\documentclass[JEP,XML,SOM,Unicode,NoFloatCountersInSection,published]{cedram} \datereceived{2020-10-15} \dateaccepted{2022-05-15} \dateepreuves{2022-06-03} \usepackage{mathrsfs} \let\mathcal\mathscr \usepackage{tikz} \theoremstyle{plain} \newtheorem{theo}{Theorem}[section] \newtheorem{prop}[theo]{Proposition} \newtheorem{lemma}[theo]{Lemma} \newtheorem{cor}[theo]{Corollary} \newtheorem{theointro}{Theorem} \newtheorem{corintro}[theointro]{Corollary} \newtheorem{propintro}[theointro]{Proposition} \theoremstyle{definition} \newtheorem{de}[theo]{Definition} \newtheorem{rk}[theo]{Remark} \newtheorem*{rk*}{Remark} \let\actson\curvearrowright \let\dunion\sqcup \let\Dunion\bigsqcup \newcommand{\cala}{\mathcal{A}} \newcommand{\calf}{\mathcal{F}} \newcommand{\calg}{\mathcal{G}} \newcommand{\calh}{\mathcal{H}} \newcommand{\calk}{\mathcal{K}} \newcommand{\calp}{\mathcal{P}} \newcommand{\calz}{\mathcal{Z}} \newcommand{\Aut}{\mathrm{Aut}} \newcommand{\lk}{\mathrm{lk}} \newcommand{\Out}{\mathrm{Out}} \newcommand{\Prob}{\mathrm{Prob}} \newcommand{\reg}{\mathrm{reg}} \newcommand{\Stab}{\operatorname{Stab}} \newcommand{\st}{\mathrm{st}} \datepublished{2022-06-09} \begin{document} \frontmatter \title{Measure equivalence classification of transvection-free right-angled Artin groups} \author[\initial{C.} \lastname{Horbez}]{\firstname{Camille} \lastname{Horbez}} \address{Université Paris-Saclay, CNRS, Laboratoire de mathématiques d'Orsay\\ 91405, Orsay, France} \email{camille.horbez@universite-paris-saclay.fr} \urladdr{https://www.imo.universite-paris-saclay.fr/~horbez/} \author[\initial{J.} \lastname{Huang}]{\firstname{Jingyin} \lastname{Huang}} \address{Department of Mathematics, The Ohio State University\\ 100 Math Tower, 231 W 18th Ave, Columbus, OH 43210, USA} \email{huang.929@osu.edu} \urladdr{https://sites.google.com/site/huangmath0/} \thanks{C.H.\ acknowledges support from the Agence Nationale de la Recherche under Grant ANR-16-CE40-0006 DAGGER} \begin{abstract} We prove that if two transvection-free right-angled Artin groups are measure equivalent, then they have isomorphic extension graphs. As a consequence, two right-angled Artin groups with finite outer automorphism groups are measure equivalent if and only if they are isomorphic. This matches the quasi-isometry classification. However, in contrast with the quasi-isometry question, we observe that no right-angled Artin group is superrigid for measure equivalence in the strongest possible sense, for two reasons. First, a right-angled Artin group~$G$ is always measure equivalent to any graph product of infinite countable amenable groups over the same defining graph. Second, when $G$ is nonabelian, the automorphism group of the universal cover of the Salvetti complex of $G$ always contains infinitely generated (non-uniform) lattices. \end{abstract} \subjclass{20F36, 20F65, 37A20, 46L36} \keywords{Right-angled Artin groups, measure equivalence} \altkeywords{Groupes d'Artin à angles droits, équivalence mesurée} \alttitle{Classification des groupes d'Artin à angles droits sans transvections pour l'équivalence mesurée} \begin{altabstract} Nous démontrons que si deux groupes d'Artin à angles droits sans transvections sont mesu\-rablement équivalents, alors ils ont des graphes d'extension isomorphes. En conséquence, deux groupes d'Artin à angles droits ayant des groupes d'automorphismes extérieurs finis sont mesurablement équivalents si et seulement s'ils sont isomorphes. Ceci coïncide avec la classification pour la quasi-isométrie. Par contre, contrairement au cas de la quasi-isométrie, un groupe d'Artin à angles droits ne peut jamais être super-rigide pour l'équivalence mesurée, pour deux raisons. D'abord, un groupe d'Artin à angles droits $G$ est toujours mesurablement équivalent à tout produit graphé de groupes moyennables infinis dénombrables sur le même graphe sous-jacent. Ensuite, lorsque $G$ est non abélien, le groupe d'automorphismes du revêtement universel du complexe de Salvetti de $G$ contient toujours des réseaux (non uniformes) qui ne sont pas de type fini. \end{altabstract} \maketitle \vspace*{-\baselineskip} \tableofcontents \mainmatter \section*{Introduction} Measured group theory studies groups through their ergodic actions on standard probability spaces. A central quest is to classify groups up to \emph{measure equivalence}, a notion introduced by Gromov \cite{Gro} as a measure-theoretic analogue of that of quasi-isometry between finitely generated groups. The definition is as follows: two countable groups $G_1$ and $G_2$ are \emph{measure equivalent} if there exists a standard measure space $\Sigma$ equipped with a measure-preserving action of $G_1\times G_2$ by Borel automorphisms, such that for every $i\in\{1,2\}$, the $G_i$-action on~$\Sigma$ is essentially free and has a finite measure fundamental domain. A typical example is that two lattices in the same locally compact second countable group are always measure equivalent. A first striking result in the theory, due to Ornstein and Weiss \cite{OW} (building on previous work of Dye \cite{Dye1,Dye2}) is that any two countably infinite amenable groups are measure equivalent. At the other extreme, strong rigidity results have been established from the viewpoint of measure equivalence, notably for lattices in higher rank Lie groups (Furman \cite{Fur}, building on earlier work of Zimmer \cite{Zim1,Zim2}) and for mapping class groups of finite-type surfaces (Kida \cite{Kid}). In \cite{MS}, Monod and Shalom used bounded cohomology techniques to obtain rigidity results for products of certain negatively curved groups. Other developments include superrigidity results for certain amalgamated free products \cite{Kid2}, for other groups related to mapping class groups \cite{CK}, or classification results within certain classes of groups, like Baumslag--Solitar groups \cite{kida2014invariants,HR}. See also \cite{Sha-survey,Gab-survey,Fur-survey} for general surveys in measured group theory. In previous work \cite{HH}, we obtained measure equivalence superrigidity results for certain classes of two-dimensional Artin groups of hyperbolic type. In the present paper, we focus on right-angled Artin groups, which have a very simple definition and played a prominent role in geometric group theory recently. We obtain a complete measure equivalence classification of right-angled Artin groups with finite outer automorphism group. One motivation for studying these groups from the viewpoint of measured group theory is that they usually tend to be significantly less rigid than certain other classes of Artin groups, as will be evidenced in Corollary~\ref{corintro:graph-products} and Theorem~\ref{theointro-non-superrigid} below. With this in mind, one hopes to explore the subtle line between rigidity and flexibility phenomena in this setting. \subsubsection*{Main classification theorem.} We recall that given a finite simple graph $\Gamma$ (\ie with no edge loops and no multiple edges between vertices), the right-angled Artin group~$G_\Gamma$ is the group defined by the following presentation: the generators are the vertices of $\Gamma$, and two generators commute if and only if the corresponding vertices of $\Gamma$ are joined by an edge. Our main theorem is the following. \begin{theointro}\label{theointro:1} Let $G_1$ and $G_2$ be two right-angled Artin groups such that $\Out(G_1)$ and $\Out(G_2)$ are finite. Then $G_1$ and $G_2$ are measure equivalent if and only if they are isomorphic. \end{theointro} Finiteness of $\Out(G_\Gamma)$ can easily be checked on the defining graph $\Gamma$; by work of Laurence \cite{laurence1995generating} and Servatius \cite{Ser}, this amounts to the two conditions below. For a vertex $v\in\Gamma$, the star of $v$, denoted by $\st(v)$, is the subgraph made of all vertices which are either equal or adjacent to $v$, together with edges of $\Gamma$ between these vertices. \begin{enumerate} \item $\Gamma$ does not contain two distinct vertices $v$ and $w$ such that the link of $v$ is contained in the star of $w$ -- this prevents the existence of transvections $v\mapsto vw$ in $\Aut(G_\Gamma)$ (we then say that $G_\Gamma$ is \emph{transvection-free}); \item $\Gamma$ contains no separating star, \ie there does not exist any vertex $v\in\Gamma$ such that $\Gamma\setminus \st(v)$ is disconnected. This prevents the existence of partial conjugations. \end{enumerate} Every right-angled Artin group has an associated \emph{extension graph}, introduced by Kim and Koberda in \cite{kim2013embedability} as the graph whose vertices are the rank one parabolic subgroups of $G_\Gamma$ (\ie conjugates of the cyclic subgroups generated by the standard generators of $G_\Gamma$), two of these being joined by an edge exactly when they commute. This graph can be viewed as an analogue of the curve graph in the context of right-angled Artin groups \cite{kim2014geometry}. Theorem~\ref{theointro:1} is in fact obtained as a consequence of the following theorem, using the additional fact that two right-angled Artin groups with finite outer automorphism groups have isomorphic extension graphs if and only if they are isomorphic \cite{Hua}. \begin{theointro}\label{theointro:main} Let $G_1$ and $G_2$ be two transvection-free right-angled Artin groups. If~$G_1$ and $G_2$ are measure equivalent, then they have isomorphic extension graphs. \end{theointro} In fact, in Theorem~\ref{theointro:1}, if we only assume that $\Out(G_1)$ is finite and that $G_2$ is transvection-free, then using \cite[Th.\,1.2]{Hua}, we deduce that $G_1$ and $G_2$ are measure equivalent if and only if they have isomorphic extension graphs, and in this case this happens if and only if $G_2$ is a finite-index subgroup of $G_1$. To our knowledge, right-angled Artin groups have not been systematically studied from the point of view of measured group theory before; however, a few things were already known concerning their measure equivalence classification. As already mentioned, all countably infinite free abelian groups (and more generally amenable groups) are measure equivalent \cite{Dye2,OW} (and not measure equivalent to nonamenable groups, see \eg \cite[Cor.\,3.2]{Fur-survey}). The behavior of measure equivalence under free products was thoroughly studied by Alvarez and Gaboriau in \cite{AG}. The invariance of the canonical decomposition of a right-angled Artin group as a direct product (corresponding to the maximal decomposition of $\Gamma$ as a join) can be proved by combining work of Monod and Shalom \cite[Th\,1.16 \& 1.17]{MS} together with a theorem of Chatterji, Fernós and Iozzi \cite[Cor.\,1.8]{CFI} stating that non-cyclic right-angled Artin groups that do not split as direct products belong to the class $\mathcal{C}_{\mathrm{reg}}$. Also, Gaboriau proved that $\ell^2$-Betti numbers can be used to obtain measure equivalence invariants \cite{Gab-betti}; they have been computed by Davis and Leary for many Artin groups including all the right-angled ones \cite{DL}. Finally, we comment that Theorem~\ref{theointro:1} was proved in our previous work on Artin groups under the extra assumption that the underlying graphs of $G_1$ and $G_2$ have girth at least $5$ \cite[Cor.\,10.3]{HH}. However, our proof in \cite{HH} relies on fact that, under this extra assumption, $G_1$ and $G_2$ are special instances of $2$-dimensional Artin groups of hyperbolic type. Our methods in \cite{HH} are specific to this class of Artin groups, and therefore cannot be used to prove Theorem~\ref{theointro:1} in general. \subsubsection*{Orbit equivalence and $W^*$-equivalence.} Our results can be reinterpreted in the language of orbit equivalence: indeed, by a result of Furman \cite{Fur2} (with an additional argument by Gaboriau \cite[Th.\,2.3]{Gab2}), two groups are measure equivalent if and only if they admit stably orbit equivalent essentially free measure-preserving ergodic actions on standard probability spaces. As will be explained with more details later in this introduction, the way we prove Theorem~\ref{theointro:main} is actually through this viewpoint -- see Theorem~\ref{theo:main-2} for a precise statement phrased in terms of measured groupoids, also including the case of non-free actions. As right-angled Artin groups are cubical, our results can in fact also be phrased using the weaker notion of (stable) $W^*$-equivalence (we refer to \cite[\S 6.2]{PV} for a detailed discussion of the notion of stable $W^*$-equivalence). \begin{corintro}\label{corintro:von-neumann} Let $G_1$ and $G_2$ be two right-angled Artin groups with finite outer automorphism groups. Assume that $G_1$ and $G_2$ have essentially free ergodic probability measure-preserving actions on standard probability spaces which are stably $W^*$\nobreakdash-equi\-valent (\ie their associated von Neumann algebras are isomorphic, or more generally one is isomorphic to an amplification of the other). Then $G_1$ and $G_2$ are isomorphic. \end{corintro} The version for $W^*$-equivalence (\ie assuming that the two von Neumann algebras are isomorphic, without passing to amplifications) can be seen directly as a consequence of \cite[Cor.\,4.2]{HHL}; we will explain in Section~\ref{sec:von-neumann} how the full statement follows from work of Popa and Vaes \cite{PV1}. \begin{rk*} In Corollary~\ref{corintro:von-neumann}, one cannot expect to reach the stronger conclusion that the given actions of $G_1$ and $G_2$ are conjugate. See indeed Remark~\ref{rk:oe-non-conj} for a construction of free, ergodic, probability measure-preserving actions that are orbit equivalent but not conjugate. \end{rk*} \subsubsection*{Comparison with quasi-isometry, and failure of superrigidity.} Incidentally, the classification results given by Theorems~\ref{theointro:1} and~\ref{theointro:main} both match the quasi-isometry classification obtained by the second named author in \cite{Hua}, generalizing earlier work of Bestvina, Kleiner and Sageev \cite{BKS}. A lot of work has revolved around the quasi-isometry classification problem for right-angled Artin groups, starting from work of Behrstock and Neumann \cite{behrstock2008quasi} for right-angled Artin groups whose defining graph is a tree, with a higher dimensional generalization in \cite{behrstock2010quasi}, and followed more recently by \cite{Hua4,Mar}, for instance. However, apart from the analogy with mapping class groups and Artin groups suggesting that having a finite outer automorphism group is often the source of further rigidity phenomena, there was \emph{a priori} no reason to expect that the measure equivalence and quasi-isometry classification should match. By contrast, they already fail to match on the most basic class of right-angled Artin groups (with infinite outer automorphism group), namely the free abelian groups. In fact, this fundamental difference on the most basic examples is the source of an extremely different situation between the quasi-isometry and measure equivalence classifications of right-angled Artin groups. This is evidenced by our next proposition (and its corollary), that we prove by adapting an argument of Gaboriau \cite{Gab-cost,Gab}, who dealt with the case of free products. We now recall the notion of graph products introduced by Green in \cite{green1990graph}. Given a finite simple graph $\Gamma$ and an assignment of a group $G_v$ to every vertex $v$ of $\Gamma$, the \emph{graph product} over $\Gamma$ with vertex groups $\{G_v\}_{v\in V\Gamma}$ is the group obtained from the free product of the groups $G_v$ by adding as only extra relations that every element of~$G_v$ commutes with every element of $G_w$ whenever $v$ and $w$ are adjacent in $\Gamma$. \begin{propintro}\label{propintro:graph-products} Let $G$ and $H$ be two groups that split as graph products over the same finite simple graph $\Gamma$, with countable vertex groups $\{G_v\}_{v\in V\Gamma}$ and $\{H_v\}_{v\in V\Gamma}$, respectively. Assume that for every vertex $v\in V\Gamma$, the groups $G_v$ and $H_v$ admit orbit equivalent free measure-preserving actions on standard probability spaces. Then $G$ and $H$ admit orbit equivalent free measure-preserving actions on standard probability spaces; in particular they are measure equivalent. \end{propintro} We mention that orbit equivalence cannot be replaced by measure equivalence in the above statement, even for free products (see \cite[\S 2.2]{Gab}). Since right-angled Artin groups are graph products where the vertex groups are isomorphic to $\mathbb{Z}$, combining the above proposition with the aforementioned theorem of Ornstein and Weiss \cite{OW} yields the following. \begin{corintro}\label{corintro:graph-products} For every right-angled Artin group $G$, every group obtained as a graph product over the defining graph of $G$ with countably infinite amenable vertex groups, is measure equivalent to $G$. \end{corintro} By choosing the vertex groups to be isomorphic to $\mathbb{Z}^n$, this gives infinitely many right-angled Artin groups that are all measure equivalent to $G$, but pairwise non-quasi-isometric. As pointed out by a referee, Corollary~\ref{corintro:graph-products} also yields a continuum of countable groups that are measure equivalent to a given right-angled Artin group (see Theorem~\ref{theo:non-rigidity}). Let us also mention that conversely, there are examples of right-angled Artin groups which are quasi-isometric but not measure equivalent. For instance, the groups $G_n=(F_3\times F_3)\ast F_n$ are all quasi-isometric \cite[Th.\,1.5]{Why} but pairwise not measure equivalent by comparison of their $\ell^2$-Betti numbers \cite[Cor.\,0.3]{Gab-betti}. Corollary~\ref{corintro:graph-products} also shows that right-angled Artin groups are never superrigid for measure equivalence in the strongest possible sense. This is in strong contrast to the situation for some non-right-angled Artin groups with finite outer automorphism groups \cite{HH}, or mapping class groups \cite{Kid}; this also strongly contrasts with the quasi-isometry superrigidity results obtained by the second named author in \cite{Hua2}. In fact, there is another source of failure of measure equivalence superrigidity of right-angled Artin groups, given by the following theorem. \begin{theointro}\label{theointro-non-superrigid} For every nonabelian right-angled Artin group $G$, the automorphism group of the universal cover of the Salvetti complex of $G$ contains an infinitely generated (non-uniform) lattice $H$ (in particular $H$ is measure equivalent to $G$, but not commensurable to $G$). \end{theointro} These lattices are constructed as follows. First we take an isometrically embedded tree in the universal cover and a non-uniform lattice $H$ acting on this tree (see \eg \cite{bass2001tree}). Then we extend this action to a larger group acting on the ambient space. This method only produces non-uniform lattices which are not finitely generated, and we do not know whether there are finitely generated examples. \subsubsection*{A word on the proof of the classification theorem.} In the remainder of this introduction, we would like to explain how we prove our main classification theorem, in the form of Theorem~\ref{theointro:main} (from which Theorem~\ref{theointro:1} follows). The general structure of our proof is inspired by Kida's strategy in the mapping class group setting \cite{Kid}. The orbit equivalence interpretation reduces our proof to a problem about measured groupoids associated to measure-preserving actions of right-angled Artin groups on standard probability spaces. Given two such actions of $G_1$ and $G_2$ on the same standard probability space $Y$ with the same orbit structure -- in other words, a measured groupoid over $Y$ coming with two cocycles, one towards $G_1$ and the other towards $G_2$, the goal is to build a canonical map that associates to every point of $Y$, an isomorphism between $\Gamma^e_1$ and $\Gamma_2^e$, where $\Gamma^e_i$ denotes the extension graph of $G_i$. To this end, given a transvection-free right-angled Artin group $G$, we characterize subgroupoids of $G\ltimes Y$ (the groupoid coming from the $G$-action on $Y$ -- or in fact, more generally, a restriction of this groupoid) that naturally correspond to stabilizers of vertices of $\Gamma^e$, or pairs of stabilizers of adjacent vertices of $\Gamma^e$, in a purely groupoid-theoretic way. In order to keep this introduction not too technical, we will not provide any groupoid-theoretic statement, however we will describe some of the ideas behind their group-theoretic analogues. Namely, we will explain how to obtain an algebraic characterization of stabilizers of vertices of $\Gamma^e$ in $G_\Gamma$ (see Proposition~\ref{prop:raag} for the version for groupoids), and an algebraic characterization of adjacency of vertices. So let $v$ be a vertex in $\Gamma^e$ -- corresponding to a cyclic parabolic subgroup $Z_v$ of~$G_\Gamma$. The stabilizer of $v$ for the $G_\Gamma$-action on $\Gamma^e$ is the centralizer of $Z_v$. Classical work of Servatius \cite{Ser} ensures that this centralizer splits as a direct product $Z_v\times Z_v^{\perp}$, where $Z_v^{\perp}$ is a parabolic subgroup of $G_\Gamma$. More precisely, if $Z_v$ is conjugate to the standard cyclic subgroup associated to a vertex $\bar v$ of $\Gamma$, then $Z_v^{\perp}$ is conjugate to the standard parabolic subgroup associated to the link of $\bar v$ in $\Gamma$. The transvection-freeness assumption ensures that $Z_v^{\perp}$ is nonabelian. In fact the following holds. \begin{enumerate}\renewcommand{\theenumi}{$*$} \item\label{propertystar} The centralizer $C_{G_\Gamma}(Z_v)$ is a nonamenable subgroup of $G_\Gamma$ that contains an infinite normal amenable subgroup. Among the subgroups of $G_\Gamma$, it is maximal with respect to this property. \end{enumerate} This is reminiscent of the characterization of centralizers of Dehn twists -- \ie stabilizers of isotopy classes of essential simple closed curves -- for finite-type mapping class groups. Property~\eqref{propertystar} is phrased on purpose in terms of \emph{amenability} and \emph{normality}, as these notions have groupoid-theoretic analogues -- in our proof, amenability is used to obtain invariant probability measures on the (compact) Roller boundary of the universal cover of the Salvetti complex, as will be explained below under the heading `Geometric tools'.\enlargethispage{.5\baselineskip}% However (contrary to the case of the curve graph of a surface), Property~\eqref{propertystar} is not enough to characterize vertex stabilizers of $\Gamma^e$: indeed $G_\Gamma$ could also contain a maximal parabolic subgroup splitting as a direct product of two nonabelian subgroups $P_1\times P_2$ with trivial center, and in this case a subgroup of the form $\langle g\rangle \times P_2$ with~$g$ generic in~$P_1$, would also satisfy Property~\eqref{propertystar}. To distinguish these two types of subgroups, we~make the following observation: when $H=C_{G_\Gamma}(Z_v)$, the centralizer~$H'$ of any infinite amenable subgroup $Z'$ not commensurable to $Z_v$, will look very different from~$H$, in the sense that one can always find a nonamenable subgroup of~$H$ that intersects $H'$ trivially. On the other hand, when $H$ is of the form $\langle g\rangle \times P_2$ as above, by choosing $g'$ to be any other generic element of $P_1$, the centralizer $H'$ of $g'$ will be much closer to $H$, and in particular every nonamenable subgroup of $H'$ will have to intersect $P_2$, whence $H$, nontrivially. See assertion~\eqref{prop:raag2b} from Proposition~\ref{prop:raag} for the groupoid-theoretic version. We also need to characterize adjacency in $\Gamma^e$. For that, the key observation is that two vertices $v,w\in \Gamma^e$ are adjacent if and only if there are only finitely many vertices of $\Gamma^e$ that are fixed by $C_{G_\Gamma}(Z_v)\cap C_{G_\Gamma}(Z_w)$. \subsubsection*{Geometric tools.} Our proof of Theorem~\ref{theointro:main} is an implementation of a groupoid version of the above strategy. This groupoid version involves a certain geometric setting (with features of negative curvature), in order to apply an argument introduced by Adams in \cite{Ada}. In the work of Kida \cite{Kid}, this came in the form of a natural partition of the (compact) Thurston boundary of the Teichmüller space into arational and nonarational laminations. In our previous work on Artin groups of hyperbolic type \cite{HH}, this came in the form of a partition of the horofunction compactification of a certain $\mathrm{CAT}(-1)$ simplicial complex on which the group is acting. In the present paper, we exploit the cubical geometry of right-angled Artin groups by considering the Roller boundary $\partial_R\widetilde{S}_\Gamma$ of the universal cover of the Salvetti complex, and its partition, introduced by Fernós in \cite{Fer}, into \emph{regular} points and \emph{non-regular} points. The three important features we exploit are the following: every non-regular point has a natural associated parabolic subgroup (see Theorem~\ref{theointro:roller} below); there is a \emph{barycenter map} that canonically associates a vertex of $\widetilde{S}_\Gamma$ to every triple of pairwise distinct regular points \cite{FLM}; the action of $G_\Gamma$ on the set of regular points (in fact on the whole $\partial_R\widetilde{S}_\Gamma$) is Borel amenable \cite{NS,Duc}. Let us explain how this geometric setting is used, by explaining why a subgroup $H\subseteq G_\Gamma$ satisfying Property~\eqref{propertystar} above (in particular, containing an infinite normal amenable subgroup $A$) preserves a parabolic subgroup. Amenability of $A$ is used to get an invariant probability measure $\nu$ on the \emph{compact} metrizable space $\partial_R\widetilde{S}_\Gamma$. Theorem~\ref{theointro:roller} below, saying that every non-regular point has a natural associated parabolic subgroup, is exploited to build a canonical $A$-invariant (whence $H$-invariant) parabolic subgroup if $\nu$ gives positive measure to the set of non-regular points. If $\nu$ is supported on the set of regular points, and if its support has cardinality at least $3$, then the aforementioned \emph{barycenter map} is used to contradict the fact that $A$ is infinite. Finally, if $\nu$ is supported on at most two points, then the amenability of point stabilizers is exploited to get a contradiction to the fact that $H$ is nonamenable -- in the groupoid version, this is replaced by the Borel amenability of the action of $G_\Gamma$ on $\partial_R\widetilde{S}_\Gamma$. Let us finally state the result we use concerning non-regular points (see Section~\ref{sec:background-roller} for relevant definitions). In this statement $\partial_R\widetilde{S}_\Gamma$ is equipped with its Borel $\sigma$-algebra. \begin{theointro}\label{theointro:roller} Let $\Gamma$ be a finite simple graph. Then we can assign to each non-regular point $\xi\in\partial_R\widetilde{S}_\Gamma$ a unique standard subcomplex $Y\subseteq\widetilde{S}_\Gamma$ such that for every combinatorial geodesic ray $r$ representing $\xi$, the subset $Y$ is the smallest standard subcomplex of $\widetilde{S}_\Gamma$ that contains a subray of $r$. This assignment is measurable and $G_\Gamma$-equivariant, and $Y$ is contained in a standard subcomplex of $\widetilde{S}_\Gamma$ which splits non-trivially as a product. \end{theointro} \subsubsection*{Open questions.} Our work raises several questions regarding the measure equivalence classification and rigidity of right-angled Artin groups. \begin{enumerate} \item Given a right-angled Artin group $G_\Gamma$ with finite outer automorphism group, what can be said of countable groups $H$ that are measure equivalent to $G_\Gamma$? For instance, do they necessarily act on a $\mathrm{CAT}(0)$ cube complex with amenable vertex stabilizers? \item A sharper notion of \emph{$L^1$-measure equivalence} was introduced by Bader, Furman and Sauer in \cite{BFS}, by imposing an integrability condition on the measure equivalence cocycle. Are there right-angled Artin groups that are superrigid for $L^1$-measure equivalence? It turns out that free abelian groups of different ranks are not $L^1$-measure equivalent to one another (see \cite{Aus}), thus suggesting that the obstruction coming from Corollary~\ref{corintro:graph-products} vanishes under this sharper notion. The question may be subdivided into two parts. First, if a finitely generated group $H$ is $L^1$-measure equivalent to $G_\Gamma$, is it commensurable to a lattice in $\Aut(\widetilde{S}_\Gamma)$? Second, are all finitely generated such lattices cocompact? \item Classify right-angled Artin groups up to measure equivalence, beyond the class of those with finite outer automorphism group. For instance, Behrstock and Neumann proved in \cite{behrstock2008quasi} that all right-angled Artin groups whose defining graph is a tree of diameter at least $3$ are quasi-isometric. Are they all measure equivalent? \end{enumerate} \subsubsection*{Organization of the paper.} The paper has four parts. The first part is mostly a background section on right-angled Artin groups. In the second part, we prove Theorem~\ref{theointro:roller} concerning the geometry of the Roller boundary of the universal cover of the Salvetti complex. The third part implements the groupoid version of the strategy explained in this introduction, and gives a proof of our main classification theorems (Theorems~\ref{theointro:1} and~\ref{theointro:main}). In Section~4, we present two sources of failure of measure equivalence superrigidity of right-angled Artin groups: graph products of countably infinite amenable groups, and non-uniform lattices in the automorphism group of the universal cover of the Salvetti complex. \subsubsection*{Acknowledgements.} We thank the referees for their careful reading of our manuscript and their helpful suggestions. \section{Right-angled Artin groups: background and complements} In the present section, we review standard facts about right-angled Artin groups and establish a few statements that we will need in the sequel. More background on right-angled Artin groups can be found in \cite{charney2007introduction}. Section~\ref{sec:review} reviews basic facts about right-angled Artin groups, their parabolic subgroups, their automorphisms, and Salvetti complexes. Section~\ref{sec:rigidity} reviews earlier work of the second named author \cite{Hua} establishing a rigidity statement for extension graphs of right-angled Artin groups. Sections~\ref{sec:transvection-free} and~\ref{sec:full-support} establish two extra statements that are used in Section~\ref{sec:me}. We will stick to the setting of right-angled Artin groups in the present paper, but would like to mention that many of the results presented here have also been generalized to broader contexts. As a specific example, several results regarding parabolic subgroups have been generalized to graph products in \cite[\S 3]{antolin2015tits}. \Subsection{Review of basic facts about right-angled Artin groups}\label{sec:review} \subsubsection*{Definition.} Given a finite simple graph $\Gamma$ with vertex set $V\Gamma$, the \emph{right-angled Artin group} with defining graph $\Gamma$, denoted by $G_\Gamma$, is given by the following presentation: \begin{center} $\langle V\Gamma$\ |\ $[v,w]=1$ if $v$ and $w$ are joined by an edge$\rangle$. \end{center} In this way $V\Gamma$ is identified to a \textit{standard generating set} for $G_\Gamma$. Given two finite simple graphs $\Gamma$ and $\Lambda$, a theorem of Droms \cite{Dro} asserts that $G_\Gamma$ and $G_\Lambda$ are isomorphic if and only if $\Gamma$ and $\Lambda$ are isomorphic. \subsubsection*{Parabolic subgroups.} Let $\Lambda\!\subseteq\!\Gamma$ be a \emph{full} subgraph, \ie two vertices of $\Lambda$ are adja\-cent in $\Lambda$ if and only if they are adjacent in $\Gamma$. Then there is an injective homomorphism $G_{\Lambda}\hookrightarrow G_{\Gamma}$, whose image is called a \emph{standard subgroup} of $G_{\Gamma}$ with \emph{type}~$\Lambda$. A~conjugate of such a subgroup is called a \emph{parabolic subgroup of $G_\Gamma$ with type $\Lambda$}. The type of a parabolic subgroup is well-defined: if $\Lambda_1,\Lambda_2\subseteq\Gamma$ are two full subgraphs, then~$G_{\Lambda_1}$ and $G_{\Lambda_2}$ are not conjugate unless $\Lambda_1=\Lambda_2$, as follows from \hbox{\cite[Prop.\,2.2]{charney2007automorphisms}}. Note that the definitions of parabolic subgroups and standard subgroups depend on the choice of a standard generating set for $G_\Gamma$. The \emph{star} of a vertex $v$ in $\Gamma$, denoted by $\st(v)$, is the full subgraph spanned by $v$ and all the vertices that are adjacent to $v$. Its \emph{link} $\lk(v)$ is defined to be the full subgraph spanned by all the vertices that are adjacent to $v$. Given two graphs $\Gamma_1$ and $\Gamma_2$, we denote by $\Gamma_1\circ \Gamma_2$ the join of $\Gamma_1$ and $\Gamma_2$. Every finite simple graph $\Gamma$ has a canonical join decomposition $\Gamma=\Gamma_0\circ \Gamma_1\circ\cdots\circ\Gamma_k$, where~$\Gamma_0$ is the maximal clique factor of $\Gamma$, and for every $i\in\{1,\dots,k\}$, the graph $\Gamma_i$ is \emph{irreducible}, \ie it does not admit any nontrivial join decomposition. We call this the \emph{de Rham decomposition} of $\Gamma$. There is also an induced \emph{de Rham decomposition} of $G_\Gamma$, namely $G_\Gamma=\mathbb Z^n\times G_{\Gamma_1}\times\cdots\times G_{\Gamma_k}$. For a full subgraph $\Lambda\subseteq \Gamma$, we define $\Lambda^{\perp}$ to be the full subgraph of $\Gamma$ spanned by the collection of all vertices of $\Gamma\setminus \Lambda$ which are adjacent to every vertex of $\Lambda$. For example, with this terminology, we have $\lk(v)=\{v\}^\perp$. The following was proved by Charney, Crisp and Vogtmann in \cite[Prop.\,2.2]{charney2007automorphisms}, building on work of Godelle \cite{God}. \begin{prop}\label{prop:normalizer} Let $\Gamma$ be a finite simple graph, and let $\Lambda\subseteq\Gamma$ be a full subgraph. Then the normalizer of $G_{\Lambda}$ in $G_{\Gamma}$ is $G_{\Lambda\circ \Lambda^{\perp}}$. \end{prop} Let now $P=gG_{\Lambda}g^{-1}$ be a parabolic subgroup. We define $P^\perp=gG_{\Lambda^\perp}g^{-1}$. This is well-defined: if we can write the parabolic subgroup $P$ in two different ways $gG_{\Lambda}g^{-1}$ and $hG_{\Lambda'}h^{-1}$, then $\Lambda=\Lambda'$ and Proposition~\ref{prop:normalizer} ensures that $gG_{\Lambda^\perp}g^{-1}=hG_{\Lambda^\perp}h^{-1}$. Moreover, the normalizer of $P$ in $G_{\Gamma}$ is $P\times P^\perp$. In the following lemma, we record basic facts about parabolic subgroups of right-angled Artin groups, which follow from the works of Duncan, Kazachkov and Remeslennikov in \cite[\S 2.2]{duncan2007parabolic}. Note that \eqref{lemma:parabolics5} and \eqref{lemma:parabolics6} generalize Proposition~\ref{prop:normalizer}. \begin{lemma}[Duncan--Kazachkov--Remeslennikov]\label{lemma:parabolics} Let $\Gamma$ be a finite simple graph. \begin{enumerate} \item\label{lemma:parabolics1} The intersection of two parabolic subgroups of $G_\Gamma$ is a parabolic subgroup. Moreover, for full subgraphs $\Gamma_1$ and $\Gamma_2$ of $\Gamma$, one has $G_{\Gamma_1}\cap G_{\Gamma_2}=G_{\Gamma_1\cap\Gamma_2}$. \item\label{lemma:parabolics2} There is a finite integer $n>0$ such that every chain $P_1\subsetneq P_2\subsetneq\cdots\subsetneq P_k$ of parabolic subgroups of $G_\Gamma$ has length $k\le n$. \item\label{lemma:parabolics3} Every element of $G_\Gamma$ is contained in a unique smallest parabolic subgroup. \item\label{lemma:parabolics4} If $P_1$ and $P_2$ are two parabolic subgroups of $G_\Gamma$ with $P_1\subseteq P_2$, then the type of~$P_1$ is contained in the type of $P_2$. \item\label{lemma:parabolics5} For every parabolic subgroup $P$ of $G_\Gamma$ and every $g\in G$, if $P\subseteq g(P\times P^\perp)g^{-1}$, then $g\in P\times P^\perp$. \item\label{lemma:parabolics6} Given any two induced subgraphs $\Gamma_1,\Gamma_2$ of $\Gamma$, and any $g\in G_\Gamma$, if \hbox{$g G_{\Gamma_1}g^{-1}\!\!\subseteq\! G_{\Gamma_2}$}, then there exists $h\in G_{\Gamma_2}$ such that $g G_{\Gamma_1}g^{-1}=h G_{\Gamma_1}h^{-1}$ (in particular $G_{\Gamma_1}\subseteq G_{\Gamma_2}$). \end{enumerate} \end{lemma} \begin{proof} The first assertion is \cite[Prop.\,2.6 \& Lem.\,2.7]{duncan2007parabolic}. Assertion~\eqref{lemma:parabolics2} follows from \cite[Prop.\,2.6]{duncan2007parabolic}, and assertion~\eqref{lemma:parabolics3} follows from \cite[Prop.\,2.8]{duncan2007parabolic}. Assertions~\eqref{lemma:parabolics4}, \eqref{lemma:parabolics5} and \eqref{lemma:parabolics6} follow from \cite[Cor.\,2.5]{duncan2007parabolic} or \cite[Prop.\,2.2]{charney2007automorphisms}. \end{proof} \begin{rk} \label{remark:parabolic in parabolic} Every parabolic subgroup $H$ of $G_\Gamma$ is itself a right-angled Artin group, with a choice of generating set induced from the generating set of $G_\Gamma$. By~Proposition~\ref{prop:normalizer}, this choice is well-defined up to conjugation by elements inside $H$. Thus it makes sense to talk about parabolic subgroups of $H$ with respect to this choice of generating set of $H$. Every parabolic subgroup of $H$ is then naturally a parabolic subgroup of $G_\Gamma$. Conversely, by Lemma~\ref{lemma:parabolics} (4) and (6), every parabolic subgroup $H'$ of $G_\Gamma$ with $H'\subseteq H$ is in fact a parabolic subgroup of $H$. Thus from now on, we will just refer to $H'$ as a parabolic subgroup, without specifying its ambient group. \end{rk} The smallest parabolic subgroup that contains an element $g\in G_\Gamma$ is called the \emph{support} of $g$. The \emph{type} of $g$ is then defined as the type of its support.\enlargethispage{.9\baselineskip}% \begin{lemma} \label{lemma:setwise vs pointwise} Let $\Gamma$ be a finite simple graph, and let $\mathbb P$ be the collection of parabolic subgroups of $G_\Gamma$, equipped with the conjugation action of $G_\Gamma$. Then for every finite set $\calf\! \subset\! \mathbb P$, the pointwise stabilizer of $\calf$ in $G_\Gamma$ coincides with the setwise stabilizer of~$\calf$. \end{lemma} \begin{proof} Let $P$ be a parabolic subgroup in the set $\calf$. Take $g\!\in\! G_\Gamma$ with \hbox{$g\calf\!=\!\calf$}. Then there exists $k\!\neq\! 0$ such that $g^k$ normalizes $P$, so $g^k\!\in\! P\times P^\perp$. Hence the sub\-group $\langle g\rangle$ has a finite-index subgroup contained in the parabolic subgroup \hbox{$P\!\times\! P^\perp$}. By \cite[Lem.\,6.4]{minasyan2012hereditary}, we have $\langle g\rangle\subseteq P\times P^\perp$. Hence $g$ fixes $P$ and the lemma follows. \end{proof} \subsubsection*{Automorphisms.} Let $\Gamma$ be a finite simple graph. Laurence \cite{laurence1995generating} and Servatius \cite{Ser} showed that the outer automorphism group $\Out(G_\Gamma)$ is generated by the outer classes of four types of automorphisms, namely \emph{inversions}, \emph{graph automorphisms}, \emph{transvections} and \emph{partial conjugations}. We say that $\Gamma$ (or $G_\Gamma$) is \emph{transvection-free} if~$\Out(G_\Gamma)$ does not contain any transvection; equivalently, there do not exist distinct vertices $v,w\in\Gamma$ such that $\lk(w)\subseteq \st(v)$. When $\Gamma$ is transvection-free, the set of parabolic subgroups of $G_\Gamma$ does not depend on the choice of standard generating set of $G_\Gamma$. It also follows from the work of Laurence and Servatius that $\Out(G_\Gamma)$ is finite if and only if $\Gamma$ is transvection-free and does not contain any separating star. \subsubsection*{The Salvetti complex.} Every right-angled Artin group $G_\Gamma$ is the fundamental group of a locally CAT(0) cube complex $S_\Gamma$, called the \emph{Salvetti complex}, obtained from a bouquet of $|V\Gamma|$ circles by gluing a $k$-cube per $k$-clique of $\Gamma$ (recall that a \emph{$k$-clique} is a complete subgraph on $k$ vertices). We refer to \cite[\S 3.6]{charney2007introduction} for more details. The 2-skeleton of $S_\Gamma$ is the presentation complex of $G_\Gamma$. For each full subgraph $\Lambda\subseteq \Gamma$, there is an isometric embedding $S_{\Lambda}\hookrightarrow S_\Gamma$. Let $\widetilde{S}_\Gamma$ be the universal cover of $S_\Gamma$, which is a CAT(0) cube complex. A \emph{standard subcomplex} of $\widetilde{S}_\Gamma$ of type $\Lambda$ is a connected component of the inverse image of $S_{\Lambda}\subseteq S_\Gamma$ with respect to the covering map $\widetilde{S}_\Gamma\to S_\Gamma$. In particular, a standard subcomplex whose type is the empty subgraph is the same as a vertex of $\widetilde{S}_\Gamma$. We collect several standard facts on $\widetilde{S}_\Gamma$. \begin{enumerate} \item Two standard subcomplexes of the same type are either disjoint or equal. If a standard subcomplex of type $\Lambda_1$ and a standard subcomplex of type $\Lambda_2$ have nonempty intersection, then their intersection is a standard subcomplex of type $\Lambda_1\cap\Lambda_2$. \item The stabilizer of a standard subcomplex is a parabolic subgroup. Each parabolic subgroup can be realized as the stabilizer of some (non-unique) standard subcomplex. \end{enumerate} \subsection{Rigidity of the extension graph}\label{sec:rigidity} Let $\Gamma$ be a connected finite simple graph. The \emph{extension graph} of $\Gamma$, denoted $\Gamma^e$, was defined by Kim and Koberda \cite{kim2013embedability} to be the graph whose vertices are the parabolic subgroups of $G_\Gamma$ isomorphic to $\mathbb Z$, where two vertices are adjacent if the corresponding parabolic subgroups commute. As follows from Proposition~\ref{prop:normalizer} and Lemma~\ref{lemma:parabolics}(6), two distinct cyclic parabolic subgroups $P_1=g_1\langle v_1\rangle g_1^{-1}$ and $P_2=g_2\langle v_2\rangle g_2^{-1}$ commute if and only if $v_1$ and $v_2$ are adjacent in $\Gamma$ and there exists $g\in G_\Gamma$ such that for every $i\in\{1,2\}$, one has $P_i=g\langle v_i\rangle g^{-1}$. The conjugation action of $G_\Gamma$ on itself induces an action $G_\Gamma\actson \Gamma^e$. The above observation about adjacency can be rephrased by saying that $\Gamma$ is a fundamental domain for the $G_\Gamma$-action on $\Gamma^e$. Now we recall the following rigidity result on extension graphs, which is a combination of \cite[Cor.\,4.16 and Lemma 4.17]{Hua}. \begin{theo}[\cite{Hua}]\label{theo:rigidity} Let $\Gamma_1$ and $\Gamma_2$ be two finite simple graphs, and assume that for every $i\in\{1,2\}$, the group $\Out(G_{\Gamma_i})$ is finite. Then $\Gamma_1^e$ and $\Gamma_2^e$ are isomorphic if and only if $\Gamma_1$ and $\Gamma_2$ are isomorphic. \end{theo} In particular, Theorem~\ref{theointro:1} from the introduction follows from Theorem~\ref{theointro:main}. \subsection{A lemma about transvection-free right-angled Artin groups}\label{sec:transvection-free} In the sequel of the paper, one way in which the transvection-freeness assumption will be used is through the following easy lemma. \begin{lemma}\label{lemma:transvection-free} Let $G=G_\Gamma$ be a transvection-free right-angled Artin group. Let $Z$ be a cyclic parabolic subgroup of $G$, and let $P$ be a nontrivial parabolic subgroup of $G$ such that $Z\times Z^{\perp}\subseteq P\times P^{\perp}$. \begin{enumerate} \item If $P$ is cyclic, then $Z=P$. \item If $P$ is noncyclic, then $P\cap Z^{\perp}$ is nonabelian. \end{enumerate} \end{lemma} \begin{proof} Up to conjugation, we can assume $P$ is standard, hence $P\times P^\perp$ is also standard. As $Z\times Z^\perp$ can be viewed as a parabolic subgroup of $P\times P^\perp$ (see Remark~\ref{remark:parabolic in parabolic}), up to conjugating $Z\times Z^\perp$ by an element in $P\times P^\perp$, we can assume without loss of generality that both $Z$ and $Z\times Z^\perp$ are standard as such conjugation brings $P$ to itself. Let $\Gamma_P\subseteq\Gamma$ be the type of $P$. Then the type of $P^\perp$ is $(\Gamma_P)^\perp$. Let $v_Z$ be the vertex of $\Gamma$ which is the type of $Z$. As $Z\times Z^{\perp}\subseteq P\times P^{\perp}$, we have $\st(v_Z)\subseteq \Gamma_P\circ (\Gamma_P)^\perp$. Let $\lk(v_Z)=\Gamma_1\circ\Gamma_2\circ\cdots\circ\Gamma_k$ be the de Rham decomposition of $\lk(v_Z)$, which induces $Z^{\perp}=Q_1\times\dots \times Q_k$. The transvection-free condition implies that no $\Gamma_i$ is a clique (in particular no $\Gamma_i$ is reduced to one vertex), as otherwise the link of $v_Z$ would be contained in the star of every vertex of $\Gamma_i$. Thus, each $Q_i$ is nonabelian. As each~$\Gamma_i$ is irreducible, either $\Gamma_i\subseteq\Gamma_P$ or $\Gamma_i\subseteq (\Gamma_{P})^\perp$. If $P$ is cyclic, then $\Gamma_P$ is reduced to a vertex $v_P$, no $\Gamma_i$ can be contained in $\Gamma_P$, and we deduce that $\lk(v_Z)\subseteq (\Gamma_P)^\perp$. So $\lk(v_Z)$ is contained in the star of $v_P$, and the transvection-free condition thus implies that $\{v_Z\}=\Gamma_P$. It follows that $Z=P$ as they are both standard, showing that the first assertion of the lemma holds. If $P$ is noncyclic, then $\lk(v_Z)\nsubseteq (\Gamma_P)^\perp$, otherwise we have $\lk(v_Z)\subseteq \st(w)$ for every vertex $w\in\Gamma_P$. Thus at least one $\Gamma_i$ is contained in $\Gamma_P$. Therefore $\Gamma_P\cap\lk(v_Z)$ contains an irreducible graph with at least two vertices, and the second assertion of the lemma follows. \end{proof} \Subsection{Full support subgroups}\label{sec:full-support} \begin{prop}\label{prop:kim-koberda} Let $\Gamma$ be a finite simple graph whose de Rham decomposition has no clique factor. Then $G_\Gamma$ contains a nonabelian free subgroup $F$ such that no nontrivial element of $F$ is contained in a proper parabolic subgroup of $G_\Gamma$. \end{prop} \begin{proof} We claim that if $\Gamma=\Gamma_1\circ\Gamma_2$, then each parabolic subgroup $P$ of $G_\Gamma$ splits as a product $P_1\times P_2$ where $P_i=P\cap G_{\Gamma_i}$ is a parabolic subgroup of $G_{\Gamma_i}$. This is clearly true if $P$ is standard, otherwise there is $g=g_1g_2\in G_{\Gamma}$ ($g_i\in G_{\Gamma_i}$) such that $gPg^{-1}$ is standard. On the other hand, $gG_{\Gamma_i}g^{-1}=g_i G_{\Gamma_i} g^{-1}_i=G_{\Gamma_i}$. Thus the claim follows. This claim implies that it suffices to prove the lemma when $\Gamma$ is not a join. In what follows, we are equipping $\widetilde{S}_\Gamma$ with the standard $\mathrm{CAT}(0)$ metric. We will use the following simple observation: if a nontrivial element $g$ belongs to a parabolic subgroup of type $\Gamma'$, then it has an axis (with respect to the action $G_\Gamma\actson \widetilde{S}_\Gamma$) contained in a standard subcomplex of type $\Gamma'$, hence every axis of $g$ is contained in a finite neighborhood of this standard subcomplex (see \cite[Chap.\,II.6]{bridson2013metric} for basic properties of axes). Let $X$ be the wedge of two circles $C_1$ and $C_2$, with the wedge point denoted by $x_0$. We will choose two words $W_1$ and $W_2$ in the standard generating set $V\Gamma$ such that \begin{enumerate} \item[(1)] each $W_i$ uses all the generators of $G_\Gamma$; \item[(2)] the map $\varphi:X\to S_\Gamma$ defined by mapping $x_0$ to the base vertex of $S_\Gamma$ and mapping each $C_i$ to the edge path in the $1$-skeleton of $S_\Gamma$ corresponding to the word~$W_i$ is a local isometry ($C_i$ is metricized so that its length is equal to the word length of~$W_i$). \end{enumerate} Then $\varphi_\ast(\pi_1 X)$ is the free subgroup satisfying our requirement, as (2) implies that any lift $\tilde \varphi:\widetilde X\to \widetilde{S}_\Gamma$ to the universal covers maps an axis of a nontrivial element $g\in \pi_1 X$ to an axis $\ell$ of $\varphi_\ast(g)$, and (1) implies that $\ell$ is not contained in any finite neighborhood of any proper standard subcomplex. Let $\Gamma^c$ be the complement graph of $\Gamma$ (\ie these two graphs have the same vertex set, two vertices in $\Gamma^c$ are adjacent if they are not adjacent in $\Gamma$). As $\Gamma$ is not a join, $\Gamma^c$ is connected. Let $u_0,v_0$ be two adjacent vertices in $\Gamma^c$. For each vertex $v\in \Gamma^c$, let $p_v$ an edge path traveling from $v_0$ to $v$ then back to $v_0$ in $\Gamma^c$. Listing consecutive vertices in $p_v$ leads to a word $W_v$ starting with $v_0$ and ending with $v_0$. Let $W$ be the product (in any order) of all $W_v$ with $v$ ranging over all vertices of $\Gamma^c$. Then the words $W_1=v_0Wu_0$ and $W_2=v^{-1}_0u_0Wu^{-1}_0$ satisfy the above two conditions. \end{proof} \begin{rk} Proposition~\ref{prop:kim-koberda} can also be proved by showing that, when $\Gamma$ does not decompose nontrivially as a join, the group $G_\Gamma$ acts nonelementarily on the graph of parabolic subgroups of $G_\Gamma$ -- having one vertex per proper parabolic subgroup of~$G_\Gamma$, where two such subgroups are joined by an edge whenever they have nontrivial intersection -- which turns out to be hyperbolic. We decided to provide an elementary proof that does not rely on a hyperbolicity statement. \end{rk} \section{The Roller boundary of the Salvetti complex} The goal of the present section is to prove Theorem~\ref{theointro:roller} from the introduction (Theorem~\ref{theo:roller} below). We start with a short review on Roller boundaries of CAT(0) cube complexes. \Subsection{Background on the Roller boundary}\label{sec:background-roller} \subsubsection*{General background.} The Roller boundary, implicit in the work of Roller \cite{Rol} and explicitly introduced in \cite{BCGNW}, gives a way of compactifying any CAT(0) cube complex. We refer to \cite{sageev2012cat} for background on CAT(0) cube complexes, including a discussion on hyperplanes and halfspaces. We now review the definition and a few facts about the Roller boundary. Let $X$ be a CAT(0) cube complex. Given a subset $Z\subseteq X$, we let $\calh(Z)$ be the set of all hyperplanes of $X$ that intersect $Z$ nontrivially. Let $\mathfrak{H}$ be the set of all halfspaces of $X$ (the boundary hyperplane of a halfspace $h$ will be denoted $\partial h$). There is an embedding of the vertex set $V(X)$ into $\{0,1\}^{\mathfrak{H}}$ (equipped with the product topology), sending a vertex $v$ to the map $\mathfrak{H}\to\{0,1\}$ that sends a halfspace $h$ to $1$ if and only if $v\in h$. The closure of the image of this embedding yields a compactification of $V(X)$. The \emph{Roller boundary} of $X$, which we denote by $\partial_RX$, is the complement of $V(X)$ in this compactification: it is compact whenever $X$ is locally compact. A point $\xi\in\partial_R X$ is thus a map $\mathfrak{H}\to\{0,1\}$, and we let $U_\xi\subseteq\mathfrak{H}$ be the set of all halfspaces sent to $1$ under this map. Let $Y\subseteq X$ be a convex subcomplex. Then there is a continuous embedding $\partial_RY\to \partial_R X$ such that a point $\xi\in\partial_R X$ lies in the image of this embedding if and only if it corresponds to a map $\mathfrak{H}\to\{0,1\}$ that sends every halfspace containing $Y$ to $1$. Thus we will identify $\partial_R Y$ with a closed subset of $\partial_R X$. When $X$ has countably many hyperplanes (\eg when $X=\widetilde{S}_\Gamma$ is the universal cover of the Salvetti complex associated to a finite simple graph $\Gamma$), the space $\{0,1\}^{\mathfrak{H}}$ is metrizable, so $\partial_RX$ is metrizable. \subsubsection*{Combinatorial geodesic rays and the Roller boundary.} We denote by $X^{(1)}$ the $1$\nobreakdash-skele\-ton of $X$, equipped with the path metric. Geodesic rays in $X^{(1)}$ are called \emph{combinatorial geodesic rays}. Let $x\in X$ be a vertex, and let $r$ be a combinatorial geodesic ray in $X^{(1)}$ originating at $x$. For every hyperplane $\mathfrak{h}$, the ray $r$ selects exactly one of the two halfspaces complementary to $\mathfrak{h}$, by considering the halfspace which virtually contains $r$ (\ie contains $r$ up to a finite segment). Thus $r$ defines a point in $\partial_R X$. Two combinatorial geodesic rays are \emph{equivalent} if they cross the same set of hyperplanes. There is a 1-1 correspondence between equivalence classes of combinatorial geodesic rays based at a given point $x\in X$, and points in $\partial_R X$ (see \eg \cite[\S A.2]{genevois2020contracting}). Now suppose that two combinatorial geodesic rays $r$ and $r'$ (potentially with different origins $x$ and $x'$) represent the same point of $\partial_R X$. By definition of $\partial_R X$, a~halfspace of $X$ virtually contains $r$ if and only if it virtually contains $r'$. Also, \hbox{every} hyperplane $\mathfrak{h}$ contained in the symmetric difference $\calh(r)\Delta\calh(r')$ separates $x$ from~$x'$. In particular $\calh(r)\Delta\calh(r')$ is finite. \subsubsection*{Regular points in the Roller boundary.} Two hyperplanes $\mathfrak{h}_1$ and $\mathfrak{h}_2$ of $X$ are \emph{strongly separated} \cite[Def.\,2.1]{behrstock2012divergence} if there does not exist any hyperplane $\mathfrak{h}$ such that for every $i\in\{1,2\}$, one has $\mathfrak{h}\cap\mathfrak{h}_i\neq\emptyset$. The notion of a regular point of the Roller boundary, introduced by Fernós in \cite[Def.\,7.3]{Fer}, is defined as follows: a point $\xi\in\partial_R X$ is \emph{regular} if given any two halfspaces $h_1,h_2\in U_\xi$, there exists a halfspace $k\in U_\xi$ such that $k\subseteq h_1\cap h_2$ and for every $i\in\{1,2\}$, the hyperplanes $\partial k$ and $\partial h_i$ are strongly separated. We denote by $\partial_{\reg}X$ the subspace of $\partial_R X$ made of regular points. \Subsection{Non-regular points in the Roller boundary of the Salvetti complex} Let $\Gamma$ be a finite simple graph. We denote by $\mathbb{S}_J$ the set of all standard subcomplexes of $\widetilde{S}_\Gamma$ that are contained in some join standard subcomplex of $\widetilde{S}_\Gamma$, \ie a standard subcomplex whose type admits a nontrivial join decomposition. The goal of the present section is to prove the following theorem. \begin{theo}\label{theo:roller} Let $\Gamma$ be a finite simple graph. Then $\partial_\reg\widetilde{S}_\Gamma$ is a Borel subset of $\partial_R\widetilde{S}_\Gamma$, and there exists a unique Borel $G_\Gamma$-equivariant map \[ \Phi:\partial_R\widetilde{S}_\Gamma\setminus\partial_\reg\widetilde{S}_\Gamma\to\mathbb{S}_J \] such that for every nonregular point $\xi\in\partial_R\widetilde{S}_\Gamma$ and every combinatorial geodesic ray $r$ representing $\xi$, the subset $\Phi(\xi)$ is the smallest standard subcomplex of $\widetilde{S}_\Gamma$ that virtually contains $r$. \end{theo} Theorem~\ref{theo:roller} is a consequence of Lemmas~\ref{lemma:confined},~\ref{lem:borel} and~\ref{lem:join subgroup} below. \begin{lemma}\label{lemma:confined} Let $\Gamma$ be a finite simple graph. For every $\xi\in\partial_R\widetilde{S}_\Gamma$, there exists a unique standard subcomplex $Y\subseteq \widetilde{S}_\Gamma$ such that for every combinatorial geodesic ray~$r$ representing $\xi$, the subcomplex $Y$ is the smallest standard subcomplex that virtually contains $r$. \end{lemma} \begin{proof} Recall that the intersection of two standard subcomplexes is again a standard subcomplex and that there is a uniform bound on the length of a strictly descending chain of standard subcomplexes. Therefore, for every combinatorial geodesic ray~$r$, there is a unique minimal standard subcomplex $Y_r\subseteq\widetilde{S}_\Gamma$ which virtually contains~$r$. We are thus left with showing that if $r$ and $r'$ are combinatorial geodesic rays representing the same point of $\partial_R \widetilde{S}_\Gamma$ (possibly with different base points), then \hbox{$Y_r=Y_{r'}$}. In this case, as observed in Section~\ref{sec:background-roller}, \begin{enumerate} \item[(1)] a halfspace of $\widetilde{S}_\Gamma$ virtually contains $r$ if and only if it virtually contains $r'$; \item[(2)] the symmetric difference of $\mathcal{H}(r)$ and $\mathcal{H}(r')$ is finite. \end{enumerate} Let $\Gamma_r$ be the type of $Y_r$. As the 1-skeleton of $\widetilde{S}_\Gamma$ is the Cayley graph of $G_\Gamma$, we~label edges of $\widetilde{S}_\Gamma$ by vertices of $\Gamma$. Then there exists a subray $r_1$ of $r$ such that the collection $L(r_2)$ of labels of edges of any further subray $r_2$ of $r_1$ satisfies $L(r_2)=V\Gamma_r$. We~define~$r'_1$ and $\Gamma_{r'}$ similarly. By (2), for any subray $r'_3$ of $r'_1$, there is a subray $r_3$ of~$r_1$ with $L(r_3)\subseteq L(r'_3)$. Thus $\Gamma_r\subseteq \Gamma_{r'}$. Similarly $\Gamma_{r'}\subseteq \Gamma_r$. Therefore $\Gamma_r=\Gamma_{r'}$, and in fact $Y_r=Y_{r'}$: indeed (1) implies that $Y_r\cap Y_{r'}\neq\emptyset$, as otherwise $Y_r$ and $Y_{r'}$ would be separated by a hyperplane (see \cite[Cor.\,13.10]{haglund2008special}). This concludes our proof. \end{proof} Lemma~\ref{lemma:confined} yields a well-defined map $\Phi:\partial_R \widetilde{S}_\Gamma\to \mathbb{S}$, where $\mathbb{S}$ is the set of all standard subcomplexes of $\widetilde{S}_\Gamma$ (including $\widetilde{S}_\Gamma$ itself). Moreover, the map $\Phi$ is equivariant with respect to the natural actions of $G_\Gamma$. \begin{rk} The map $\Phi$ can be reinterpreted as follows: Lemma~\ref{lemma:confined} shows that given any two standard subcomplexes $Y_1,Y_2\subseteq\widetilde{S}_\Gamma$, one has $\partial_R Y_1\cap\partial_R Y_2=\partial_R(Y_1\cap Y_2)$ (viewed as subsets of $\partial_R\widetilde{S}_\Gamma$), and the map $\Phi$ then sends a point $\xi\in\partial_R\widetilde{S}_\Gamma$ to the smallest standard subcomplex $Y$ such that $\xi\in\partial_RY$. \end{rk} We equip the countable set $\mathbb{S}$ with the discrete topology. \begin{lemma} \label{lem:borel} The map $\Phi$ is Borel. \end{lemma} \begin{proof} Let $Y\in \mathbb {S}$ be a standard subcomplex, and let $x\in \widetilde{S}_\Gamma$. Then a point $\xi\in\partial_R\widetilde{S}_\Gamma$ belongs to $\Phi^{-1}(Y)$ if and only if it is represented by a combinatorial geodesic ray based at $x$ which is virtually contained in $Y$, but not virtually contained in any proper standard subcomplex of $Y$. Thus $\Phi^{-1}(Y)=\partial_R Y\setminus(\bigcup_{Z\in \mathbb {S}, Z\subsetneq Y}\partial_R Z)$. The lemma follows as the subspaces $\partial_R Z$ and $\partial_R Y$ are closed subsets of $\partial_R \widetilde{S}_\Gamma$. \end{proof} Now we recall some standard facts to prepare for the next lemma. As $\widetilde{S}^{(1)}_\Gamma$ is the Cayley graph of $G_\Gamma$ with respect to its standard generating set, we label each edge of $\widetilde{S}_\Gamma$ by a vertex of $\Gamma$. Recall that an edge is \emph{dual} to a hyperplane if this hyperplane intersects the edge in its midpoint. Note that if two edges are dual to the same hyperplane, then they have the same label. Hence each hyperplane has a well-defined label. Note that two hyperplanes having the same label are either equal or disjoint. Let $h$ be a hyperplane in $\widetilde{S}_\Gamma$. Then the smallest subcomplex of $\widetilde{S}_\Gamma$ containing $h$ splits as a product $h\times [0,1]$. Note that both $h\times\{0\}$ and $h\times\{1\}$ are standard subcomplexes of type $\lk(v)$, where $v\in \Gamma$ is the label of $h$. Thus it is natural to consider \emph{standard subcomplexes} of $h$ and their types, which correspond to standard subcomplexes of $h\times\{0\}$ (or $h\times\{1\}$). Let $X_1$ and $X_2$ be standard subcomplexes of $\widetilde{S}_\Gamma$. It is a standard fact that one can find a standard subcomplex $B\subseteq X_1$ such that $\calh(B)=\calh(X_1)\cap \calh(X_2)$. Actually, we can take $B$ to be the subset of $X_1$ made of all points whose distance to $X_2$ is equal to $d(X_1,X_2)$. Then $B$ is a standard subcomplex of $X_1$ (see \cite[Lem.\,3.1]{Hua}) and $\calh(B)=\calh(X_1)\cap \calh(X_2)$ (see \eg \cite[Lem.\,2.14]{Hua3}). By the discussion in the previous paragraph, the fact still holds if we require $X_1$ and $X_2$ to be hyperplanes. \begin{lemma}\label{lem:join subgroup} Let $\Gamma$ be a finite simple graph. Let $r$ be a combinatorial geodesic ray in $\widetilde{S}_\Gamma$. Then $r$ represents a point in $\partial_\reg\widetilde{S}_\Gamma$ if and only if $r$ is not virtually contained in a join standard subcomplex of $\widetilde{S}_\Gamma$. \end{lemma} \begin{proof} First we show that if $r$ is virtually contained in a join standard subcomplex~$X$ representing $\xi\in \partial_R \widetilde{S}_\Gamma$, then $\xi$ is not regular. To see this, let $X_1$ and $X_2$ be two unbounded standard subcomplexes of $\widetilde{S}_\Gamma$ such that $X$ splits as $X=X_1\times X_2$. We assume without loss of generality that $r\subset X$, and the image of $r$ under the natural projection $X\to X_1$ is unbounded. Take halfspaces $h_1,h_2\in U_\xi$ such that $\partial h_i\in \mathcal H(X_1)\cap\mathcal H(r)$. By definition of a regular point, it suffices to show that if $k$ is a halfspace with $k\in U_\xi$ and $k\subset h_1\cap h_2$, then $\partial k$ and $\partial h_i$ are not strongly separated for any $i\in\{1,2\}$. Note that $\partial k\cap X\neq \emptyset$, as otherwise we will have $r\subset k$, and therefore $r\subset h_1$, contradicting that $\partial h_1\in \mathcal H(r)$. As $\mathcal H(X)=\mathcal H(X_1)\sqcup\mathcal H(X_2)$, we have $\partial k\in \mathcal H(X_1)$ or $\partial k\in \mathcal H(X_2)$. The latter is not possible as it will imply $\partial k\cap\partial h_1\neq\emptyset$, contradicting $k\subset h_1$. Now $\partial k\in \mathcal H(X_1)$ implies that $\partial k$ and $\partial h_1$ cannot be strongly separated, because any hyperplane in $\mathcal H(X_2)$ will intersect both~$\partial k$ and $\partial h_1$.\enlargethispage{\baselineskip}% Now we suppose that $r$ does not represent a regular point. Let $\{\mathfrak h_i\}_{i\in\mathbb{N}}$ be an infinite collection of hyperplanes crossed by $r$ such that for every $i\in\mathbb{N}$, the hyperplane $\mathfrak h_{i}$ separates $\mathfrak h_{i-1}$ from $\mathfrak h_{i+1}$ (such a collection can be found by looking at hyperplanes with the same label, though this is a general fact for geodesic rays in finite dimensional CAT(0) cube complexes). By \cite[Prop.\,7.5]{Fer}, up to passing to a subcollection, we can assume that for all $i,j\in\mathbb{N}$, the hyperplanes $\mathfrak h_i$ and $\mathfrak h_j$ are not strongly separated. For every $i\in\mathbb{N}$, let $B_i\subseteq\mathfrak h_1$ be a standard subcomplex of $\mathfrak h_1$ with $\calh(B_i)=\calh(\mathfrak h_1)\cap \calh(\mathfrak h_i)$. For every $i_10$ such that for all sufficiently large $n\in\mathbb{N}$, one has $f(n)\le g(Cn)$). Moreover we can assume that $A_t$ is torsion-free \cite[Rem.\,1.1]{BE}. Let $v\in V\Gamma$ be a vertex, and for every $t\in (\alpha,1)$, let $G_t$ be the graph product over~$\Gamma$ such that the vertex group $G_v$ is $A_t$ and all other vertex groups are $\mathbb Z$. Then $G_t$ is torsion-free \cite[Cor.\,5.9]{antolin2015tits}. By Corollary~\ref{cor:flexibility}, all groups $G_t$ are measure equivalent, so we are left with proving that they are pairwise not commensurable (hence not commensurable up to finite kernel). Given $t\in (\alpha,1)$, we claim that every amenable subgroup $H\subseteq G_t$ is contained in an amenable subgroup $K$ which is either finitely generated and virtually abelian, or equal to the direct product of $A_t$ and a finitely generated virtually abelian subgroup of $G_t$. We prove this by induction on the number of vertices of $\Gamma$. Let $\Gamma=\Gamma_0\circ \Gamma_1\circ\cdots\circ\Gamma_k$ be the de Rham decomposition of $\Gamma$ with $\Gamma_0$ being the clique factor. The claim is trivial if $\Gamma=\Gamma_0$. Now we assume $\Gamma\neq\Gamma_0$. If this decomposition is non-trivial, we know the claim is true for each $G_{\Gamma_i}$ by induction. As $G_{\Gamma}=\oplus_{i=0}^kG_{\Gamma_i}$, the claim follows by considering the projection of $H$ to each $G_{\Gamma_i}$. If the de Rham decomposition has only one factor, then \cite[Th.\,4.1]{antolin2015tits} ensures that either $H$ is virtually cyclic (and we are done), or else $H$ is contained in $G_{\Gamma'}$ for some proper subgraph $\Gamma'$ of $\Gamma$, in which case the claim follows by induction. Now take $t