JSJ V: Fioravanti's JSJ decomposition space

 Today's post is based on a talk given by Melvin Weiss. We will briefly discuss the JSJ decompositions constructed by Fioravanti. Due to the technical length of this we will be rather brief.

Let us recall the setup of classical JSJ theory for one-ended hyperbolic groups. We have the theory of $(\mathcal{A}, \mathcal{H})$-splittings where vertex groups are either rigid or quadratically hanging. When we take $\mathcal{A}$ to be the set of cyclic subgroups we can choose the splitting to be canonical, i.e. invariant under the Out(G) action in the following sense: for each automorphism $\varphi\in\mathrm{Aut}(G)$, there exists $\Phi\in\mathrm{Aut}(T)$ such that $\Phi(gx)=\varphi(g)\Phi(x)$ for all $g\in G$ and $x\in T$.

This is however unsatisfactory from the viewpoint of understanding automorphisms of virtually compact special groups. Fioravanti gives the following two reasons:

Canonical JSJs are available for more general groups $G$ (the canonical tree of cylinders), but always under the assumption that abelian subgroups of $G$ do not interact with each other in complicated ways (the typical example is that of a relatively hyperbolic group). By contrast, many special groups (e.g. freely indecomposable RAAGs) are thick in the sense of Behrstock-Drutu-Mosher, which is a strong negation of the kind of property needed for canonical trees of cylinders to work.

Cyclic splittings of hyperbolic groups encode all their automorphisms: recall that if $G$ is a torsion-free, $1$--ended hyperbolic group that does not split over $\mathbb{Z}$ as an amalgamated product or HNN extension, then $Out(G)$ is finite. This property again badly fails for special groups: there are right-angled Artin groups $A_{\Gamma}$ that do not split over any abelian subgroups, but nevertheless have infinite $\mathrm{Out}(A_{\Gamma})$.

Hence we need a stronger theory. For a virtually compact special group $G$, define $\mathcal{ZZ}(G)$ to be the union of cyclic subgroups and centralisers, and let $\mathcal{H}$ be a family of subgroups containing all singular subgroups, i.e. maximal subgroups virtually splitting as direct products of infinite groups.
Fioravanti constructs a  $(\mathcal{ZZ}(G),\mathcal{H}(G))$ splitting where the vertex groups are $(\mathcal{Z}(G), \mathcal{H})$ rigid or optimal QH, which means that the vertex group is isomorphic to $\pi_1(\Sigma)$ for some surface which isn't a pair of pants such that each incident edge group is either trivial or $Q-$conjugate to the entire fundamental group of a connected component of the boundary, plus an extra technical condition which we omit. The splitting is again canonical.

The reason for making all the singular subgroups elliptic is because in the proof we need to use the canonicity of the JSJ space in the classical sense, and the first point tells us that JSJ decompositions are bad at dealing with large amounts of abelianness. 

At some point we will need to restrict actions, so let us say what we mean now:

If a subgroup $\mathcal{O}\leq\mathrm{Out}(G)$ preserves the $G$--conjugacy class of a subgroup $H\leq G$, we can define a restriction $\mathcal{O}|_H\leq\mathrm{Out}(H)$: one considers the subgroup $U\leq\mathrm{Aut}(G)$ formed by automorphisms $\varphi$ with $\varphi(H)=H$ and outer class in $\mathcal{O}$, then one defines $\mathcal{O}|_H$ as the image of the composition $U\rightarrow \mathrm{Aut}(H)\rightarrow \mathrm{Out}(H)$ given by first restricting to $H$ and then projecting to outer automorphisms. The conjugation action $N_G(H)\curvearrowright H$ also determines a subgroup $C^G_H\leq\mathrm{Out}(H)$, and we always have $C^G_H\lhd\mathcal{O}|_H$. For each $\phi\in\mathcal{O}$, the possible restrictions of $\phi$ to $H$ form a canonical coset $\phi|_H\cdot C^G_H$ within $\mathcal{O}|_H$.

The precise statement is the following:

Theorem 1: Let $\mathcal{O}\leq\mathrm{Out}(G)$ be a subgroup, and let $G$ be $1$--ended relative to an $\mathcal{O}$--invariant collection $\mathcal{H}$ that contains $\mathcal{S}(G)$. Then there exists an $\mathcal{O}$--invariant $(\mathcal{ZZ}(G),\mathcal{H})$--tree $G\curvearrowright T$ with the following properties.

1. The $G$--stabiliser of each vertex of $T$ is of one of two kinds: (a) an optimal quadratically hanging subgroup relative to $\mathcal{H}$ (in fact we can assume these are convex cocompact and root closed), or (b) a convex-cocompact root-closed subgroup of $G$ that is $(\mathcal{Z}(G),\mathcal{H})$--rigid in G.
2. Each edge $e\subseteq T$ with $G$--stabiliser not in $\mathcal{Z}(G)$ is incident to a type (a) vertex.
3. If the family $\mathcal{H}$ contains all cyclic subgroups of $G$ whose conjugacy class has finite $\mathcal{O}$--orbit, then each type (b) vertex group $V$ is $(\mathcal{Z}(V),\mathcal{H}|_V)$--rigid in itself.

The proof strategy is as follows:

1. Split along centralisers, then the cyclic subgroups, $\mathcal{O}$ invariantly as a $(\mathcal{Z}(G),\mathcal{H})$-tree with only $(\mathcal{Z}(G),\mathcal{H})$-rigid vertex groups,
2. For all vertex groups V,

• apply classical JSJ tree of cylinders to get a $O|_V$-invariant $(\mathcal{C}(V, \mathcal{H}|_V)$ splitting $V \curvearrowright S$ of $(\mathcal{C}(V, \mathcal{H}|_V)$-rigid or QH vertex groups.
• Modify $S$ near the QH vertices to ensure optimality and modify further to ensure convex cocompact and root closed (this latter part is just fiddling and not particularly deep).
• Blow up $T$ at each $V$ by the corresponding splitting of $V$ to get a refinement $\pi:T' \to T$ such that $\pi^{-1}(V)=S$ and each other vertex has preimage a point. This procedure can be done $\mathcal{O}-$invariantly if $S$ was $O|_V$ invariant. The local splittings being invariant can be bootstrapped to the whole splitting being globally invariant.

Denote by $\mathcal{Z}_S(G)$ the non-cyclic centralisers of $G$. The main step in proving theorem 1 is the following:

 Theorem 2: Under the assumptions of theorem 1, there exists an $\mathcal{O}-$invariant $(\mathcal{Z}_S(G), \mathcal{H})-$splitting of $G$ with only $(\mathcal{Z}_S(G), \mathcal{H})-$rigid vertex groups.

 The general procedure to show theorem 2 is the following: given $\mathcal{O}-$invariant collections $\mathcal{F, K}$ of subgroups of $G$ consider a $(\mathcal{F}, \mathcal{F \cup K}$) tree with $G \curvearrowright T$ unconditionally accessible. One can probably forget about the word unconditionally on a first reading, but for completeness here is a definition:
 A splitting $G \curvearrowright T$ is irredundant if $T$ has no degree-2 vertices, except for those where the vertex-stabiliser swaps the two incident edges; in other words, T was not obtained from a smaller splitting of G simply by subdividing an edge. Reduced splittings are irredundant, but the converse does not hold. $G$ is accessible over a family of subgroups $\mathcal{F}$ if there exists a number $N(G)$ such that any irredundant splitting of $G$ over $\mathcal{F}$ has at most $N(G)$ orbits of edges.
 The accessibility result from the previous post can be upgraded to unconditional accessibility.
 Define $\mathcal{F}_{int}$ to be the set of all subgroups which arise as intersections of elements of $\mathcal{F}$. We need the following facts:

1. We can refine $T$ to an $(\mathcal{F_{int}}, \mathcal{K})$ tree $T'$ with $H$ elliptic in $T'$ iff the whole $\mathcal{O}$ orbit of $H$ is elliptic w.r.t $T$. 
2. Under suitable additional assumptions, we can replace $T'$ by $T''$ with the same elliptic subgroups as $T$ such that $T''$ is $\mathcal{O}-$minimal.

 For special groups: consider a collection of $\mathcal{O}$-invariant, minimal irredundant $(\mathcal{Z}_S(G), \mathcal{H})-$splittings. Let $T$ be maximal in that collection. Suppose that some vertex group $V$ is non-rigid.
 This would imply that there is some $(\mathcal{Z}_S(G), \mathcal{H})-$tree in which $V$ isn't elliptic. $V \curvearrowright M$ some minimal subtree.
 Use facts 1,2 to refine $V \curvearrowright M$ into an $O|_V$ invariant $(\mathcal{Z}_S(G)|_V, \mathcal{H}|_V)-$ splitting. blow up $T$ by this splitting of $V.$ This gives us a new $\mathcal{O}$ invariant splitting because the splitting of $V$ is invariant. This gives a larger splitting than $T$, which is a contradiction.