** RAAG subgroups of RAAGs **

Joint work with Beatrice Pozzetti.

RAAGs have plenty of interesting subgroups. Weird Bestvina-Brady beasts. Every 3-manifold group virtually embeds in a RAAG.

Carl Droms proved that if graph does not contain a length 3 path or a 4-cycle as a full subgraph, then every finitely generated subgroup of the RAAG is a RAAG.

What about other RAAGs?

**1. Motivation **

I am interested in (outer) automorphism groups of RAAGs. Among interesting examples are free and free abelian groups. They are dratsically different. For instance, is not linear (Formanek-Procesi), whereas is. has property FA (fixed points on trees, in fact, -trees) (Culler-Vogtmann). However, a finite index of acts on a tree (it maps onto , Lubotzky), so it is not Kazhdan.

**Question**. Does , , have property (T) ?

Here is Lubotzky’s argument. Consider the morphism of to kiling two generators, let be its kernel. Consider the subgroup sending to . This has finite index in . acts on the double cover of the bouquet of 3 circles that correspinds to . This graph has connectivity 5. So acts on , and commutes with the . The -action has a eigenspace, whence an action of on .

Grunewald-Lubotzky generalized this. Given a finite quotient of , with finite index kernel , the centralizer in the action on the homology of covering graph preserves eigenspaces.

**Question**. Analogously, can you find interesting representations of ?

The answer depends on the structure of graph . There is a partial ordering on (equivalence classes) vertices: say if . Say if and . Equivalence classes split among abelian and non-abelian classes (depending wether they commute or not).

Proposition 1 (Charney-Vogtmann)There is a finite index subgroup and epimorphisms

if is maximal. If is not maximal, need to compose with one last .

So, if is abelian or if , we get actions on trees.

Grunewald-Lubotzky’s argument uses induction based on the fact that subgroups of free groups are free. This raises the question wether finite index normal subgroups of RAAGs are RAAGs.

**2. Diameter of a finite quotient **

Build a graph whose vertices are equivalence classes of vertices of . It embeds in .

Let finite be an epimorphism, . Let be the maximal distance in between equivalence classes which are mapped to nontrivial elements of .

means that everything is mapped to 1 but elements of one single equivalence class. In this case, is a RAAG. Indeed, use the Salvetti complex (union of tori). descends to . The corresponding covering merely opens the bouquet of circles corresponding to into a graph.

Let us assume that and is a tree. Then is a tree. Let and be two vertices representing distinct classes, not mapped to 1.

Proposition 2If , then is a RAAG. Otherwise

- either or is a leaf. Then is a RAAG.
- Otherwise, is not a RAAG.

On can see that is not always coherent.