Rigidity and flexibility for handlebody groups
Our paradigm is Mostow rigidity: isomorphic cocompact lattices in must be conjugate.
Consider mapping class groups, say in genus to avoid technicalities.
Theorem 1 (Ivanov) Isomorphic finite index subgroups of MCG are conjugate. In fact, MCG is its own commensurator.
There are many variants, for instance involving self maps of Teichmüller space.
Our goal is to study where MCG is replaced with a (underestimated) subgroup of MCG.
2. Handlebody groups
Let be a handlebody of genus . We fix an identification of with a surface . Let be the subgroup of MCG of classes of homeos of which extend to homeos of . It is isomorphic to the mapping class group of as a 3-manifold.
All handlebody groups are conjugate.
2.2. Crash course
Dehn twists along meridians (simple closed curves on that bound a disk in ) belong to .
Dehn twists along simple closed curves that do not bound do not belong to . Indeed, it maps meridians to non meridians. This shows that is not normal and has infinite index.
In the Torelli exact sequence
is mapped to the stabilizer of a Lagrangian subspace in . Conversely, every symplectic matrix in stabilizing is in the image of (Hirose).
It follows that there exist elements of the Torelli group which are not conjugated into (Jorgensen).
Beware that Dehn twists around meridians do not generate . In fact they generate the kernel of the following sequence (Luft):
3. Rigidity result
Theorem 2 (Hensel) Let be a finite index subgroup. Any monomorphism is conjugate to the standard embedding .
Corollary 3 is its own commensurator.
Improves on Korhmaz-Schleimer: .
4. Flexibility result
Super-rigidity seems hard. For instance, one does not know if a finite index subgroup of MCG can surject onto .
Theorem 4 (Hensel) For every , there exists , a finite index subgroup and a monomorphism which is not conjugate into .
These examples come from covers.
5. Proof of Ivanov’s theorem
Recall the scheme of proof of Mostow rigidity. Think of the given isomorphism as a coarse geometric data: a quasi-isometry of hyperbolic space. This induces a boundary homeomorphism. Find extra symmetry (quasiconformality), show homeo is actually conformal. This is a candidate for extension into a hyperbolic isometry.
In our case, the coarse object is a simplicial automorphism of the curve graph. The extra symmetry. Show that such automorphisms are induced by elements of MCG.
5.1. Characterization of Dehn twists
Dehn twists have large centralizers, the centers of centralizers are small, so large (resp. small) that this characterizes powers of Dehn twists.
Lemma 5 (Ivanov) Isomorphisms between finite index subgroups of MCG map powers of Dehn twists to powers of Dehn twists.
It follows that an isomorphism defines an assignment curves to curves.
5.2. The curve graph
Its vertices correspond to isotopy classes of essential scimple closed curves. Edges correspond to disjointness (up to isotopy).
Since disjointness corresponds to commutation of Dehn twist, an isomorphism defines a simplicial map on the curve graph .
Theorem 6 .
It follows that isomorphic is a conjugation.
6. Case of handlebody group
Something of centralizers and centers of centralizers remains.
Lemma 7 Monomorphisms between finite index subgroups of MCG map powers of Dehn twists around meridians to powers of Dehn twists.
Thus monomorphism induces an assignment meridians to curves, hence an superinjection of the disk graph to .
Proposition 8 Any superinjection is conjugate to the standard embedding.
This uses the fact that superinjections of to itself are surjective.
This might seem to give us a control on the subgroup only. No, elements of are indeed determined by their effect on meridians.
Unknown: does Torelli group have non-standard embeddings into MCG ?
7. Flexible examples
Take a meridian on , take a 3-fold cover. This extends to a branched cover between handlebodies, branched over a disk.