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.

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, {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)<Out(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.


About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
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