Notes of Karen Vogtmann’s Cambridge lecture 13-04-2017

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 ${\Gamma}$ does not contain a length 3 path or a 4-cycle as a full subgraph, then every finitely generated subgroup of the RAAG ${A_\Gamma}$ is a RAAG.

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, ${Out(F_n)}$ is not linear (Formanek-Procesi), whereas ${Out({\mathbb Z}^n)}$ is. ${Out(F_n)}$ has property FA (fixed points on trees, in fact, ${{\mathbb R}}$-trees) (Culler-Vogtmann). However, a finite index of ${Out(F_3)}$ acts on a tree (it maps onto ${Sl(2,{\mathbb Z})}$, Lubotzky), so it is not Kazhdan.

Question. Does ${Out(F_n)}$, ${n\geq 3}$, have property (T) ?

Here is Lubotzky’s argument. Consider the morphism of ${F_3}$ to ${{\mathbb Z}/2}$ kiling two generators, let ${K}$ be its kernel. Consider the subgroup ${Aut(F_3,K)}$ sending ${K}$ to ${K}$. This has finite index in ${Aut(F_3)}$. ${{\mathbb Z}/2}$ acts on the double cover of the bouquet of 3 circles that correspinds to ${K}$. This graph has connectivity 5. So ${Aut(F_3,K)}$ acts on ${{\mathbb Z}^5}$, and commutes with the ${{\mathbb Z}/2}$. The ${{\mathbb Z}/2}$-action has a ${{\mathbb Z}^2}$ eigenspace, whence an action of ${Aut(F_3,K)}$ on ${{\mathbb Z}^2}$.

Grunewald-Lubotzky generalized this. Given a finite quotient ${Q}$ of ${F_n}$, with finite index kernel ${K}$, the centralizer ${Aut^Q(F_n,K)}$ in the action on the homology of covering graph preserves eigenspaces.

Question. Analogously, can you find interesting representations of ${Aut(A_\Gamma)}$?

The answer depends on the structure of graph ${\Gamma}$. There is a partial ordering on (equivalence classes) vertices: say ${v\leq w}$ if ${lk(v)\subset st(w)}$. Say ${v\sim w}$ if ${v\leq w}$ and ${w\leq v}$. 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 ${Out^0(A_\Gamma) and epimorphisms

$\displaystyle \begin{array}{rcl} E_v:Out^0(A_\Gamma)\rightarrow Out^0(A_{\Gamma\setminus[v]}),\quad R_v:Out^0(A_\Gamma)\rightarrow Out^0(A_{[v]}) \end{array}$

if ${[v]}$ is maximal. If ${[v]}$ is not maximal, need to compose ${E_w}$ with one last ${R_v}$.

So, if ${[v]}$ is abelian or if ${|[v]|\leq 3}$, 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 ${\Gamma_0}$ whose vertices are equivalence classes of vertices of ${\Gamma}$. It embeds in ${\Gamma}$.

Let ${\pi:A_\Gamma\rightarrow Q}$ finite be an epimorphism, ${K=\mathrm{ker}(\pi)}$. Let ${d_\pi}$ be the maximal distance in ${\Gamma_0}$ between equivalence classes which are mapped to nontrivial elements of ${Q}$.

${d_\pi=0}$ means that everything is mapped to 1 but elements of one single equivalence class. In this case, ${K}$ is a RAAG. Indeed, use the Salvetti complex ${S_\Gamma}$ (union of tori). ${\pi}$ descends to ${A_{[v]}}$. The corresponding covering merely opens the bouquet of circles corresponding to ${A_{[v]}}$ into a graph.

Let us assume that ${d_\pi=1}$ and ${\Gamma}$ is a tree. Then ${\Gamma_0}$ is a tree. Let ${v}$ and ${w}$ be two vertices representing distinct classes, not mapped to 1.

Proposition 2 If ${\langle \pi v\rangle \cap \langle \pi w\rangle=\{1\}}$, then ${K}$ is a RAAG. Otherwise

• either ${v}$ or ${w}$ is a leaf. Then ${K}$ is a RAAG.
• Otherwise, ${K}$ is not a RAAG.

On can see that ${A_\Gamma}$ is not always coherent.