idea what the hyperbolic groups arising are. Philosophically, this isn't unprecedented. A similar thing happened in the proof of virtual Haken: it was shown to be true assuming all hyperbolic groups are residually finite, but it was sufficient to prove that hyperbolic and cubulated groups are residually finite, which Agol did. I also want to mention previous work on killing things in MCGs. Dahmani and later Dahmani--Guirardel--Osin figured out how to kill high powers of lots of things, but all the things they could kill lived in the top level of the hierarchy, whereas here what we wanted was to be able to kill things in the lowest level.