Gabriel Pallier’s notes of Alexandre Martin’s Cambridge lecture 13-01-2017



Notes from talk 28 at NPCW01, Cambridge, January 13th 2017. A. Martin (Universität Wien), joint work with A. Genevois.

\paragraph{Reminders about graph products} Let {\Gamma} be a finite simplicial graph. For every {v \in V \Gamma}, let {G_v} be a group, and form the graph product

\displaystyle  G = \frac{\ast_{v \in V \Gamma} G_v}{\left\langle \left\langle \left\{ [g,g'] : g \in G_v, g' \in G_{v'}, (v,v') \in E \Gamma \right\} \right\rangle \right\rangle},

where {\langle \langle - \rangle \rangle} denotes taking the normal closure.

When {\Gamma} is discrete, this is a free product ; when {\Gamma} is complete, a direct product.

If for all {v \in V\Gamma}, {G_v = \mathbb{Z}} (resp. {G_v = \mathbb{Z}_2}) then {G} is a right angled Artin (resp. Coxeter) group.

\paragraph{Problem} Understand the structure and geometry of {\mathrm{Aut}(\Gamma)}. Today’s “baby” case : {\Gamma} is a long cycle (with length { n \geqslant 9}). The strategy is to construct a curve complex {Y}, and an adequate action {G \curvearrowright Y} such that {\mathrm{Aut}(G)} also acts on {Y}.

Recall that the Davis complex of a graph product is defined as follows:

  • {vertices are cosets {g G_{\Gamma'}} where {\Gamma'} is a clique of {\Gamma} and {G_{\Gamma'}} denotes the graph product over {\Gamma'}.}
  • {there is an edge from {g_{\Gamma''}} to {g_{\Gamma'}} whenever {\Gamma''} is a subclique of {\Gamma'} and {\vert \Gamma' \setminus \Gamma'' \vert = 1}.}
  • {the cubes in the graph are filled.}

This yields a {\mathrm{CAT}(0)} cube complex.

In our case, the Davis complex is the cubical subdivision of an {n}-gonal complex {X}. The stabilizers are as follows:

  • {Stabilizers of vertices are conjugated to {G_{v_i, v_{i+1}}}.}
  • {Stabilizers of edges are conjugated to {G_{v_i}}‘s.}
  • {Stabilizers of the whole polygon is trivial}

Strict fundamental domains are single polygons. Define a new complex {\Delta_X} from {X}, as follows: start form a graph whose vertices represent the wall-trees, and edges intersections of wall-trees. Then fill the embedded {n}-cycles.

{\Delta_X} is a {C'(1/4) -T(4)} complex.

Recall that {C'(1/4)} is the small cancellation condition whereas {T(4)} means that the links of vertices have girth at least {4}.

{G \curvearrowright \Delta_X} is weakly acylindrical. More precisely, write {v_T} for the vertex of {\Delta_X} induced by a wall-tree {T} of {X}; then, for any pait of vertices {v_T}, {v_{T'}} in {\Delta_X}

  • {If {d(v_T, v_{T'}) \geqslant 3} then {\mathrm{Stab}_{\Delta_X} (v_T) \cap \mathrm{Stab}_{\Delta_X}(v_{T'}) = \lbrace 1 \rbrace}.}
  • {If {d(v_T, v_{T'}) = 2 } then {\mathrm{Stab}_{\Delta_X} (v_T) \cap \mathrm{Stab}_{\Delta_X}(v_{T'})} is a subgroup in a conjugate of a {G_{v_i}} (Observe that those are not self-normalizing).}
  • {If {d(v_T, v_{T'}) = 1 } then {\mathrm{Stab}_{\Delta_X} (v_T) \cap \mathrm{Stab}_{\Delta_X}(v_{T'})} is conjugated to a {G_{v_i, v_{i+1}}} (Those are self-normalizing).}
  • {If {d(v_T, v_{T'}) = 0 } then {\mathrm{Stab}_{\Delta_X} (v_T) = \mathrm{Stab}_{\Delta_X}(v_{T'})} and are conjugated to a {G_{v_{i-1}, v_i, v_{i+1}}}.}

There exists an action {\mathrm{Aut}(G) \curvearrowright \Delta_X}.

Proof: First make {G} act on {\Delta_X^{(0)}} ; then observe by the weak acylindricity property that the edges are preserved (via a discussion on the distance between the images of a pair of adjacent vertices). \Box

The automorphism group of {G} admits the following decomposition:

\displaystyle  \mathrm{Aut}(G) = \mathrm{Inn}(G) \rtimes \mathrm{Out}(G) \simeq G \rtimes \left( \prod_{v \in V \Gamma} \mathrm{Aut}(G_v) \rtimes \underline{\mathrm{Aut}}(\Gamma) \right).

(Beware: Here {\underline{\mathrm{Aut}}(\Gamma)} denotes the automorphism group of {\Gamma} labelled with the isomorphism classes of the {G_v}‘s.)

Assume that all the {G_v} are finitely generated. Then {\mathrm{Aut}(G)} is an acylindrical group.

Assume {H \curvearrowright Y}, where {Y} is a finite-dimensional and irreducible {\mathrm{CAT}(0)} cube complex. Assume further that this action is essential, without fixed points at {\infty}, and that one can find hyperplanes {\widehat{ h_1}}, {\widehat{h_2} \subset Y} such that {\mathrm{Stab}(\widehat{h_1}) \cap \widehat{h_2}} is finite. Then {H} is either virtually cyclic or acylindrically hyperbolic.



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