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

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Notes from talk 28 at NPCW01, Cambridge, January 13th 2017. A. Martin (UniversitÃ¤t 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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