We will need the following theorem which comes from the work of Monod–Shalom (see <cit.>). Its formulation comes from [71].

[Monod–Shalom, see <cit.>]
Let $G,H$ be countable groups. Let $G\actson X$ and $H\actson Y$ be free ergodic measure-preserving actions on standard probability spaces. Let $f:U\to V$ be a stable orbit equivalence from $G\actson X$ to $H\actson Y$ (where $U\subseteq X$ and $V\subseteq Y$ are positive measure Borel subsets). Let $c:G\times X\to H$ be an SOE cocycle associated to $f$.

Let $A\unlhd G$ and $B\unlhd H$ be normal subgroups acting ergodically on $X,Y$, and assume that for every $x\in U$, one has $f((A\cdot x)\cap U)= (B\cdot f(x))\cap V$.

Then there exist a group isomorphism $\alpha:G/A\to H/B$, and a cocycle $c':G\times X\to H$ such that

* $c'$ is $H$-cohomologous to $c$;

* for every $g\in G$ and almost every $x\in X$, if $x,g\cdot x\in U$, then $c'(g,x)=c(g,x)$;

* for every $g\in G$ and almost every $x\in X$, one has $c^\prime(g,x)\in\alpha(gA)$.

§.§.§ End of the proof

We now complete our proof of Theorem <ref>. We follow the proof of <cit.> very closely, with only a few departures in the presentation (in particular, we have decided to treat the cases corresponding to <cit.> and <cit.> in a harmonized way).

We will prove the following: if $G\actson X$ and $H\actson Y$ are vertexwise ergodic free, probability measure-preserving actions, and if $c:G\times X\to H$ is an SOE cocycle, then there exists a vertex group $G_v\subseteq G$ such that $c_{|G_v\times X}$ is $H$-cohomologous to a group homomorphism $G_v\to H$. In view of Lemma <ref>, this will be enough to conclude our proof.

By collapsing some subgraphs if necessary, we can write $G$ and $H$ as graph products over finite simple graphs $\bar\Gamma_G$ and $\bar\Gamma_H$ that satisfy the following two conditions:

* $\bar\Gamma_G$ and $\bar\Gamma_H$ are joins (possibly with only one join factor) of strongly reduced graphs in the sense of Definition <ref>, de:strongly-reduced.

* The graph $\bar\Gamma_G$ does not contain any vertex $v$ such that $\bar\Gamma_G=\st(v)$, and neither does $\bar\Gamma_H$.

The vertex groups for the graph product structure of $G$ over $\bar\Gamma_G$ contain those for its graph product structure over $\Gamma_G$, so in particular the action of $G\actson X$ is still vertexwise ergodic with respect to the graph product structure over $\bar\Gamma_G$ (and likewise for $H$). Therefore $\bar\Gamma_G$ and $\bar\Gamma_H$ still satisfy all assumptions from Theorem <ref>. Thus, from now on, we will assume without loss of generality that $\Gamma_G$ and $\Gamma_H$ are joins of  strongly reduced graphs to start with.

Let $U\subseteq X$ and $V\subseteq Y$ be the positive measure Borel subsets defined as the source and range of the stable orbit equivalence $f$. In other words $f$ is a measure-scaling isomorphism $U\to V$ that sends orbits (restricted to $U$) to orbits (restricted to $V$). After renormalizing the measures on $U$ and $V$ to probability measures, the groupoid $(G\ltimes X)_{|U}$ is isomorphic (via $f$) to $(H\ltimes Y)_{|V}$. We denote by $\calg$ this common groupoid, which naturally comes equipped with two cocycles $\rho_G:\calg\to G$ and $\rho_H:\calg\to H$, which are action-like in view of Lemma <ref>, lemma:action-like-1 and Remark <ref>, rk:action-like-restriction. We view $\calg$ as a measured groupoid over $U$.

Our proof has three steps.
Step 1. We show that there exist

* parabolic subgroups $P_G^1,P_G^2\subseteq G$, and $P_H^1,P_H^2\subseteq H$,

* normal infinite parabolic subgroups $C_G^i\unlhd P_G^i$ and $C_H^i\unlhd P_H^i$ for every $i\in\{1,2\}$, such that

* $C_G^1$ and $C_G^2$ intersect trivially and commute,

* $C_H^1$ and $C_H^2$ intersect trivially and commute,

* for every $i\in\{1,2\}$, one has $P_G^i=C_G^i\times (C_G^i)^{\perp}$ and $P_H^i=C_H^i\times (C_H^i)^{\perp}$,

* a positive measure Borel subset $W\subseteq U$, such that

* for every $i\in\{1,2\}$, one has $\rho_G^{-1}(P_G^i)_{|W}=\rho_H^{-1}(P_H^i)_{|W}$,

* for every $i\in\{1,2\}$, one has $\rho_G^{-1}(C_G^i)_{|W}=\rho_H^{-1}(C_H^i)_{|W}$.

The proof of Step 1 decomposes into three cases.

Case 1: The graph $\Gamma_G$ decomposes non-trivially as a join.

Write $\Gamma_G=\Gamma_1\circ\dots\circ\Gamma_k$, where each $\Gamma_j$ does not split further as a join – notice that no $\Gamma_j$ is reduced to one vertex, as $\Gamma_G$ is not contained in the star of a vertex by assumption. Lemma <ref> ensures that $\Gamma_H$ also splits as a join: we write $\Gamma_H=\Lambda_1\circ\dots\circ\Lambda_n$ in such a way that no $\Lambda_j$ further splits as a join. Here again our assumption on $\Gamma_H$ ensures that no $\Lambda_j$ is reduced to one vertex.

We let $P_G^1=P_G^2=G$, and $P_H^1=P_H^2=H$, and we let $C_G^1$ and $C_G^2$ be two different factors of $G$ (so $C_G^1$ and $C_G^2$ intersect trivially and commute).

We now apply Lemma <ref>, lemma:factor-2. This lemma gives a characterization of being of factor type with no reference to any action-like cocycle, and therefore it ensures that there exists a positive measure Borel subset $W\subseteq U$, and distinct factors $C_H^1,C_H^2$ of $H$, such that $\rho_G^{-1}(C_G^i)_{|W}=\rho_H^{-1}(C_H^i)_{|W}$ for every $i\in\{1,2\}$. This completes Step 1 in this case.

Case 2: The graph $\Gamma_G$ does not decompose non-trivially as a join, and $G$ contains a maximal product parabolic subgroup $P_G$ with trivial clique factor.

We let $P_G^1=P_G^2:=P_G$, and let $C_G^1$ and $C_G^2$ be two distinct factors of $P_G$ (so $C_G^1$ and $C_G^2$ commute and intersect trivially). Notice that the maximality of $P_G$ ensures that $P_G=C_G^i\times (C_G^i)^{\perp}$ for every $i\in\{1,2\}$.

By Lemma <ref>, the graph $\Gamma_H$ does not decompose non-trivially as a join either. And since $\Gamma_G$ and $\Gamma_H$ do not contain isolated cliques (because they are connected), there is no subgroupoid of $\calg$ of isolated clique type for either $\rho_G$ or $\rho_H$, even in restriction to a positive measure subset. Lemma <ref>, lemma:maximal-product therefore implies that we can find a positive measure Borel subset $V\subseteq U$, and a maximal product parabolic subgroup $P_H\subseteq H$, such that $\rho^{-1}(P_G)_{|W}=\rho_H^{-1}(P_H)_{|W}$. We let $P_H^1=P_H^2:=P_H$.

We now apply Lemma <ref>, lemma:factor. This lemma gives a characterization of being of factor type with no reference to any action-like cocycle, and therefore it ensures that, up to replacing $W$ by a positive measure Borel subset, there exist two distinct factors $C_H^1,C_H^2$ of $P_H$ such that $\rho^{-1}(C_H^i)_{|W}=\rho^{-1}(C_G^i)_{|W}$ for every $i\in\{1,2\}$. And the maximality of $P_H$ ensures that $P_H=C_H^i\times (C_H^i)^{\perp}$ for every $i\in\{1,2\}$.  This completes the proof of Step 1 in this case.

Case 3: The graph $\Gamma_G$ does not decompose non-trivially as a join, and $G$ does not contain any maximal product parabolic subgroup $P_G$ with trivial clique factor.

In this case, Lemma <ref> ensures that $G$ contains two distinct maximal product parabolic subgroups $P_G^1,P_G^2$ whose clique factors $C_G^1,C_G^2$ are non-trivial and commute; in this case $C_G^1\cap C_G^2=\{1\}$ by Lemma <ref>. For every $i\in\{1,2\}$, the maximality of $P_G^i$ ensures that $P_G^i=C_G^i\times (C_G^i)^{\perp}$.

Again by Lemma <ref>, the graph $\Gamma_H$ does not decompose non-trivially as a join either.

Since $\Gamma_G$ has no isolated clique, Lemma <ref>, lemma:maximal-product gives a characterization of being of maximal product type that does not depend on any action-like cocycle. Since $\Gamma_H$ has no isolated clique, Lemma <ref>, applied to the cocycle $\rho_H$, therefore ensures that there exist maximal product parabolic subgroups $P_H^1,P_H^2$ and a positive measure Borel subset $W\subseteq U$ such that for every $i\in\{1,2\}$, we have $\rho_G^{-1}(P_G^i)_{|W}=\rho_H^{-1}(P_H^i)_{|W}$. Notice that, since $P_G^1\neq P_G^2$, we have $\rho_G^{-1}(P_G^1)_{|W}\neq \rho_G^{-1}(P_G^2)_{|W}$ (see Lemma <ref>, lemma:inclusion-parabolics), and therefore $P_H^1\neq P_H^2$.

By Lemma <ref>, the fact that $P_G^i$ has a non-trivial clique factor is characterized by a groupoid-theoretic property of $\rho_G^{-1}(P_G^i)$ which is independent from the action-like cocycle. It follows that $P_H^1$ and $P_H^2$ also have non-trivial clique factors, which we denote respectively by $C_H^1$ and $C_H^2$. By Lemma <ref>, we have $C_H^1\cap C_H^2=\{1\}$.

Likewise, Lemma <ref>, which characterizes the clique factor by a groupoid-theoretic property, ensures that, up to replacing $W$ by a positive measure Borel subset, we have $\rho_G^{-1}(C_G^i)_{|W}=\rho_H^{-1}(C_H^i)_{|W}$ for every $i\in\{1,2\}$.

Since $C_G^1$ and $C_G^2$ commute, we deduce that $\rho_G^{-1}(C_G^1)_{|W}$ and $\rho_G^{-1}(C_G^2)_{|W}$ normalize each other, and thus the groups $C_H^1$ and $C_H^2$ commute (as in the proof of Lemma <ref>, lemma:adjacency).

Finally, the maximality of $P_H^1$ and $P_H^2$ as product parabolic subgroups ensures that $P_H^1=C_H^1\times (C_H^1)^{\perp}$ and $P_H^2=C_H^2\times (C_H^2)^{\perp}$.

Step 2. We show that for every $i\in\{1,2\}$, there exist

* a group homomorphism $\alpha_i:P_G^i\to P_H^i$, which restricts to an isomorphism from $C_G^i$ to $C_H^i$, with $(\alpha_1)_{|C_G^1\times C_G^2}=(\alpha_2)_{|C_G^1\times C_G^2}$, and

* measurable maps $\varphi_i:X\to H$ and $\kappa_i:P_G^i\times X\to C_H^i$,

such that
\begin{equation}\label{eq:cocycle-2}
c(g,x)=\varphi_i(g \cdot x)\alpha_i(g)\kappa_i(g,x)\varphi_i(x)^{-1}, ~~~\forall g\in P_G^i, ~\text{a.e.~} x\in X.
\end{equation}

Informally, this means that we have untwisted the cocycle on $P_G^i$ to a group homomorphism, “up to an error in $C_H^i$”, for every $i\in\{1,2\}$.

Up to replacing $c$ by a cohomologous cocycle, we can assume that $c:G\times X\to H$ is an SOE cocycle associated to $f_{|W}$, and that whenever $x,g\cdot x\in W$, the element $c(g,x)$ is the unique element $h\in H$ such that $f(g\cdot x)=h\cdot f(x)$. Recall also that $P_G^i$ acts ergodically on $X$, and that $P_G^i$ and $P_H^i$ have the same orbits in restriction to $W$ (because $\rho_G^{-1}(P_G^i)_{|W}=\rho_H^{-1}(P_H^i)_{|W}$). It follows that $c$ is cohomologous to a cocycle $c_i:G\times X\to H$ having the following two properties (see e.g. the discussion at the end of Section <ref>, sec:cocycles):

* $(c_i)_{|P_G^i\times X}$ is an SOE cocycle associated to $f_{|W}$, viewed as a stable orbit equivalence between the actions $P_G^i\actson X$ and $P_H^i\actson Y$,

* $c(g,x)=c_i(g,x)$ whenever $x,g\cdot x\in W$.

For every $i\in\{1,2\}$, the normal subgroups $C_G^i\unlhd P_G^i$ and $C_H^i\unlhd P_H^i$ act ergodically on $X$ and on $Y$. In addition $C_G^i$ and $C_H^i$ have the same orbits in restriction to $W$ (because $\rho_G^{-1}(C_G^i)_{|W}=\rho_H^{-1}(C_H^i)_{|W}$). We can therefore apply Theorem <ref> to the stable orbit equivalence $f_{|W}$, and deduce that, up to replacing $c_i$ by a cohomologous cocycle, we can assume that the following hold:

* there is a group isomorphism $\bar\alpha_i:P_G^i/C_G^i\to P_H^i/C_H^i$ such that for every $g\in P_G^i$ and almost every $x\in X$, one has $c_i(g,x)\in\bar\alpha_i\left(gC_G^i\right)$;

* one has $c_i(g,x)=c(g,x)$ whenever $x,g \cdot x\in W$.

Let $r_i:P_H^i\to (C_H^i)^{\perp}$ be the retraction (sending $C_H^i$ to $\{1\}$). Let
\[c'_i=(r_i\circ c_i)_{|(C_G^i)^{\perp}\times X},\]
a cocycle with values in $(C_H^i)^{\perp}$. Through the natural isomorphisms $P_G^i/C_G^i\to (C_G^i)^{\perp}$ and $P_H^i/C_H^i\to (C_H^i)^{\perp}$ (given by the choice of the unique representative), $\bar\alpha_i$ yields an isomorphism $\alpha_i:(C_G^i)^{\perp}\to (C_H^i)^{\perp}$ such that

\begin{equation}\label{eq:alpha}
c'_i(g,x)=\alpha_i(g),~~~\forall g\in (C_G^i)^{\perp},~ \text{a.e~} x\in X.
\end{equation}

Recall that $C_G^1\subseteq (C_G^2)^{\perp}$ (because the parabolic subgroups $C_G^1$ and $C_G^2$ intersect trivially and commute), and $C_H^1\subseteq (C_H^2)^\perp$. We will now prove the following claim.

We have $\alpha_2(C_G^1)=C_H^1$.

We first prove that $\alpha_2(C_G^1)\subseteq C_H^1$. So let $g\in C_G^1$. By Poincaré recurrence, for almost every $x\in W$, there exists $n>0$ such that $g^n\cdot x\in W$. Since $x,g^n\cdot x\in W$, our definition of the cocycles $c_i$ ensures that  $c_2(g^n,x)=c(g^n,x)$. Since $C_G^1$ and $C_H^1$ have the same orbits in restriction to $W$, we deduce that $c_2(g^n,x)\in C_H^1$, so $c_2(g^n,x)=c'_2(g^n,x)$. Since $g^n$ belongs to $C_G^1$, which is contained in $(C_G^2)^{\perp}$, our definition of $\alpha_2$ (see Equation (<ref>)) shows that $c_2(g^n,x)=\alpha_2(g)^n$. It follows that $\alpha_2(g)^n\in C_H^1$, and Lemma <ref>, lemma:root implies that $\alpha_2(g)\in C_H^1$. This proves that $\alpha_2(C_G^1)\subseteq C_H^1$.

Conversely, let us now prove that $C_H^1\subseteq\alpha_2(C_G^1)$. So let $h\in C_H^1$. By Poincaré recurrence again, for almost every $y\in W$, there exists $m>0$ such that $h^m\cdot f(y)\in f(W)$. Since the orbits of $C_G^1$ and of $C_H^1$ coincide in restriction to $W\approx f(W)$, we can find $g\in C_G^1$ such that

$c_2(g,y)=h^m$. We deduce that

\begin{equation*}
\alpha_2(g)=c'_2(g,y)=r_2\circ c_2(g,y)= r_2(h^m)=h^m.
\end{equation*}

Since $\alpha_2:(C_G^2)^{\perp}\to (C_H^2)^{\perp}$ is an isomorphism, we can also find $g_0\in (C_G^2)^\perp$ such that $\alpha_2(g_0)=h$. Then $\alpha_2(g_0^m)=\alpha_2(g)$, so $g_0^m=g$. Lemma <ref> implies that $g_0\in C_G^1$, and we are done.

The above claim shows that $\alpha_2$ restricts to an isomorphism from $C_G^1$ to $C_H^1$. Symmetrically $\alpha_1$ restricts to an isomorphism from $C_G^2$ to $C_H^2$.

We now extend $\alpha_1$ to $P_G^1$ by letting $(\alpha_1)_{|C_G^1}=(\alpha_{2})_{|C_G^1}$. Likewise, we extend $\alpha_2$ to $P_G^2$ by letting $(\alpha_2)_{|C_G^2}=(\alpha_1)_{|C_G^2}$. With this extension,
for every $i\in\{1,2\}$, we have

\begin{equation}\label{eq:cocycle}
c_i(g,x)\in\alpha_i(g)C_H^i \quad \forall g\in P_G^i, \ \text{a.e.}\ x\in X.
\end{equation}

* if $g\in C_G^i$, then $c_i(g,x)$ belongs to $\bar\alpha_i(C_G^i)=C_H^i$, and by our extension of $\alpha_i$ to $C^i_G$ and <ref> we get $\alpha_i(g)\in C_H^i$;

* if $g\in (C_G^i)^{\perp}$, this comes from the fact that $c_i(g,x)\in \bar\alpha_i(gC^i_G)$ and the definition of $\alpha_i$;

* in general, every element $g\in P_G^i$ can be decomposed as $g=g_1g_2$ with $g_1\in (C_G^i)^{\perp}$ and $g_2\in C_G^i$, and Equation (<ref>) follows from the cocycle relation.

At this point, for every $i\in\{1,2\}$, using Equation (<ref>) together with the fact that $c$ and $c_i$ are $H$-cohomologous, we have measurable maps $\varphi_i:X\to H$ and $\kappa_i:P_G^i\times X\to C_H^i$ such that for every $g\in P_G^i$ and a.e. $x\in X$, we have

\begin{equation}\label{eq:cocycle-4}
c(g,x)=\varphi_i(g \cdot x)c_i(g,x)\varphi_i(x)^{-1},
\end{equation}

and $c_i(g,x)=\alpha_i(g)\kappa_i(g,x)$, so <ref> is verified.

Step 3. We show that $c_{|C_G^1\times X}$ is $H$-cohomologous to the homomorphism $\alpha_1$. This will be done by comparing the two ways of untwisting the cocycle $c$ given by Equation (<ref>), for $i=1$ and $i=2$.

Let $X^*=\sqcup_{j\in J}X_j$ be a partition of a conull Borel subset $X^*\subseteq X$ into at most countably many Borel subsets, such that for every $j\in J$, the maps $\varphi_1$ and $\varphi_2$ are constant when restricted to $X_j$, with respective values denoted by $\gamma_{1,j},\gamma_{2,j}$, and Equation (<ref>) holds everywhere on $X^*$.

Let $h\in C_H^1$ be such that $h$ does not belong to any proper parabolic subgroup of $C_H^1$, and neither do its proper powers (see e.g. the proof of <ref> for the existence of $h$). Let $g\in C_G^1$ such that $\alpha_1(g)=\alpha_2(g)=h$. For every $j\in J$, let $x\in X_j$ and $k_j>0$ be such that $g^{k_j}\cdot x\in X_j$ (these exist by Poincaré recurrence). Since $C_G^1\subseteq P_G^1\cap P_G^2$, we can use Equation (<ref>) with both $i=1$ and $i=2$ to write

\begin{equation}\label{eq:cocycle-3}
\end{equation}

Let $r_2:H\to C_H^2$ be the natural retraction (see the last paragraph of Section <ref>, sec:first-facts). Since $C_H^1\subseteq (C_H^2)^{\perp}$ (by Step $1$), we have $r_2(C_H^1)=\{1\}$. In particular remark that $\alpha_1(g), \alpha_2(g),\kappa_1(g^{k_j},x)\in C^1_H$. By applying $r_2$ to the rightmost equality of Equation (<ref>), we deduce that $\kappa_2(g^{k_j},x)=1$ for every $x\in X^*$.

Let $\beta_j=\gamma_{2,j}^{-1}\gamma_{1,j}$. Using that $\kappa_2(g^{k_j},x)=1$, Equation (<ref>) rewrites as

\[\beta_{j}\alpha_1(g)^{k_j}\kappa_1(g^{k_j},x)\beta_j^{-1}=\alpha_2(g)^{k_j}.\]

Therefore $\beta_j C_H^1\beta_j^{-1}\cap C_H^1$ contains $\alpha_2(g)^{k_j}$, so our choice of $g$ ensures that $\beta_j C_H^1\beta_j^{-1}\cap C_H^1$ cannot be a proper parabolic subgroup of $C_H^1$. It follows that $\beta_jC_H^1\beta_j^{-1}\supseteq C_H^1$. The second point of Proposition <ref> implies that in fact  $\beta_jC_H^1\beta_j^{-1}=C_H^1$, i.e. $\beta_j$ belongs to the normalizer of $C_H^1$.

In other words, for every $x\in X_j$, the constant value of $\varphi_2(x)^{-1}\varphi_1(x)$ belongs to $C_H^1\times (C_H^1)^{\perp}$. Since this is true for every $j\in J$, there exist measurable maps $\eta_1:X\to C_H^1$ and $\mu_1:X\to (C_H^1)^{\perp}$ such that for every $x\in X^*$, one has $\varphi_2(x)=\varphi_1(x)\eta_1(x)\mu_1(x)$. Now, by rewriting Equation (<ref>) with $i=2$, we deduce that for every $x\in X^*$ and every $g\in P_G^1$, one has
\[c(g,x)=\varphi_1(g \cdot x)\eta_1(g \cdot x)\mu_1(g \cdot x)\alpha_2(g)\kappa_2(g,x)\mu_1(x)^{-1}\eta_1(x)^{-1}\varphi_1(x)^{-1},\]
in other words

\begin{equation}\label{eq:4}
\varphi_1(g\cdot x)^{-1}c(g,x)\varphi_1(x)=\eta_1(g\cdot x)\mu_1(g \cdot x)\alpha_2(g)\kappa_2(g,x)\mu_1(x)^{-1}\eta_1(x)^{-1}.
\end{equation}

Now, when $g\in C_G^1$, using Equations (<ref>) and (<ref>), the leftmost term belongs to $C_H^1$. We now apply the retraction $r_1:H\to C_H^1$ (which sends $(C_H^1)^{\perp}$ to $\{1\}$, so in particular sends $C_H^2$ to $\{1\}$), to Equation (<ref>). We deduce that
\[\varphi_1(g \cdot x)^{-1}c(g,x)\varphi_1(x)=\eta_1(g \cdot x)\alpha_2(g)\eta_1(x)^{-1}, \quad \forall g\in C_G^1,~\text{a.e.}~x\in X .\]
This proves that there exists a measurable map $\psi:X\to H$ and a homomorphism $\alpha_1:C_G^1\to H$ such that for every $g\in C_G^1$ and a.e. $x\in X$, one has
\[c(g,x)=\psi(g \cdot x)\alpha_1(g)\psi(x)^{-1}.\]
In other words, we have untwisted the cocycle $c$ to a group homomorphism on some vertex group of $G$. An application of Lemma <ref> then concludes our proof of Theorem <ref>.

§.§ An application to von Neumann algebras

When combined with a uniqueness theorem for Cartan subalgebras established by Chifan and Kunnawalkam Elayavalli in the framework of Popa's deformation/rigidity theory, Theorem <ref> has the following consequence at the level of von Neumann algebras. We refer to <ref>, sec:von-neumann for the needed definitions.

Let $\Gamma_G,\Gamma_H$ be two connected finite simple graphs, with no vertex joined to every other vertex. Let $G,H$ be graph products with countably infinite torsion-free vertex groups over $\Gamma_G,\Gamma_H$, respectively. Let $G\actson X$ and $H\actson Y$ be two free, vertexwise ergodic, measure-preserving actions on standard probability spaces. Assume that either

* $G\actson X$ is vertexwise weakly mixing, or

* $G\actson X$ is vertexwise totally ergodic, or

* $H$ has the unique root property.

If the $\mathrm{II}_1$ factors $L(G\actson X)$ and $L(H\actson Y)$ are isomorphic (or more generally, if $L(H\actson Y)$ is isomorphic to an amplification $L(G\actson X)^t$ with $t\in (0,1]$), then the two actions are conjugate.

By <cit.>, the subalgebra $L^\infty(Y)$ is, up to unitary conjugation, the unique Cartan subalgebra of $L(H\actson Y)$. Let $U\subseteq X$ be a Borel subset of measure $t$, and let $\calr$ be the restriction to $U$ of the orbit equivalence relation for the $G$-action on $X$. Then $L(G\actson X)^t\simeq L(\calr)$, and $L^\infty(U)$ is a Cartan subalgebra. So there exists an isomorphism $\theta:L(H\actson Y)\to L(G\actson X)^t$ sending $L^\infty(Y)$ to $L^\infty(U)$. By [113, 43], this implies that the orbit equivalence relation of the $H$-action on $Y$ is isomorphic to $\calr$. In particular, there exists a stable orbit equivalence with compression $\frac{1}{t}\ge 1$ from $G\actson X$ to $H\actson Y$. Theorem <ref> thus implies that the two actions are conjugate.

As in Corollary <ref>, we could also derive the triviality of the fundamental group of any action $G\actson X$ which is either vertexwise weakly mixing (or vertexwise totally ergodic), with $G$ torsion-free, or vertexwise ergodic, with $G$ having the unique root property. Likewise, the associated von Neumann algebra $L(G\actson X)$ has trivial fundamental group.

§ SUPERRIGIDITY

The goal of this section is to show the following theorem.

Let $\Gamma$ be a connected finite simple graph, with no vertex joined to every other vertex. Let $G$ be a graph product with countably infinite vertex groups over $\Gamma$, satisfying the unique root property, and let $H$ be an arbitrary countable group. Let $G\actson X$ and $H\actson Y$ be two free, ergodic, measure-preserving actions on standard probability spaces. Assume that

* one of the following two options holds:

* either $G\actson X$ is vertexwise ergodic, and $G$ has the unique root property,

* or $G\actson X$ is vertexwise totally ergodic, and $G$ is torsion-free,

* and $H\actson Y$ is mildly mixing.

If the actions $G\actson X$ and $H\actson Y$ are stably orbit equivalent, then they are virtually conjugate.

Recall that a group is ICCICC (infinite conjugacy classe) (standing for infinite conjugacy classesInfinite conjugacy classes (ICC)) if the conjugacy class of every non-trivial element of $G$ is infinite. We will need the following lemma.

If $G$ is a graph product over a finite simple graph $\Gamma$ such that no vertex is joined to every other vertex, then $G$ is ICC.

Write $\Gamma=\Gamma_1\circ\dots\circ\Gamma_k$, where each $\Gamma_j$ is irreducible (possibly $k=1$). It is enough to prove that each $G_{\Gamma_j}$ is ICC. So let $j\in\{1,\dots,k\}$. Our assumption that $\Gamma$ is not equal to the star of one of its vertices implies that $\Gamma_j$ is not a clique. By <cit.>, the group $G_{\Gamma_j}$ is acylindrically hyperbolic (equivalently by <cit.>, the group $G$ contains a proper infinite hyperbolically embedded subgroup in the sense of <cit.>). In addition, by <cit.>, the group $G_{\Gamma_j}$ has no non-trivial finite normal subgroup. So by <cit.>, the group $G_{\Gamma_j}$ is ICC. So $G$ is ICC, as desired.

We are now in position to prove Theorem <ref>.

We first assume that the action $G\actson X$ is vertexwise ergodic and $G$ has the unique root property. In the terminology from <cit.>, vertexwise ergodicity of $G\actson X$ is exactly $\calf$-ergodicity, with $\calf$ the collection of all vertex groups of $G$. And Theorem <ref> shows that $(G,\calf)$ is strongly cocycle rigid in the sense of <cit.>. Since $G$ is ICC (Lemma <ref>), the conclusion therefore follows from <cit.>.

We now assume that the action $G\actson X$ is vertexwise totally ergodic and $G$ is torsion-free. In the terminology from <cit.>, vertexwise total ergodicity is exactly $\calf'$-ergodicity, with $\calf'$ the collection of all infinite subgroups of vertex groups of $G$. As above, Theorem <ref> shows that $(G,\calf')$ is strongly cocycle rigid in the sense of <cit.>, and the conclusion follows from <cit.>.

Again, using the uniqueness of the Cartan subalgebra, up to unitary conjugacy, for $L(G\actson X)$, Theorem <ref> can be reformulated as a $W^*$-rigidity statement.

Let us finally mention the following corollary to Theorem <ref>, which states the rigidity of graph product actions within the class of mixing group actions.

Let $G,H$ be two torsion-free countable groups. Let $G\actson X$ and $H\actson Y$ be two mixing measure-preserving actions on standard probability spaces. Assume that $G$ splits as a graph product over a connected finite simple graph $\Gamma$, with no vertex joined to every other vertex.

If the two actions $G\actson X$ and $H\actson Y$ are orbit equivalent, then they are conjugate.

Being mixing, the two actions are in particular totally ergodic (and in particular the action $G\actson X$ is vertexwise totally ergodic) and mildly mixing. They are also essentially free by <cit.>. Therefore Theorem <ref> applies and shows that the two actions are virtually conjugate. But $G$ and $H$ have no finite normal subgroup, and an induced action of the form $\Ind_{G'}^G(X)$ is never mixing if $G'\subsetneq G$ (because $G'$ does not act ergodically on $\Ind_{G'}^G(X)$). Therefore the actions $G\actson X$ and $H\actson Y$ are conjugate.

Measure equivalence rigidity of some graph products

This chapter deals with measure equivalence rigidity of graph products. Contrary to the previous chapter, we will not make any ergodicity assumption on the couplings we consider; but on the other hand, we will impose some strong rigidity conditions on the vertex groups.

A general question is the following: to what extent is it true that if the vertex groups are rigid in measure equivalence, then the graph product is rigid in measure equivalence?

We will not answer this question in full generality, but we will give a sufficient criterion in Section <ref>, and an example in Section <ref>. Our example is, to our knowledge, the first example of a group which is rigid in measure equivalence, but not in quasi-isometry, among torsion-free groups. On this matter, let us make two remarks.

* It is not true that a graph product of quasi-isometrically rigid groups is quasi-isometrically rigid. A basic observation is that in a graph product $G$, changing a vertex group $G_v$ by $G_v\times\mathbb{Z}/2\mathbb{Z}$ can often yield a group which is quasi-isometric but not commensurable to $G$. So the phenomena established in this chapter are radically different from their counterparts in quasi-isometry.

* In the converse direction, examples of groups that are rigid in quasi-isometry but not in measure equivalence exist. First, some amenable groups are quasi-isometrically rigid, while countably infinite amenable groups are all measure equivalent by the Ornstein–Weiss theorem [102]. Quasi-isometrically rigid amenable groups include $\mathbb{Z}^n$ for every $n\in\mathbb{N}$ (Pansu [105]) or solvable Baumslag–Solitar groups (Farb–Mosher [44]). Examples of non-amenable groups that are rigid in quasi-isometry but not in measure equivalence include many right-angled Artin groups, see work of Huang [75] on the quasi-isometry side.

When dealing with the above question, it is in fact useful to impose a very strong form of rigidity on the vertex groups $G_v$, and require at least that for every $v\in V\Gamma$, every self measure equivalence coupling of $G_v$ factors through the tautological coupling on $G_v$ itself, by left-right multiplication. This will be an important assumption in our criterion in Section <ref>.

To illustrate the subtleness of the question, let us make the following observation in the context of $L^1$-measure equivalence: it is possible to have a graph product $G$ of countably infinite groups over a finite simple graph with no transvection and no partial conjugation (and not reduced to a point), where all vertex groups are rigid in $L^1$-measure equivalence in the sense that every countable group that is $L^1$-measure equivalent to $G_v$ is virtually isomorphic to $G_v$, but $G$ is not. Indeed, it is enough to take the vertex groups to be isomorphic to $\mathbb{Z}$, and observe that some right-angled Artin groups are not rigid in $L^1$-measure equivalence [70], while $\mathbb{Z}$ is <cit.>.

§ AN ME-RIGIDITY CRITERION FOR GRAPH PRODUCTS

The goal of the present section is to prove the following theorem. The definition of Monod–Shalom's class $\Creg$ is given just below the statement.

Let $\Gamma$ be a finite simple graph with no transvection and no partial conjugation, with $\Aut(\Gamma)=\{1\}$. For every $v\in V\Gamma$, let $G_v$ be a countably infinite group, and let $G$ be the graph product over $\Gamma$ with vertex groups $(G_v)_{v\in V\Gamma}$. Assume that the following four hypotheses hold.

$(\mathrm{H}_1)$ For every vertex $v\in V\Gamma$, there exists an extension $\hat{G}_v=G_v\rtimes F_v$ with $F_v$ finite, such that for every self measure equivalence coupling $\Omega_v$ of $G_v$, there exists a Borel $(G_v\times G_v)$-equivariant map $\Omega_v\to \hat{G}_v$.

$(\mathrm{H}_2)$ For every vertex $v\in V\Gamma$, the group $G_v$ is torsion-free, has no proper finite-index subgroup, and belongs to the class $\calc_\reg$.

$(\mathrm{H}_3)$ The vertex groups are pairwise not measure equivalent.

$(\mathrm{H}_4)$ For every clique $C$ with $VC=\{v_1,\dots,v_k\}$, the image of every injective homomorphism $G_{v_1}\to G_{v_1}\times\dots\times G_{v_k}$ is equal to $G_{v_1}\times \{1\}\times\dots\times\{1\}$.

Then every torsion-free countable group which is measure equivalent to $G$, is in fact isomorphic to $G$.

As usual, throughout the section, all the stated equivariances are valid almost everywhere.

The class $\mathcal{C}_{\mathrm{reg}}$, introduced by Monod–Shalom in [96], is the class of all countable groups $G$ such that $\mathrm{H}^2_{\mathrm{b}}(G,\ell^2(G))\neq\{0\}$. Of particular importance for us is the following example.

Recall that a group is acylindrically hyperbolic in the sense of Osin [101] if it admits a non-elementary acylindrical action on a hyperbolic space (see also Appendix <ref>, sec:acyl-hyp). By a theorem of Hamenstädt <cit.>, all acylindrically hyperbolic groups belong to the class $\Creg$.

Hypothesis $(\mathrm{H}_2)$ implies in particular that all vertex groups are ICC (i.e. every non-trivial conjugacy class in $G_v$ is infinite), since a countable torsion-free $\calc_\reg$ group is always ICC by <cit.>.

*Outline of the proof, and organization of the section.

Recall that the group $G$ acts on its right-angled building $\bD_G$ (see Section <ref>, sec:def-building for its definition). We first prove that the $G$-stabilizer $G_u$ of every vertex $u\in V\bD_G$ is rigid in measure equivalence in a strong sense: every self-coupling of $G_u$ factors through the action on (some finite-index extension of) $G_u$ by left-right multiplication. When $u$ is a rank $1$ vertex, the group $G_u$ is conjugate to a vertex group for the graph product structure of $G$, and rigidity is exactly the contents of Hypothesis $(\mathrm{H}_1)$. Stabilizers of higher rank vertices split as direct products of stabilizers of rank $1$ vertices, and we exploit an argument from the successive works of Monod–Shalom [97] and Kida [83] to establish the measure equivalence rigidity of these direct products. This step is carried in Section <ref>.

Let now $H$ be a torsion-free group which is measure equivalent to $G$. We know from Part <ref> of the present work that every self-coupling of $G$ factors through $\Aut(\bD_G)$. By a standard technique, originating from the work of Furman [48], we can transfer the rigidity from self-couplings to arbitrary couplings, and deduce that $H$ also acts on $\bD_G$, and that (through this action) every measure equivalence coupling $\Omega$ between $G$ and $H$ factors through $\Aut(\bD_G)$. This is proved in Section <ref>. A key technical ingredient in the proof is to establish a strong ICC property for the Polish group $\Aut(\bD_G)$ – our proof of this fact relies on previous work of Huang and the second-named author [70].

In Section <ref>, we then analyze this $H$-action on $\bD_G$. In a way that is similar to our work in Part <ref>, we prove that for every vertex $u\in V\bD_G$, the coupling $\Omega$ (between $G$ and $H$) restricts to a measure equivalence coupling between $G_u$ and $H_u$. And the combination of Hypotheses $(\mathrm{H}_1)$ and $(\mathrm{H}_2)$ then forces $H_u$ to be isomorphic to $G_u$ (the facts that $H_u$ is torsion-free and that $G_u$ has no proper finite-index subgroup are used to obtain an actual isomorphism, not a virtual one). In other words $H$ acts on $\bD_G$ with the same vertex stabilizers as $G$. And, by crucially using our assumption that all groups considered have no proper finite-index subgroup, we also prove in Section <ref> that the $H$-action on $\bD_G$ shares a common fundamental domain (which is a subcomplex) with the action of $G$.

In the terminology of Bridson–Haefliger [15], this says that $H$ has the structure of a simple complex of groups over the fundamental domain $Y$ for the $G$-action on $\bD_G$. In other words $H$ is exactly the group which is obtained from the free product of all $H$-stabilizers of cells in $Y$, by taking the quotient by the relations coming from the natural inclusions of stabilizers (coming from face inclusions of the cells). Recall that each cell stabilizer splits as a direct product of some of the vertex groups $G_v$, and there remains to understand the inclusion homomorphisms. This is where we crucially use our Hypothesis $(\mathrm{H}_4)$: it forces all inclusion homomorphisms to be factor inclusions. In other words $H$ has exactly the presentation of a graph product, and therefore $H$ is isomorphic to $G$.

§.§ Self-couplings of clique subgroups

Using the fact that all vertex groups belong to the class $\calc_\reg$, we extend Assumption $(\mathrm{H}_1)$ to all self-couplings of clique subgroups of $G$ (and not just to self-couplings of vertex groups). This is the contents of the following proposition, which was essentially established by Kida <cit.>, relying on work of Monod–Shalom [97]. We provide a brief argument as Kida's statements were formulated in the setting of mapping class groups.

Let $G_1,\dots,G_n$ be torsion-free countable groups in the class $\Creg$, and for every $i\in\{1,\dots,n\}$, let $\hat{G}_i$ be a finite-index extension of $G_i$.  Let $G=G_1\times\dots\times G_n$, and let $\hat G=\hat{G}_1\times\dots\times\hat{G}_n$. Assume that

* for every $i\in\{1,\dots,n\}$, and every self measure equivalence coupling $\Omega_i$ of $G_i$, there exists a $(G_i\times G_i)$-equivariant Borel map $\Omega_i\to \hat{G}_i$, and

* for any distinct $i,j\in\{1,\dots,n\}$, the groups $G_i$ and $G_j$ are not measure equivalent.

Then for every self measure equivalence coupling $\Omega$ of $G$, there exists a $(G\times G)$-equivariant Borel map $\Omega\to \hat{G}$.

By considering the ergodic decomposition, every self measure equivalence coupling of $G$ can be disintegrated into ergodic self measure equivalence couplings, see the second point of <cit.>. Therefore, in view of <cit.> (applied to $G=H$, with $\Omega$ in place of their space $X$, and $\hat G$ in place of their space $Y$), it is enough to treat the case where the action of $G\times G$ on $\Omega$ is ergodic. From now on we work under this extra assumption.

Using an argument from the proof of <cit.>, Kida proved in <cit.> that for every $i\in\{1,\dots,n\}$, there exists a standard measure space $\Omega_i$, equipped with

* a measure-preserving action of $G_i\times G_i$;

* a measurable projection $p_i:\Omega\to\Omega_i$ which is both $(G_i\times G_i)$-equivariant, and $(G_j\times G_j)$-invariant for every $j\neq i$;

* a $(G_i\times G_i)$-invariant Borel partition $\Omega_i^*=\dunion_k\Omega_{i,k}$ of a $(G_i\times G_i)$-invariant conull Borel subset $\Omega_i^*$ into at most countably many subspaces $\Omega_{i,k}$, such that each $\Omega_{i,k}$ is a self measure equivalence coupling of $G_i$.

(Notice that Kida has a space $\Omega_i$ which is invariant under $G_i\times G_{t(i)}$ for some bijection $t:\{1,\dots,n\}\to\{1,\dots,n\}$, but our second assumption implies that $t(i)=i$.)

For every $i\in\{1,\dots,n\}$ and every $k\in\mathbb{N}$, let $\theta_{i,k}:\Omega_{i,k}\to \hat{G}_i$ be a $(G_i\times G_i)$-equivariant Borel map, given by our first assumption. Combining these maps as $k$ varies in $\mathbb{N}$ yields a $(G_i\times G_i)$-equivariant Borel map $\theta_i:\Omega_i\to \hat{G}_i$. Then the map $(\theta_1\circ p_1,\dots,\theta_n\circ p_n):\Omega\to \hat{G}$ gives the desired almost $(G\times G)$-equivariant Borel map.

§.§ Factoring the coupling through the right-angled building

The goal of this section is to prove the following proposition (we refer to Section <ref>, sec:def-building,  for the definition of the right-angled building $\bD_G$).

Let $G$ be a graph product with countably infinite vertex groups, over a finite simple graph $\Gamma$ with no transvection and no partial conjugation. Let $H$ be a countable group, and let $\Omega$ be a measure equivalence coupling between $G$ and $H$. Let $\bD_G$ be the right-angled building of $G$.

Then there exist a homomorphism $\iota: H\to\Aut(\bD_G)$ and a $(G\times H)$-equivariant Borel map $\theta:\Omega\to\Aut(\bD_G)$, where the action of $G\times H$ on $\Aut(\bD_G)$ is via $(g,h)\cdot f=gf\iota(h)^{-1}$.

We already know that self-couplings of $G$ factor through $\Aut(\bD_G)$ (Corollary <ref>, cor:factor-through-building). Our proof relies on a general strategy to transfer this fact to an information about arbitrary couplings between $G$ and $H$. This strategy was initiated by Furman in [48], and we will use a version found in [84] (or alternatively [13]). This argument requires establishing a strong ICC property for the Polish group $\Aut(\bD_G)$, which is the contents of our next lemma.

Let $G$ be a graph product of countably infinite groups over a finite simple graph $\Gamma$, such that no vertex $v\in V\Gamma$ satisfies $\Gamma=\st(v)$. Let $\bD_G$ be its right-angled building.

Then the only Borel probability measure on $\Aut(\bD_G)$ which is invariant by the conjugation by any element of $G$ is the Dirac mass at the identity.

The conclusion of the lemma precisely says that $\Aut(\bD_G)$ is strongly ICC relative to $G$ in the sense of Bader–Furman–Sauer <cit.> (or that the inclusion  $G\subseteq\Aut(\bD_G)$ is strongly ICCICC!Strongly ICCStrongly ICC in the terminology of <cit.>).

This was proved when $G$ is the right-angled Artin group $A_\Gamma$ over $\Gamma$ in <cit.>. We now explain how to reduce the proof to the case of a right-angled Artin group. We will use the language of $\mathrm{CAT}(0)$ cube complexes and their half-spaces, for which we refer to [110]. We will also need the notion of the regular boundary of a $\mathrm{CAT}(0)$ cube complex $X$, a natural compact subspace of its Roller boundary equal to the closure of the set of regular points in the sense of Fernós in <cit.>, also denoted $S(X)$ in [87]. All the necessary background for this proof was also reviewed in <cit.>.

Let $\mathbb{D}_\Gamma$ be the right-angled building of the right-angled Artin group over $\Gamma$. Recall from Lemma <ref>, lemma:building, that there exists an isomorphism from $\bD_G$ to $\bD_\Gamma$ that sends $G$-orbits to $A_\Gamma$-orbits.

Step 1. Assuming that $\Gamma$ is irreducible (and not reduced to one vertex), we will prove that the $G$-action on the regular boundary $R(\bD_G)$ is minimal and strongly proximal in the sense of Furstenberg [47], i.e. it is minimal and the $G$-orbit of any probability measure on $R(\bD_G)$ contains a Dirac mass in its weak-$*$ closure.

By <cit.>, the building $\mathbb{D}_G\approx\mathbb{D}_\Gamma$ is irreducible (it does not split as a product), and

* $A_\Gamma$ acts non-elementarily, i.e. with no global fixed point, and no finite orbit in the visual boundary $\partial_\infty\mathbb{D}_\Gamma$, and

* $A_\Gamma$ acts essentially, in the sense of Caprace–Sageev [31], i.e. no $A_\Gamma$-orbit stays in a bounded neighborhood of a half-space of $\mathbb{D}_\Gamma$.

The first of these properties (non-elementarity) was in fact proved by Caprace–Hume <cit.> in the more general context of graph products, so it applies to the $G$-action on $\bD_G$. The second property (essentiality) for the $G$-action on $\bD_G$ follows from the same property for the $A_\Gamma$-action on $\bD_\Gamma$ as these two actions have the same orbits.

It thus follows from <cit.> that the $G$-action on the regular boundary $R(\bD_G)$ is minimal and strongly proximal, as claimed.

Step 2. We now conclude the proof of the lemma. Write $\Gamma=\Gamma_1\circ \dots\circ \Gamma_k$ as a join, where each $\Gamma_j$ is irreducible (possibly with $k=1$). Notice that no $\Gamma_j$ is reduced to a vertex, by our assumption that $\Gamma$ is not contained in the star of a vertex. Let $G=G_1\times\dots\times G_k$ be the corresponding direct product decomposition. For every $j\in\{1,\dots,k\}$, let $\bD_j$ be the right-angled building of $G_j$, and let $R=R(\bD_1)\times\dots\times R(\bD_k)$. By Step 1, for every $j\in\{1,\dots,k\}$, the action of $G_j$ on $R(\bD_j)$ is minimal and strongly proximal, and it is faithful in view of <cit.>. Let now $\mu$ be a probability measure on $\Aut(\bD_G)$ which is invariant under conjugation by every element of $G$. By <cit.>, we can pushforward $\mu$ to a probability measure $\bar\mu$ on $\Homeo(R)$ which is invariant under the conjugation by every element of $G$. It thus follows from <cit.> that $\bar\mu$ is the Dirac mass at the identity. Using again that the map $\Aut(\bD_G)\to \Homeo(R)$ is injective (<cit.>), it follows that $\mu$ is the Dirac mass at the identity.

We are now in position to prove Proposition <ref>.

Let $\iota_G:G\to\Aut(\bD_G)$ be the natural inclusion. By Corollary <ref>, cor:factor-through-building, for every self measure equivalence coupling $\Sigma$ of $G$, there exists a $(G\times G)$-equivariant Borel map $\Sigma\to\Aut(\bD_G)$. Combined with Lemma <ref>, this precisely says that $G$ is coupling rigid with respect to $(\Aut(\bD_G),\iota_G)$ in the sense of Kida <cit.>.
The conclusion thus follows from <cit.>.

§.§ Step 2: A common fundamental domain for G and H on the Davis complex

In the whole section, we will work in the following setting.

General setting. Let $\Gamma$ be a finite simple graph with no transvection and no partial conjugation, with $\Aut(\Gamma)=\{1\}$, and $G$ be a graph product with countably infinite vertex groups over $\Gamma$ that satisfies Assumptions $(\mathrm{H}_1)$, $(\mathrm{H}_2)$ and $(\mathrm{H}_3)$ from Theorem <ref> (Assumption $(\mathrm{H}_4)$ will only be used in Section <ref>). Let $H$ be a torsion-free countable group which is measure equivalent to $G$, and let $(\Omega,m)$ be a measure equivalence coupling between $G$ and $H$. Proposition <ref> gives us

* a homomorphism $\iota:H\to\Aut(\bD_G)$;

* a $(G\times H)$-equivariant Borel map $\theta:\Omega\to\Aut(\bD_G)$.

In particular $\iota$ gives an action of $H$ on $\bD_G$, and from now on we will work with this action of $H$.

§.§.§ The $H$-action is type preserving

Recall from Section <ref> that vertices of $\bD_G$ correspond to left cosets of the form $gG_\Upsilon$, for some $g\in G$ and some clique subgraph $\Upsilon\subseteq\Gamma$. The rank of a vertex is the cardinality of $V\Upsilon$. We also define the typeType!Type of a vertex of $\bD_G$ of a vertex of $\bD_G$ as the corresponding subgraph $\Upsilon$. More generally, given a cube $\sigma\subseteq\bD_G$, we define the typeType!Type of a cube in $\bD_G$ of $\sigma$ as the set consisting of all types of vertices of $\sigma$. Notice that the type of $\sigma$ is determined by the type of its smallest and largest vertices, for the partial order given by  inclusion (when viewing vertices as cosets in $G$).

The fact that $\Aut(\Gamma)=\{1\}$ will be used in the form of the following lemma.

The actions of $G$ and of $\Aut(\bD_G)$ on $\bD_G$ have the same orbits of cubes. In particular, the action of $\Aut(\bD_G)$ on $\bD_G$ preserves types of cubes.

Since the $G$-action on $\bD_G$ preserves types of vertices, it preserves types of cubes, and therefore it is enough to prove that $G$ and $\Aut(\bD_G)$ have the same orbits of cubes.

Step 1: The actions of $G$ and $\Aut(\bD_G)$ on $\bD_G$ have the same orbits of vertices.

When $G=A_\Gamma$ is the right-angled Artin group over $\Gamma$, this was proved in <cit.>: indeed the group $\hat{G}$ appearing in <cit.> splits as a semi-direct product $G\rtimes\Aut(\Gamma)$ (see its definition in the first paragraph of <cit.>), so it coincides with $G$ under our assumption that $\Aut(\Gamma)=\{1\}$.

The general case now follows from the fact that there always exists an isomorphism $\bD_G\to\bD_\Gamma$ (where $\bD_\Gamma$ is the right-angled building for $A_\Gamma$), sending $G$-orbits of vertices to $A_\Gamma$-orbits of vertices (Lemma <ref>, lemma:building).

Step 2: The actions of $G$ and $\Aut(\bD_G)$ on $\bD_G$ have the same orbits of cubes.

Let $\sigma,\sigma'\subseteq\bD_G$ be two cubes in the same $\Aut(\bD_G)$-orbit; we aim to prove that they are in fact in the same $G$-orbit. Let $v_{\min},v_{\max}$ be the minimal and maximal vertices of $\sigma$, and let $v'_{\min},v'_{\max}$ be the minimal and maximal vertices of $\sigma'$. Since $\Aut(\bD_G)$ preserves types of vertices (as a consequence of Step 1), the vertices $v_{\min}$ and $v'_{\min}$ are in the same $\Aut(\bD_G)$-orbit. Using Step 1 again, they are in fact in the same $G$-orbit: there exists $g\in G$ such that $v'_{\min}=gv_{\min}$. Likewise, the vertices $v_{\max}$ and $v'_{\max}$ are in the same $\Aut(\bD_G)$-orbit. In particular they have the same type, which we denote by $\Lambda$ (a subgraph of $\Gamma$). So $v_{\max}$ (resp. $v'_{\max}$) is the unique coset of $G_{\Lambda}$ that contains the coset of $G$ corresponding to $v_{\min}$ (resp. to $v'_{\min}=gv_{\min}$). Therefore $v'_{\max}=gv_{\max}$, and this in enough to ensure that $\sigma'=g\sigma$.

Let us record the following consequence of Lemma <ref>.

Let $\sigma\subseteq\bD_G$ be a cube, and let $v$ be the unique vertex of $\sigma$ of minimal rank. Then the stabilizers of $v$ and of $\sigma$ in $\Aut(\bD_G)$ coincide.

Since $v$ is the unique vertex of $\sigma$ of its type and the action of $\Aut(\bD_G)$ preserves types of vertices (Lemma <ref>), we have $\Stab_{\Aut(\bD_G)}(\sigma)\subseteq\Stab_{\Aut(\bD_G)}(v)$.

The converse inclusion follows from the observation that $\sigma$ is the unique cube of its type that contains $v$ as its minimal vertex. Since $\Aut(\bD_G)$ preserves types of cubes (Lemma <ref>), this implies that the $\Aut(\bD_G)$-stabilizer of $v$ also preserves $\sigma$.

§.§.§ Stabilizers for the actions of $G$ and $H$ are measure equivalent

Given a cube $\sigma\subseteq V\bD_G$, we let $\mathrm{Stab}_G(\sigma),\mathrm{Stab}_H(\sigma)$ and $\Stab_{\Aut(\bD_G)}(\sigma)$ be the respective stabilizers of $\sigma$ for the actions of $G$, of $H$, and of $\Aut(\bD_G)$ on $\bD_G$. We let $\Omega_\sigma:=\theta^{-1}(\Stab_{\Aut(\bD_G)}(\sigma))$, where we recall that $\theta:\Omega\to\Aut(\bD_G)$ is our given $(G\times H)$-equivariant Borel map. Then $\Omega_\sigma$ is a Borel subset of $\Omega$ which is $(\mathrm{Stab}_G(\sigma)\times \mathrm{Stab}_H(\sigma))$-invariant. The following lemma is a special case of <cit.>.

For every cube $\sigma\subseteq V\bD_G$, the space $\Omega_\sigma$ is a measure equivalence coupling between $\mathrm{Stab}_G(\sigma)$ and $\mathrm{Stab}_H(\sigma)$.

In view of Corollary <ref>, it is enough to prove the lemma when $\sigma$ is a vertex, so we will assume so.

By Lemma <ref>, the groups $G$ and $\Aut(\bD_G)$ have the same orbits of vertices for their actions on $\bD_G$. The lemma therefore follows from <cit.>, applied with $K=\bD_G$, with $L=\Aut(\bD_G)$, with $\mathsf{G}=G$, and with $\mathsf{H}=H$.

As a particular case, we obtain the following statement.

The $H$-stabilizer of every rank $0$ vertex of $\bD_G$ is trivial.

Let $v\in V\bD_G$ be a rank $0$ vertex. By Lemma <ref>, the stabilizer $\mathrm{Stab}_H(v)$ is measure equivalent to $\mathrm{Stab}_G(v)$, which is trivial. Therefore $\mathrm{Stab}_H(v)$ is finite, and in fact trivial because $H$ is torsion-free.

In fact we will deduce more generally that $\mathrm{Stab}_G(\sigma)$ and $\mathrm{Stab}_H(\sigma)$ are isomorphic for every cube $\sigma\subseteq\bD_G$, in view of the following lemma.

Let $\sigma\subseteq \bD_G$ be a cube. Let $K$ be a torsion-free countable group which is measure equivalent to $\mathrm{Stab}_G(\sigma)$.

Then $K$ is isomorphic to $\mathrm{Stab}_G(\sigma)$.

The group $\mathrm{Stab}_G(\sigma)$ is isomorphic to $G_\Upsilon=G_{v_1}\times\dots\times G_{v_k}$ for some clique $\Upsilon$ of $\Gamma$ with $V\Upsilon=\{v_1,\dots,v_k\}$. We also write $\hat G_\Upsilon:=\hat G_{v_1}\times\dots\times\hat G_{v_k}$. Recall that each $G_{v_i}$ is ICC (Remark <ref>), in particular $G_\Upsilon$ is ICC.

Hypotheses $(\mathrm{H}_1)$,
$(\mathrm{H}_2)$ and $(\mathrm{H}_3)$, together with Proposition <ref>, ensure that for every self measure equivalence coupling $\Omega_\Upsilon$ of $G_\Upsilon$, there exists a $(G_\Upsilon\times G_\Upsilon)$-Borel map $\Omega_\Upsilon\to \hat{G}_\Upsilon$. Therefore <cit.> implies that there exists a homomorphism $\pi_\sigma:K\to \hat G_\Upsilon$ with finite kernel, and finite-index image. Since $K$ is torsion-free, $\pi_\sigma$ is in fact injective. Then $\pi_\sigma(K)\cap G_\Upsilon$ is a finite-index subgroup of $G_\Upsilon$, and is therefore equal to $G_\Upsilon$ in view of Hypothesis $(\mathrm{H}_2)$. Recall also from Hypothesis $(\mathrm{H}_1)$ that for every $i\in\{1,\dots,k\}$, we have a semi-direct product $\hat{G}_{v_i}=G_{v_i}\rtimes F_{v_i}$, with $F_{v_i}$ finite. Then the image of $\pi_\sigma(K)$ in each $F_{v_i}$ is trivial: otherwise $\pi_\sigma(K)$ would intersect a lift of $F_{v_1}\times\dots\times F_{v_k}$ to $\hat G_{\Upsilon}$ non-trivially, contradicting that $K$ is torsion-free. Therefore $\pi_\sigma(K)=G_\Upsilon$, which shows that $K$ is isomorphic to $\mathrm{Stab}_G(\sigma)$.

§.§.§ A common strict fundamental domain for the actions of $G$ and $H$

Let $Y_G\subseteq\bD_G$ be the standard fundamental domain for $G$, spanned by all vertices of the form $G_\Upsilon$, where $\Upsilon\subseteq\Gamma$ is a clique. Notice that $Y_G$ is a strict fundamental domainStrict fundamental domain for the $G$-action on $\bD_G$, i.e. it is a subcomplex that meets every $G$-orbit in exactly one point. We will prove in Lemma <ref> below that $Y_G$ is also a strict fundamental domain for the $H$-action on $\bD_G$. We start with a lemma.

For every rank $1$ vertex $v\in VY_G$, every measure equivalence coupling between $\Stab_G(v)$ and $\Stab_H(v)$ has coupling index $1$.

In view of Lemmas <ref> and <ref>, the groups $\Stab_G(v)$ and $\Stab_H(v)$ are isomorphic, so it is enough to show that every self measure equivalence coupling of $\Stab_G(v)$ has coupling index $1$.

Recall that since $v\in VY_G$, the stabilizer $\Stab_G(v)$ is a vertex group of $G$ (for its graph product structure). Hypothesis $(\mathrm{H}_1)$ gives us a finite-index extension $\widehat{\Stab}_G(v)$.

We first observe that every self measure equivalence coupling $\hat\Sigma_v$ of $\widehat{\Stab}_G(v)$ has coupling index $1$. Indeed, since $\Stab_G(v)$ has finite index in $\widehat{\Stab}_G(v)$, the space $\hat\Sigma_v$ is also a self measure equivalence coupling of $\Stab_G(v)$. So Hypothesis $(\mathrm{H}_1)$ yields a Borel $(\Stab_G(v)\times\Stab_G(v))$-equivariant map $\theta_v:\hat\Sigma_v\to \widehat{\Stab}_G(v)$. Since $\Stab_G(v)$ is ICC (Remark <ref>) and normal in $\widehat{\Stab}_G(v)$, it follows from <cit.> that $\theta_v$ is in fact $(\widehat{\Stab}_G(v)\times\widehat{\Stab}_G(v))$-equivariant. Then $\theta_v^{-1}(\{1\})$ is a common Borel fundamental domain for the two actions of $\widehat{\Stab}_G(v)$ on $\hat \Sigma_v$, so $\hat{\Sigma}_v$ has coupling index $1$.

Let now $\Sigma_v$ be a self measure equivalence coupling of $\Stab_{G}(v)$. Starting from $\Sigma_v$ and from a measure equivalence coupling between $\Stab_G(v)$ and $\widehat{\Stab}_G(v)$ of coupling index $[\widehat{\Stab}_G(v):{\Stab}_G(v)]$, by composition of couplings, one obtains a self measure equivalence coupling of $\widehat{\Stab}_G(v)$ with the same coupling index as $\Sigma_v$ (see e.g. Item c on page 300 of [51], below Definition 2.1). It thus follows from the previous paragraph that the coupling index of $\Sigma_v$ is $1$, as desired.

The subcomplex $Y_G$ is a strict fundamental domain for the $H$-action on $\bD_G$.

Assume towards a contradiction that $Y_G$ is not a strict fundamental domain for the $H$-action on $\bD_G$. This means that

* either two distinct points $x,y$ of $Y_G$ belong to the same $H$-orbit,

* or there is a point in $\bD_G$ which is not in the $H$-orbit of any point of $Y_G$.

We start by excluding the first option. Assuming that $x,y\in Y_G$ belong to the same $H$-orbit, we will prove that $x=y$. Let $\sigma_x,\sigma_y$ be the cubes of minimal dimension that contain $x,y$; these are contained in $Y_G$ because $Y_G$ is a subcomplex of $\bD_G$. And $\sigma_x,\sigma_y$ also belong to the same $H$-orbit, in particular they belong to the same $\Aut(\bD_G)$-orbit. In view of Lemma <ref>, the cubes $\sigma_x,\sigma_y$ are in the same $G$-orbit. Since $Y_G$ is a strict fundamental domain for the $G$-action, it follows that $\sigma_x=\sigma_y$. But since automorphisms of $\bD_G$ preserve types of vertices, every automorphism that preserves $\sigma_x$ acts as the identity on $\sigma_x$. It follows that $x=y$, as desired. The first option is excluded.

We now assume that there exists a point $x$ in $\bD_G$ which is not in the $H$-orbit of any point of $Y_G$, and aim for a contradiction. Let $\sigma$ be a cube that contains $x$ and contains a rank $0$ vertex $v$. Then $\sigma$ is not in the $H$-orbit of any cube of $Y_G$. Denoting by $v_0$ the unique rank $0$ vertex of $Y_G$, it follows that $v$ is not in the $H$-orbit of $v_0$.

The rank $0$ vertices $v_0$ and $v$ can be joined by an edge path, such that vertices along the path alternate between rank $0$ and rank $1$ vertices (see <ref> for an example). Let $v_1$ be the last rank $0$ vertex along this path that belongs to the $H$-orbit of $v_0$, and let $v_2$ be the next rank $0$ vertex along this path. Then $v_1$ and $v_2$ are rank $0$ vertices of $\bD_G$ in distinct $H$-orbits which are adjacent to a common rank $1$ vertex $w$.

Illustration of a path going from $v_0=e$ to $v=g_1g_2$ (where $g_i\in G_{v_i}$, for $i\in \{1,2\}$) such that vertices on the path alternate between rank $1$ and $0$.

Recall from Lemma <ref> that

\[\Omega_w:=\theta^{-1}(\Stab_{\Aut(\bD_G)}(w))\]
is a measure equivalence coupling between the stabilizers $\Stab_G(w)$ and $\Stab_H(w)$. For every $i\in\{1,2\}$, let
\[X_{w,i}:=\theta^{-1}(\{f\in\Stab_{\Aut(\bD_G)}(w)\mid f(v_i)=v_1\}).\]
The action of $\Stab_G(w)$ on the set of rank $0$ vertices adjacent to $w$ is nothing but the action of $\Stab_G(w)$ by left translations on itself. In particular it is free and transitive, so for every $i\in\{1,2\}$, the set $X_{w,i}$ is a Borel fundamental domain for the action of $\Stab_G(w)$ on $\Omega_w$. In particular $m(X_{w,1})=m(X_{w,2})$.

For every $i\in\{1,2\}$, the group $\Stab_H(v_i)$ is trivial (Corollary <ref>).
Therefore, since $v_1$ and $v_2$ are not in the same $\Stab_H(w)$-orbit, the set $X_{w,1}\cup X_{w,2}$ meets almost every $\Stab_H(w)$-orbit in $\Omega_w$ at most once, and therefore it is contained in a Borel fundamental domain $Y_w$ for the action of $\Stab_H(w)$ on $\Omega_w$. In particular $m(Y_w)\ge 2 m(X_{w,1})$, so the coupling index satisfies
\[[\Stab_G(w):\Stab_H(w)]_{\Omega_w}\ge 2.\]
This is a contradiction to Lemma <ref>, which completes our proof.

§.§ Exploiting the structure of simple complex of groups

In this final step, we will exploit the structure of $H$ as a simple complex of groups in the sense of Bridson–Haefliger <cit.> to get a fine control on the nature of the group $H$. We continue with the same general setting as in Section <ref>, and assume in addition that the group $G$ satisfies Hypothesis $(\mathrm{H}_4)$ from Theorem <ref>.

Let $\mathfrak{P}$ be the poset of cubes of $\bD_G$, ordered by inclusion. Then $\bD_G$ has a stratification indexed by $\mathfrak{P}$, whose strata are exactly the cubes of $\bD_G$: here stratificationStratification is understood in the sense of <cit.>, i.e.

* $\bD_G$ is the union of its strata;

* the intersection of two strata, if non-empty, is again a stratum;

* for every point in $\bD_G$, the intersection of all strata that contain $x$ is again a stratum (namely, it is the cube of smallest dimension that contains $x$).

The actions of $G$ and of $H$ on $\bD_G$ are strata-preserving, i.e. every element sends every stratum bijectively onto a stratum. And in view of Lemma <ref>, the subcomplex $Y_G$ is a strict fundamental domain for the strata-preserving actions of $G$ and $H$ in the sense of <cit.>: every cube of $\bD_G$ has a unique translate in $Y_G$.

In addition $\bD_G$ is simply connected (it is even $\mathrm{CAT}(0)$). Therefore, we can apply <cit.> and deduce that $H$ is the direct limit (amalgam) of the isotropy subgroups of the cubes contained in $Y_G$, in the sense of <cit.>. The meaning of this is the following. Let
\[\mathfrak{Q}=\{\sigma\in\mathfrak{P}\mid \sigma\subseteq Y_G\}.\]
As before, for every cube $\sigma\subseteq Y_G$, we let $\mathrm{Stab}_H(\sigma)$ be the stabilizer of $\sigma$ for the $H$-action on $\bD_G$. Also, given cubes $\tau\subseteq\sigma$ both contained in $Y_G$, we let $\psi_{\tau\sigma}:\mathrm{Stab}_H(\sigma)\to \Stab_H(\tau)$ be the inclusion. Then $H$ is isomorphic to the group obtained from the free product of the groups $\mathrm{Stab}_H(\sigma)$, with $\sigma$ ranging in $\mathfrak{Q}$, by imposing as only extra relations that $\psi_{\tau\sigma}(h)=h$, whenever $\tau\subseteq\sigma$ both belong to $\mathfrak{Q}$ and $h\in \mathrm{Stab}_H(\sigma)$.

We are now in position to prove the main theorem of this section.

Let $v_1,\dots,v_n$ be the rank $1$ vertices in $Y_G$. Let $\sigma\subseteq Y_G$ be a cube, and let $w$ be its minimal vertex. Let $v_{i_1},\dots,v_{i_k}$ be the rank $1$ vertices that satisfy $v_{i_j}\le w$ (where the partial order $\le$ is just the inclusion of the cosets represented by these vertices). We say that $v_{i_1},\dots,v_{i_k}$ are the rank $1$ vertices below $\sigma$ (notice that in the case where $\sigma$ contains the rank $0$ vertex of $Y_G$, there are no rank $1$ vertices below $\sigma$). For every $j\in\{1,\dots,k\}$, there is a cube $\sigma'_j$ in $Y_G$ that contains $v_{i_j}$ as its minimal vertex, and also contains $w$. So Corollary <ref> implies that, for every $j\in\{1,\dots,k\}$, we have

\[\Stab_{\Aut(\bD_G)}(v_{i_j})=\Stab_{\Aut(\bD_G)}(\sigma'_j)\subseteq\Stab_{\Aut(\bD_G)}(\sigma)=\Stab_{\Aut(\bD_G)}(w). \]

Therefore the $H$-stabilizer of $v_{i_j}$ is equal to $\Stab_H(\sigma'_j)$, and is contained in $\Stab_H(w)=\mathrm{Stab}_H(\sigma)$.

By Lemmas <ref> and <ref>, we already know that the stabilizer $\mathrm{Stab}_H(\sigma)$ is abstractly isomorphic to $\mathrm{Stab}_G(\sigma)$, which splits as $\mathrm{Stab}_G(\sigma)=G_{v_{i_1}}\times\dots\times G_{v_{i_k}}$ (here we naturally identify the rank $1$ vertices $v_{i_1},\dots,v_{i_k}$ with vertices of the defining graph $\Gamma$, and the groups $G_{v_{i_j}}$ are exactly vertex groups for the graph product structure of $G$).

In addition, for every $j\in\{1,\dots,k\}$, we have an inclusion map $\Stab_H(v_{i_j})\to \mathrm{Stab}_H(\sigma)$, and $\Stab_H(v_{i_j})$ is isomorphic to $G_{v_{i_j}}$ (<ref>). It thus follows from Hypothesis $(\mathrm{H}_4)$ that the image of $\Stab_H(v_{i_j})$ in $\mathrm{Stab}_H(\sigma)$ is exactly equal to

\[ \{1\}\times\dots\times\{1\}\times G_{v_{i_j}}\times\{1\}\times\dots\times\{1\},\]

under the above identification between $\mathrm{Stab}_H(\sigma)$ and $\mathrm{Stab}_G(\sigma)$. Therefore $\mathrm{Stab}_H(\sigma)$ splits as the direct product $\Stab_H({v_{i_1}})\times\dots\times  \Stab_H({v_{i_k}})$, in such a way that the inclusion maps $\Stab_H({v_{i_j}})\to  \mathrm{Stab}_H(\sigma)$ are just factor inclusions.

In other words, given any finite collection of rank $1$ vertices $v_{i_1},\dots,v_{i_k}$ spanning a clique when viewed as vertices of $\Gamma$, the subgroup generated by $\Stab_H(v_{i_1}),\dots, \Stab_H(v_{i_k})$ is equal to their direct product, and it is also equal to the stabilizer of any cube $\sigma$ such that the collection of rank $1$ vertices below $\sigma$ is exactly $\{v_{i_1},\dots,v_{i_k}\}$. From this, it follows that whenever $\sigma,\tau$ are two cubes with $\tau\subseteq\sigma$, the inclusion map $\mathrm{Stab}_H(\sigma)\to \Stab_H(\tau)$ is a factor inclusion.

It follows that the direct limit of the groups $\Stab_H(v_i)$ is nothing but their graph product over $\Gamma$, and it follows from <cit.> that this direct limit is isomorphic to $H$. Since $\Stab_H(v_i)$ is isomorphic to $G_{v_i}$, this shows that $H$ is isomorphic to $G$.

§ AN ME-RIGID GROUP WHICH IS NOT QI-RIGID

In this section, we provide an example of a graph product that satisfies all assumptions from Theorem <ref>. Our example will involve Higman groups [72], defined for $k\ge 4$ as
\[\Hig_k=\langle a_1,\dots, a_k \mid a_ia_{i+1}a_i^{-1}=a_{i+1}^{2} \quad \forall i~(\text{mod~} k)\rangle.\]

Our construction will enable us to prove the following theorem, which we recall from the introduction.

Let $\Gamma$ be a finite simple graph with no transvection and no partial conjugation, such that $\Aut(\Gamma)=\{1\}$. Let $(k_v)_{v\in V\Gamma}$ be a sequence of integers at least $5$, such that $k_v$ does not divide $k_w$ whenever $v\neq w$. Let $G$ be the graph product over $\Gamma$ with vertex groups $(\Hig_{k_v})_{v\in V\Gamma}$. Then

* If a torsion-free countable group $H$ is measure equivalent to $G$, then $H$ is isomorphic to $G$.

* There exists an infinite family of torsion-free countable groups that are all quasi-isometric to $G$, and pairwise not commensurable.

§.§ Homomorphisms between Higman groups

The following proposition is an elaboration on work of Martin [92], who proved the second assertion in the case where $k=4$. We will follow his proof very closely.

Let $k,k'\ge 4$. Then

* There exists a non-trivial homomorphism $\Hig_{k'}\to\Hig_k$ if and only if $k$ divides $k'$.

* Every non-trivial homomorphism $\Hig_k\to\Hig_k$ is an automorphism.

We let $\{a_1,\dots,a_k\}$ be a standard generating set of $\Hig_{k}$, and $\{a'_1,\dots,a'_{k'}\}$ be a standard generating set of $\Hig_{k'}$ (as in their standard presentations).

If $k$ divides $k'$, then the map sending the generator $a'_i$ of $\Hig_{k'}$ to the generator $a_{i}$ of $\Hig_k$ (with the index $i$ considered modulo $k$), extends to a non-trivial homomorphism from $\Hig_{k'}$ to $\Hig_k$.

Conversely, assume that there exists a non-trivial homomorphism  $\theta:\Hig_{k'}\to\Hig_k$. We will prove that $k$ divides $k'$, and that if $k=k'$ then $\theta$ is an isomorphism, which will establish the proposition. Most of the proof follows <cit.> very closely. Martin's proof was written when $k=k'=4$, but the arguments readily extend to all $k,k'\ge 4$; we will provide a guide through the proof.

First, observe that $\theta$ cannot send any generator of $\Hig_{k'}$ to $1$, as otherwise the defining relations of $\Hig_{k'}$ would imply that every generator is sent to $1$, contradicting that $\theta$ is non-trivial.

As in <cit.> or <cit.>, the group $\Hig_k$ (resp. $\Hig_{k'}$) acts on a $\mathrm{CAT}(0)$ polygonal complex $X_k$ (resp. $X_{k'}$) with strict fundamental domain a $2$-dimensional $k$-gon (resp. a $2$-dimensional $k'$-gon). For both actions, stabilizers of $2$-dimensional cells are trivial, edge stabilizers are isomorphic to $\mathbb{Z}$ (conjugate to the cyclic subgroups generated by the generators $a_i$ or $a'_i$), and vertex stabilizers are isomorphic to $\BS(1,2)$ (conjugate to the subgroups of the form $\langle a_i,a_{i+1}\rangle$, or $\langle a'_i,a'_{i+1}\rangle$, with indices understood modulo $k$ or $k'$). Moreover, for every edge $e=vw$ of $X_k$, exactly one of the homomorphisms $\Stab_G(e)\hookrightarrow \Stab_G(v)$ and $\Stab_G(e)\hookrightarrow \Stab_G(w)$ has undistorted image. As in <cit.>, we orient the edge towards the vertex $w$ such that the inclusion $\Stab_G(e)\hookrightarrow \Stab_G(w)$ is undistorted. If $e$ is oriented towards $w$, and $e'$ is any other edge incident on $w$, then $\Stab_G(e)\cap \Stab_G(e')=\{1\}$: this is the contents of <cit.>. In particular, as in <cit.>, the fixed point set $\mathrm{Fix}(h)$ of any non-trivial element of $\Hig_k$ is always contained in the outward star of a vertex, i.e. in the union of all edges emanating from that vertex ($\mathrm{Fix}(h)$ can be empty or reduced to one vertex).

As in <cit.> (with the same proof), the homomorphism $\theta$ sends every subgroup $\langle a'_i,a'_{i+1}\rangle$ into a unique subgroup of the form $g \langle a_j,a_{j+1}\rangle g^{-1}$. In particular $\theta$ induces a map $\theta_\ast:VX_{k'}\to VX_k$. As in <cit.>, the map $\theta_\ast$ extends to an orientation-preserving map from the $1$-skeleton $X^1_{k'}$ to the $1$-skeleton $X^1_k$. And $\theta_\ast$ does not collapse any edge to a vertex (see the second paragraph of the proof of <cit.>).

Consider the vertices $v_1,\dots,v_{k'}$ of $X_{k'}$ corresponding to the subgroups $\langle a'_i,a'_{i+1}\rangle$ with $i$ varying in $\{1,\dots,k'\}$ (with indices considered modulo $k'$). They are aligned along a directed $k'$-gon $P_{k'}$. Therefore, there images under $\theta_\ast$ are aligned along an (immersed) oriented $k'$-gon $P$ in $X_k$. The projection of $P$ under the quotient map $X^1_{k}\to X^1_{k}/G$ is an immersed oriented $k'$-gon inside an oriented $k$-gon. This implies that $k$ divides $k'$, and proves the first part of the proposition.

We now prove the second assertion of the proposition, so from now on we assume that $k=k'$, and continue with the above notations. Since the case where $k=4$ was proved by Martin [92], we will assume that $k\ge 5$.

The map $\theta_\ast$ sends the boundary of any $2$-cell in $X_k$ to an immersed $k$-gon. And an immersed $k$-gon in $X_k$ is always the boundary of a $2$-cell in view of <cit.>. So $\theta_\ast$ sends the fundamental $k$-gon $P_k\subseteq X_k$ to a translate of $P_k$. Therefore, up to post-composing $\theta$ by an inner automorphism, we can assume that $\theta_\ast(P_k)=P_k$. In particular the map $(\theta_\ast)^k=(\theta^k)_\ast$ fixes $P_k$ pointwise. It follows from <cit.> (whose proof remains valid for every $k\ge 5$) that $\theta^k$ is the identity, in particular $\theta$ is an automorphism.

§.§ Measure equivalence rigidity of G

We are now in position to complete our proof of the first part of Theorem <ref>, theo:qi-me.

We apply the criterion provided by Theorem <ref> to the graph product $G$.

Hypothesis $(\mathrm{H}_1)$ is given by <cit.>.

For Hypothesis $(\mathrm{H}_2)$, notice that the vertex groups, being Higman groups, are torsion-free (as recorded for example in <cit.>) and have no proper finite-index subgroups (as proved by Higman [72]). They are acylindrically hyperbolic (see <cit.> or <cit.>), and therefore they belong to the class $\calc_\reg$ (see Example <ref>). This verifies Hypothesis $(\mathrm{H}_2)$ from Theorem <ref>.

For Hypothesis $(\mathrm{H}_3)$, notice that if $\Hig_{k_v}$ is measure equivalent to $\Hig_{k_w}$, then <cit.> implies that $\Hig_{k_v}$ and $\Hig_{k_w}$ are virtually isomorphic (i.e. isomorphic up to taking the quotient by a finite normal subgroup, and passing to a finite-index subgroup). Since Higman groups are torsion-free and have no proper finite-index subgroup, it follows that $\Hig_{k_v}$ is isomorphic to $\Hig_{k_w}$. In view of Proposition <ref>, this is impossible unless $v=w$. This checks Hypothesis $(\mathrm{H}_3)$.

Finally, Hypothesis $(\mathrm{H}_4)$ is a consequence of Proposition <ref>, as we are assuming that $k_v$ does not divide $k_w$ whenever $v\neq w$.

The conclusion thus follows from Theorem <ref>.

§.§ Flexibility in quasi-isometry

Recall that two finitely generated groups $G,H$ are bi-Lipschitz equivalent if there exists a bi-Lipschitz bijection from $G$ to $H$, equipped with word metrics associated to finite generating sets.

Let $G,H$ be two graph products of finitely generated groups over a finite simple graph $\Gamma$. Assume that for every $v\in V\Gamma$, the vertex groups $G_v$ and $H_v$ are bi-Lipschitz equivalent.

Then $G$ and $H$ are bi-Lipschitz equivalent.

Through normal forms, a family of bi-Lipschitz bijections from $G_v$ to $H_v$ induces a bi-Lipschitz bijection from $G$ to $H$.

Let $v\in V\Gamma$. For every $n\in\mathbb{N}$, let $G_n$ be the graph product over $\Gamma$ whose vertex groups are the same as $G$, except that $G_v$ is replaced by $G_v\times\mathbb{Z}/n\mathbb{Z}$.

The group $G_v$ is non-amenable (it is even acylindrically hyperbolic <cit.>). Therefore, for every $n\in\mathbb{N}$, the groups $G_v$ and $G_v\times\mathbb{Z}/n\mathbb{Z}$ are bi-Lipschitz equivalent [117], so Lemma <ref> ensures that all groups $G_n$ are bi-Lipschitz equivalent and therefore quasi-isometric.

On the other hand, since $G_v$ has no proper finite-index subgroup [72], when $n\neq m$ the groups $G_v\times\mathbb{Z}/n\mathbb{Z}$ and $G_v\times\mathbb{Z}/m\mathbb{Z}$ are never strongly commensurable (and they are not strongly commensurable to any $G_w$ either, as can be seen using Proposition <ref>). It thus follows from Theorem <ref>, theo:classification-commensurability, that $G_n$ and $G_m$ are not commensurable for $n\neq m$.

Finally, the groups $G_n$ are not torsion-free, but they all admit a finite-index torsion-free subgroup. Indeed, a normal form argument shows that every torsion element of $G_n$ is $G_n$-conjugate to an element of the subgroup $\{e\}\times \mathbb{Z}/n\mathbb{Z} \subseteq G_v\times\mathbb{Z}/n\mathbb{Z}$. Let $\theta_n:G_n\to\mathbb{Z}/n\mathbb{Z}$ be the homomorphism which coincides with the second projection on the vertex group $G_v\times\mathbb{Z}/n\mathbb{Z}$, and is the identity on every vertex group $G_w$ with $w\neq v$. Then the kernel of $\theta_n$ is a torsion-free finite-index subgroup $G_n^0$ of $G_n$.

The groups $G_n^0$ are then torsion-free, and they are all quasi-isometric to $G$, but pairwise non-commensurable.

§ ON THE ISOMORPHISM PROBLEM FOR GRAPH PRODUCTS

We refer to Definition <ref>, de:strongly-reduced for the definition of a strongly reduced graph. The main goal of this appendix is to prove the following theorem, see also Theorem <ref> below for a version including the case of direct products.

Let $G,H$ be graph products over strongly reduced finite simple graphs $\Gamma_G,\Gamma_H$, not reduced to a vertex.

Then for every isomorphism $f:G\to H$ and every untransvectable vertex $v\in V\Gamma_G$, there exists an untransvectable vertex $w\in V\Gamma_H$ such that $f(G_v)$ is conjugate to $H_w$.

When combined with <cit.>, Theorem <ref> has the following consequence to the isomorphism problem for graph products.

Let $G,H$ be graph products over strongly reduced finite simple graphs $\Gamma_G,\Gamma_H$, not reduced to a vertex, with $\Gamma_G$ transvection-free.

If $G$ and $H$ are isomorphic, then there exists an isomorphism $\sigma:\Gamma_G\to \Gamma_H$ such that for every $v\in V\Gamma_G$, the vertex groups $G_v$ and $H_{\sigma(v)}$ are isomorphic.

Let $f:G\to H$ be an isomorphism. Since $\Gamma_G$ is transvection-free, every vertex $v\in V\Gamma_G$ is untransvectable. Hence, Theorem <ref> implies that $f$ sends every vertex group $G$ to a conjugate of a vertex group of $H$. The conclusion therefore follows from <cit.>.

Another consequence of Theorem <ref>, regarding the acylindrical hyperbolicity of $\Aut(G)$, will be derived in Section <ref>.

[Comparison with the framework of measured groupoids]
In the case where the defining graphs $\Gamma_G$ and $\Gamma_H$ are strongly reduced and transvection-free, Theorem <ref> ensures that $f$ sends every vertex group to a conjugate of a vertex group. When $G$ and $H$ are countable, this can be seen as an analogue of the Vertex Recognition Property from our measured group-theoretic framework (Definition <ref>). Notice however that Theorem <ref> has two advantages compared to the work in Part <ref>, namely:

* it allows for finite vertex groups, and also for uncountable vertex groups;

* even when the graph is not transvection-free, it enables to recognize untransvectable vertices; in this respect, this is similar to our work in Section <ref>, though in the latter section vertex groups were assumed to be amenable.

We will say more on the comparison between the group-theoretic and groupoid-theoretic frameworks in Remark <ref>.

[Comparison with Genevois's work]
We now compare our assumptions in Theorem <ref> with those of <cit.>. We have the extra assumption that $\Gamma_G$ and $\Gamma_H$ are strongly reduced. On the other hand, we do not assume that the vertex groups are graphically irreducible (a group $G_v$ is graphically irreducibleGraphically irreducible if whenever it decomposes as a graph product over a finite simple graph, it has to be over a clique <cit.>).

Both assumptions (strongly reduced graphs and graphically irreducible vertex groups) cannot be removed simultaneously. In a sense, we decompose $G,H$ minimally as non-trivial graph products, while Genevois decomposes them maximally as graph products.

None of the theorems implies the other. In particular, our theorem allows for infinitely generated vertex groups, which sometimes do not admit any maximal decomposition as a graph product over a finite simple graph (e.g. $F_\infty$). In fact, it is also unknown whether or not a finitely generated group always admits a decomposition as a graph product where every vertex group is graphically irreducible.

§.§ Proof of Theorem <ref>

Our proof of Theorem <ref> will follow the one in Section <ref> quite closely, the main novelty being the use of thick free factors in Sections <ref> and <ref>. It can be read independently from Section <ref>, and in fact can serve as a warm-up before reading its groupoid-theoretic version. To emphasize the parallel between this appendix and Section <ref>, the group-theoretic counterparts of Properties $\Pprod,\Padm$, etc. from Section <ref> will be called $\Qprod,\Qadm$, etc.

The goal is to characterize untransvectable vertex groups by a purely group-theoretic property that will be preserved under group isomorphisms.

§.§.§ Thick free factors

A free factorFree factorFactor!Free Factor of a group $G$ is a subgroup $A\subseteq G$ such that there exists $B\subseteq G$ with $G=A\ast B$. By Bass–Serre theory, $A$ is a free factor of $G$ if and only if there exists an action of $G$ on a simplicial tree $T$ with trivial edge stabilizers, such that $A$ is equal to the stabilizer of a vertex of $T$.

Notice for future use that if $A$ is a free factor of $G$, and $P$ is a subgroup of $G$ that contains $A$, then $A$ is a free factor of $P$: indeed, letting $T$ be a $G$-tree with trivial edge stabilizers with one vertex stabilizer equal to $A$, the tree $T$ is also a $P$-tree with trivial edge stabilizers with one vertex stabilizer equal to $A$.

A group is freely indecomposableFreely indecomposable if it does not admit any action on a simplicial tree with trivial edge stabilizers with no global fixed point. In particular, the only free factors of a freely indecomposable group $G$ are $\{1\}$ and $G$.

Let $G$ be a graph product over a finite simple graph $\Gamma$. Let $\Lambda_1,\dots,\Lambda_k$ be the connected components of $\Gamma$ that contain at least two vertices, and let $v_1,\dots,v_\ell$ be the isolated vertices of $\Gamma$. Then

\[G=G_{\Lambda_1}\ast\dots\ast G_{\Lambda_k}\ast G_{v_1}\ast\dots\ast G_{v_\ell}.\]

A subgroup of $G$ which is conjugate to $G_{\Lambda_j}$ for some $j\in\{1,\dots,k\}$ will be called a thick free factorThick free factorFree factor!Thick free factorFactor!Thick free factor of $G$.

Notice that thick free factors of $G$ are indeed free factors of $G$, and they are freely indecomposable. Conversely, we make the following observation.

Let $G$ be a graph product over a finite simple graph $\Gamma$, and let $L\subseteq G$ be a freely indecomposable free factor of $G$.

Then either $L$ is a thick free factor, or else $H$ is conjugate to a subgroup of $G_v$ for some isolated vertex $v\in V\Gamma$.

Write $G=G_{\Lambda_1}\ast\dots\ast G_{\Lambda_k}\ast G_{v_1}\ast\dots\ast G_{v_\ell}$, where the $G_{\Lambda_j}$ are thick, and each $v_j$ is an isolated vertex. Let $T$ be the Bass–Serre tree of this decomposition of $G$ as a free product: this is a $G$-tree with trivial edge stabilizers.

Being freely indecomposable, the group $H$ fixes a vertex of $T$. So

* either $L$ is conjugate to a subgroup of $G_{v_j}$ for some isolated vertex $v_j$ (and we are done),

* or else $L$ is non-trivial and conjugate to a subgroup of a thick free factor $G_{\Lambda_j}$ of $G$.

We assume that the latter case holds, and without loss of generality that $H\subseteq G_{\Lambda_j}$. Since $L$ is a free factor of $G$, it is also a free factor of $G_{\Lambda_j}$. And since $G_{\Lambda_j}$ is freely indecomposable and $L$ is non-trivial, we have $L=G_{\Lambda_j}$.

§.§.§ A combinatorial lemma

Our starting point for proving Theorem <ref> is the following variation over Lemma <ref>, lemma:graph.

Let $\Gamma$ be a strongly reduced finite simple graph not reduced to one vertex, and let $G$ be a graph product over $\Gamma$. Let $B\subseteq G$ be a parabolic subgroup. The following two assertions are equivalent.

* $B$ is conjugate to $G_v$ for some untransvectable vertex $v\in V\Gamma$.

* There exist $n\geq 1$ and a chain of parabolic subgroups

\[G=F_0\supseteq L_1\supseteq P_1\supseteq F_1\supseteq\dots\supseteq L_n\supseteq P_n\supseteq F_n=B\]

such that

* for every $j\in\{1,\dots,n\}$, $L_j$ is a thick free factor of $F_{j-1}$ whose type is not a clique,

* for every $j\in\{1,\dots,n\}$, $P_j$ is a maximal product parabolic subgroup of $L_{j}$,

* for every $j\in\{1,\dots,n-1\}$, the clique factor of $P_j$ is trivial, and $F_j$ is a factor of $P_j$,

* $F_n$ is the clique factor of $P_n$,

* for every non-trivial subgroup $W\subseteq F_n$, one has $N_G(W)\subseteq N_G(F_n)$.

In our proof, we will use the following fact: every subgroup $A\subseteq G$ is contained in a unique smallest parabolic subgroup $\tilde{A}$, called the parabolic supportParabolic!Parabolic support of $A$ (it is also the intersection of all parabolic subgroups that contain $A$). And one has $N_G(A)\subseteq N_G(\tilde{A})$.

Remark for future use, that it implies that if $N$ is a subgroup of $G$ that normalizes $A$, then $\tilde{N}$ normalizes $\tilde{A}$.

We first prove that $(1)\Rightarrow (2)$.

Without loss of generality we will assume that $B=G_v$, with $v$ untransvectable.
Consider subgroups $F_j,P_j$ given by Lemma <ref>. In particular they satisfy Assertions (c) and (d). Since $P_j$ is a maximal product parabolic subgroup of $F_{j-1}$, it is contained in a thick free factor $L_j$ of $F_{j-1}$. And $P_j$ is a maximal product parabolic subgroup of $L_j$. Additionally, since $P_j$ is not of isolated clique type, the type of $L_j$ is not a clique. So Assertions (a) and (b) are also satisfied. Finally, Assertion (e) follows from the fact that $F_n=G_v$ is the parabolic support of any non-trivial subgroup $W\subseteq F_n$.

We now prove that $(2)\Rightarrow (1)$.

Assume that we have subgroups $F_j,L_j,P_j$ as in the statement. We will prove that

* the subgroups $F_j$ and $P_j$ satisfy the three conditions from Lemma <ref>, and

* $F_n=B$ is conjugate to a vertex group $G_v$.

Once these two facts are proved, the untransvectability of $v$ follows from Lemma <ref>.

We first prove the second fact. Assertion (d) ensures that $F_n$ is conjugate to $G_{\Upsilon}$, for some complete subgraph $\Upsilon\subseteq\Gamma$. If $|V\Upsilon|\ge 2$, the fact that $\Gamma$ is strongly reduced ensures that $\Upsilon$ is not collapsible, so there exists a vertex $w\in V\Upsilon$ such that $N_G(G_w)\nsubseteq N_G(G_\Upsilon)$. This contradicts Assertion (e). Therefore $|V\Upsilon|=1$, in other words $F_n$ is conjugate to a vertex group.

We now prove the first fact. The second and third conditions from Lemma <ref> are exactly Assertions (c) and (d). We focus on the first condition, so let $j\in \{1,\dots,n\}$. Assertions (a) and (b) ensure that $P_j$ is a product parabolic subgroup of $F_{j-1}$. We observe that as such, it is maximal. Indeed, if $P_j\subseteq P'$ with $P'\subseteq F_{j-1}$ a product parabolic subgroup, then the type of $P'$ is a connected subgraph of the type of $F_{j-1}$, and therefore $P'$ must be contained in the free factor $L_j$. There remains to prove that $P_j$ is not of isolated clique type. If $j\in\{1,\dots,n-1\}$, this follows from the fact that the clique factor of $P_j$ is trivial. If $j=n$, since $L_n$ is a thick free factor, $P_n$ is not conjugate to a vertex group. And since the clique factor of $P_n$ is $F_n$, which is conjugate to a vertex group (by the previous paragraph), the type of $P_n$ cannot be a clique on at least $2$ vertices either. So $P_n$ is not of isolated clique type.

In view of Lemma <ref>, we are left with giving a group-theoretic characterization of thick free factors, maximal products and their factors and clique factors.

[More on the comparison with the groupoid framework]
As a follow-up to Remark <ref>, let us continue our comparison between the group-theoretic and groupoid-theoretic frameworks, and explain why we reach a stronger conclusion in this appendix than in Sections <ref> and <ref> (namely, a characterization of all conjugates of untransvectable vertex groups, without assuming $\Gamma$ to be transvection-free, and with no assumption on the vertex groups).

In the zoom-in process described in Lemmas <ref> and <ref>, it can happen that the type $\Upsilon_j$ of one of the factors $F_j$ is disconnected. A delicate situation for us is when one of the connected components of $\Upsilon_j$ is a clique (possibly just one vertex). In Section <ref>, when this happens, Property $\Pprod$ did not allow us to recognize maximal products in $F_j$ from subgroup(oid)s that are contained in an isolated clique factor. This is why in Step 4 of the proof of Lemma <ref> (the key lemma that characterizes subgroupoids of vertex type), we had to use an extra argument, which crucially relied on the fact that $\Gamma$ was transvection-free, to distinguish actual subgroupoids of vertex type from those that are only contained in a subgroupoid of clique type.

In Section <ref>, the situation was slightly different: since we were assuming that all vertex groups were amenable, the isolated clique subgroups were always amenable, which enabled us to easily distinguish them from the other maximal product subgroups.

In this appendix, the notion of a thick free factor is the crucial tool to discard isolated vertex groups that arise in the process. This is why it is important for us to introduce the groups $L_j$ in Lemma <ref>, that we were not using in Part <ref> of this work.

§.§.§ Maximal products are preserved

The following lemma is a variation on Lemma <ref>, lemma:maximal-product-nonamenable.

Let $G$ be a non-abelian graph product over a connected finite simple graph $\Gamma_G$ which is not reduced to a vertex.

If $P_1,P_2$ are parabolic subgroups such that $P=P_1\times P_2$ is a maximal product parabolic subgroup, then either $P_1$ or $P_2$ is non-cyclic.

We first treat the case where $\Gamma_G$ is a clique. Since $G$ is non-abelian, either $P_1$ or $P_2$ is non-cyclic.

We now assume that $\Gamma_G$ is not a clique. Let $\Lambda$ be the type of $P$. Without loss of generality, we will assume that $P=G_{\Lambda}$, and that $\Lambda=\Lambda_1\circ\Lambda_2$, with $P_1=G_{\Lambda_1}$ and $P_2=G_{\Lambda_2}$.

We first prove that $\Lambda$ is not a clique.

* If $\Lambda=\Gamma_G$, by assumption its type is not a clique.

* Otherwise, since $\Gamma_G$ is connected, we can find two adjacent vertices $v,w$, with $v\in V\Lambda$ and $w\notin V\Lambda$. Then $G_{\st(v)}$ is a parabolic subgroup of product type, which is different from $P$ because it contains $G_w$. By maximality of $P$, we have $P\nsubseteq G_{\st(v)}$, which implies that $\Lambda\nsubseteq\st(v)$. So $\Lambda$ is not a clique.

We can therefore assume without loss of generality that the type $\Lambda_1$ of $P_1$ contains two non-adjacent vertices $v,w$. Then $P_1$ contains a subgroup isomorphic to $G_v*G_w$, and is therefore non-cyclic.

The following definition can be seen as a group theoretic analogue of Property $\Pprod$ (see <ref>, de:pprod).

[Property $\Qprod$]Property!$\Qprod$
Let $G$ be a group, and $P\subseteq G$ be a subgroup. We say that the pair $(G,P)$ satisfies Property $\Qprod$ if it verifies the following assertions.

$\Qprod_1$ There exists a non-trivial element $g\in P$, and a non-cyclic subgroup $N\subseteq P$, such that $N$ centralizes $g$ and is normal in $P$.

$\Qprod_2$ $P$ is maximal for inclusion among all subgroups $P'\subseteq G$ that satisfy Property $\Qprod_1$.

This property is stable under group isomorphisms. That is, if $f:G\to H$ is an isomorphism between two groups, and if $P\subseteq G$ is a subgroup such that $(G,P)$ satisfies Property $\Qprod$, then $(H,f(P))$ satisfies Property $\Qprod$. All properties considered in the present section will be stable under group isomorphisms.

The following lemma is a group theoretic analogue of <ref>, lemma:maximal-product.

Let $G$ be a non-abelian graph product over a connected finite simple graph $\Gamma_G$ which is not reduced to a vertex. Let $P\subseteq G$ be a subgroup.

Then $P$ is a maximal product parabolic subgroup if and only if $(G,P)$ satisfies Property $\Qprod$.

Step 1. Let us first prove that if $P$ is a maximal product parabolic subgroup, then $(G,P)$ satisfies Property $\Qprod_1$.

Write $P=P_1\times P_2$ as a product of two parabolic subgroups, with $P_1$ non-cyclic (see Lemma <ref>). Then Property $\Qprod_1$ is satisfied by letting $g\in P_2$ be any non-trivial element, and letting $N=P_1$.

Step 2. Let us now prove that if $(G,P)$ satisfies Property $\Qprod_1$, then $P$ is contained in a product parabolic subgroup.

So let $g\in P$ and $N\subseteq P$ be as in Property $\Qprod_1$. Let $\tilde A\subseteq G$ be the parabolic support of $g$, and let $\tilde{N}$ be the parabolic support of $N$. Notice that $\tilde A$ and $\tilde{N}$ are non-trivial (because $g$ and $N$ are non-trivial). Then $\tilde{N}\subseteq \tilde A\times \tilde A^{\perp}$, and $P\subseteq \tilde{N}\times\tilde{N}^{\perp}$.

* If $\tilde{N}^\perp\neq\{1\}$, then $P$ is contained in $\tilde{N}\times \tilde{N}^\perp$, which is a product of two non-trivial subgroups.

* Now assume that $\tilde{N}^\perp=\{1\}$, then $P\subseteq \tilde{N}\subseteq \tilde{A}\times \tilde{A}^\perp$.

* If $\tilde{A}$ is conjugate to a vertex group, then $\tilde{A}^{\perp}\neq\{1\}$ because $\Gamma_G$ is connected and not reduced to one vertex. So we are done in this case.

* If $\tilde{A}$ splits non-trivially as a product, we are done.

* Otherwise, we can apply <cit.>, which says that if $\tilde{A}$ is not conjugate to a vertex group and does not split as a product, then $Z_G(g)=\langle g\rangle\times \tilde{A}^{\perp}$. Since the centralizer of $g$ is non-cyclic, we deduce that $\tilde{A}^{\perp}\neq\{1\}$, and again we are done.

Step 3. We now prove that if $P$ is a maximal product parabolic subgroup, then $(G,P)$ satisfies Property $\Qprod_2$.

So let $P'\subseteq G$ be a subgroup that satisfies Property $\Qprod_1$, with $P\subseteq P'$. Step 2 implies that $P'$ is contained in a product parabolic subgroup $Q$. Then $P\subseteq Q$, and by maximality $P=Q$. In particular $P=P'$, as desired.

Step 4. We are left with proving that if $(G,P)$ satisfies Property $\Qprod$, then $P$ is a maximal product parabolic subgroup.

By Step 2, $P$ is contained in a product parabolic subgroup $Q$. Up to increasing $Q$, we will assume that $Q$ is a maximal product parabolic subgroup, so Step 1 ensures that $Q$ satisfies Property $\Qprod_1$. And Property $\Qprod_2$ for $P$ thus ensures that $P=Q$, so $P$ is a maximal product parabolic subgroup.

§.§.§ Thick free factors are preserved

[Property $\Qisol$]Property!$\Qisol$
Let $F$ be a group, and let $L\subseteq F$ be a subgroup. We say that the triple $(G,F,L)$ satisfies Property $\Qisol$ if the following three properties hold.

$\Qisol_1$ $L$ is a free factor of $F$.

$\Qisol_2$ $L$ is freely indecomposable.

$\Qisol_3$ There exists a non-trivial subgroup $B\subseteq L$ such that $N_G(B)\nsubseteq N_G(L)$.

Let $G$ be a graph product over a strongly reduced finite simple graph $\Gamma$, let $F\subseteq G$ be a parabolic subgroup, and let $L\subseteq F$ be a subgroup.

Then $L$ is a thick free factor of $F$ if and only if $(G,F,L)$ satisfies Property $\Qisol$.

Without loss of generality, we will assume that $F=G_{\Lambda}$, for some induced subgraph $\Lambda\subseteq\Gamma$.

We first assume that $L$ is a thick free factor – in particular $L$ is a free factor and is freely indecomposable, so Properties $\Qisol_1$ and $\Qisol_2$ hold. To prove $\Qisol_3$, without loss of generality, we will assume that $L=G_\Upsilon$, for some induced subgraph $\Upsilon\subseteq\Lambda\subseteq\Gamma$ which is a connected component of $\Lambda$ on at least two vertices. Since $\Gamma$ is strongly reduced, there exists $v\in V\Upsilon$ such that $\lk_{\Gamma}(v)\nsubseteq\Upsilon\circ\Upsilon^{\perp}$. Letting $B=G_v$ shows that Property $\Qisol_3$ holds.

Conversely, we now assume that $(G,F,L)$ satisfies Property $\Qisol$. Since $L$ is a freely indecomposable free factor of $G_\Lambda$, Lemma <ref> ensures that

* either $L$ is thick, and we are done,

* or else there exists an isolated vertex $v\in V\Lambda$, such that $L$ is contained in a conjugate of $G_v$.

We assume that we are in the second situation, and without loss of generality that $L\subseteq G_v$, and we aim for a contradiction. Since $L$ is a free factor of $G$ contained in $G_v$, we deduce that $L$ is a free factor of $G_v$, i.e. $G_v$ splits as $G_v=L\ast L'$ for some subgroup $L'\subseteq G_v$ (possibly with $L'=\{1\}$).

We now observe that $G$ splits as a graph product over a graph $\tilde\Gamma$, with one vertex group equal to $L$: indeed,

* if $L=G_v$ and $L'=\{1\}$, we simply let $\tilde\Gamma=\Gamma$;

* otherwise, $\tilde\Gamma$ is obtained from $\Gamma$ by replacing $v$ by two non-adjacent vertices $w,w'$, and joining both $w$ and $w'$ by an edge to all vertices of $\Gamma$ that are adjacent to $v$; then $G$ is a graph product over $\tilde{\Gamma}$ with $G_w=L$ and $G_{w'}=L'$.

Now, if $B\subseteq L$ is a non-trivial subgroup, then $L$ is the parabolic support of $B$ for the graph product structure over $\tilde{\Gamma}$. Therefore $N_G(B)\subseteq N_G(L)$, which contradicts Property $\Qisol_3$.

§.§.§ Special subproducts

Let $P=\mathsf{F}_1\times\dots\times\mathsf{F}_n$ be a product parabolic subgroup of $G$, written in such a way that no $\mathsf{F}_j$ further splits as a direct product of parabolic subgroups. In particular the clique factor of $P$ is decomposed as the product of factors that are conjugate to vertex groups. Recall from <ref>, sec:factors that a subproduct is a subgroup of $P$ equal to the product of finitely many factors $\mathsf{F}_j$ (possibly zero or one). A subproduct $S$ of $P$ is special if $N_G(S)\nsubseteq N_G(P)$.

The following definition is the group theoretic analogue of <ref>, de:pspec.

[Property $\Qspec$]Property!$\Qspec$
Let $G$ be a group, and $S\subseteq P$ be subgroups of $G$. We say that the triple $(G,P,S)$ satisfies Property $\Qspec$ if $S\unlhd P$, and $N_G(S)\nsubseteq N_G(P)$.

Notice that if $G$ is a graph product and $P$ is a product parabolic subgroup of $G$, and if $S$ is a special subproduct of $P$, then $(G,P,S)$ satisfies Property $\Qspec$. Conversely, we have the following statement (analogous to <ref>, lemma:spec).

Let $G$ be a graph product over a finite simple graph $\Gamma$. Let $P\subseteq G$ be a product parabolic subgroup, and $S\subseteq P$ be a subgroup.

If $(G,P,S)$ satisfies Property $\Qspec$, then the parabolic support of $S$ is a special subproduct of $P$.

Let $\tilde{S}$ be the parabolic support of $S$. By assumption $S\unlhd P$, so $\tilde{S}\unlhd P$, and therefore $P\subseteq \tilde{S}\times \tilde{S}^{\perp}$. So $\tilde{S}$ is a subproduct of $P$. In addition, since $N_G(S)\nsubseteq N_G(P)$, we have $N_G(\tilde{S})\nsubseteq N_G(P)$, i.e. $\tilde{S}$ is special. In summary $S$ is contained in the special subproduct $\tilde{S}$.

§.§.§ Clique-inclusive co-factors

Recall from Section <ref>, sec:factors that if $P$ is a product parabolic subgroup of $G$, which splits as $P=C_0\times F_1\times\dots\times F_k$, where $C_0$ is the clique factor, and no $F_j$ further decomposes as a direct product of two parabolic subgroups, then a clique-inclusive co-factor is a subgroup of $P$ of the form $C_0\times F_1\times\dots\times\hat{F}_j\times\dots\times F_k$, for some $j\in\{1,\dots,k\}$. By convention, if $P=C_0$, there is no clique-inclusive co-factor.
The following definition is analogous to <ref>, de:padm.

[Property $\Qadm$]Property!$\Qadm$
Let $G$ be a group, and $Q\subseteq P$ be subgroups. We say that the triple $(G,P,Q)$ satisfies Property $\Qadm$Property!$\Qadm$ if the following three properties hold.

$\Qadm_1$ The subgroup $Q$ is normal in $P$.

$\Qadm_2$ There exists a non-trivial subgroup $B\subseteq P$, which is normalized both by $Q$ and by a subgroup $N\subseteq G$ such that

(a) we have $N\nsubseteq N_G(P)$;

(b) whenever $S\subseteq P$ is a subgroup such that $B\subseteq S$, and such that $(G,P,S)$ satisfies Property $\Qspec$, we have $N\nsubseteq N_G(S)$.

$\Qadm_3$ The subgroup $Q$ is maximal among all subgroups $Q'\subseteq P$ such that $(G,P,Q')$ satisfies Properties $\Qadm_1$ and $\Qadm_2$.

We now distinguish two cases depending on whether the type of $P$ is a clique or not.

Let $G$ be a graph product over a finite simple graph $\Gamma$. Let $P\subseteq G$ be a product parabolic subgroup whose type is a clique.

Then there does not exist any subgroup $Q\subseteq P$ such that $(G,P,Q)$ satisfies Property $\Qadm$.

It is enough to prove that whenever $B,N$ are subgroups of $P$, with $B$ non-trivial and normalized by $N$, one of the Assertions (a) and (b) from $\Qadm_2$ fails.

So let $B,N$ be as above, and let $\tilde{B}$ be the parabolic support of $B$. Notice that $N\subseteq N_G(B)\subseteq N_G(\tilde{B})$. Notice also that $\tilde{B}$ is a subproduct of $P$, because the type of $P$ is a clique.

If $\tilde{B}$ is non-special, then $N_G(\tilde{B})=N_G(P)$, and therefore $N\subseteq N_G(P)$, so Assertion (a) fails.

If $\tilde{B}$ is special, then $(G,P,\tilde{B})$ satisfies Property $\Qspec$, and the fact that $N\subseteq N_G(\tilde{B})$ shows that Assertion (b) fails.

The following lemma is analogous to <ref>, lemma:admissible.

Let $G$ be a graph product over a strongly reduced finite simple graph $\Gamma$. Let $P\subseteq G$ be a product parabolic subgroup whose type is not a clique. Let $Q\subseteq P$ be a subgroup.

Then $Q$ is a clique-inclusive co-factor of $P$ if and only if $(G,P,Q)$ satisfies Property $\Qadm$.

Step 1. We first prove that if $Q$ is a clique-inclusive co-factor of $P$, then $(G,P,Q)$ satisfies Properties $\Qadm_1$ and $\Qadm_2$.

We clearly have $Q\unlhd P$, i.e. $\Qadm_1$ holds.

To prove Property $\Qadm_2$, write $P=C_0\times F_1\times\dots\times F_k$, where $C_0$ is the clique factor, and no $F_j$ further decomposes as a direct product of two parabolic subgroups. Without loss of generality, we will assume that $P=G_\Lambda$ for some induced subgraph $\Lambda\subseteq \Gamma$, and each $F_j$ is equal to $G_{\Lambda_j}$, where $\Lambda_j$ is a join factor of $\Lambda$ that is not reduced to a point. Notice that $k\ge 1$ because the type of $P$ is not a clique. There thus exists $j\in\{1,\dots,k\}$ such that  $Q=C_0\times F_1\times\dots\times\hat{F}_j\times\dots\times F_k$.

Since $\Gamma$ is strongly reduced, the subgraph $\Lambda_j$ is not collapsible. Therefore, there exists a vertex $v\in V\Lambda_j$ which is joined to a vertex $w\notin V\Lambda_j$, such that $w$ is not joined to all vertices of $\Lambda_j$. Notice in particular that

\[G_w \subseteq G^\perp_v \quad \text{and}
\quad G_w\nsubseteq F_j\times F^\perp_j \quad \text{and}
\quad G_w\nsubseteq P\times P^\perp.\]

Let $B=G_v$. Then $B\subseteq P$ is non-trivial and normalized by $Q$. Let $N=G_v\times G_v^{\perp}$, which normalizes $B$. Then $G_w\subseteq N$, so in particular $N\nsubseteq N_G(P)$, which verifies $\Qadm_2$(a).

For $\Qadm_2$(b), let $S\subseteq P$ be a subgroup such that $B\subseteq S$, and such that $(G,P,S)$ satisfies Property $\Qspec$. Lemma <ref> ensures that the parabolic support $\tilde{S}$ of $S$ is a special subproduct of $P$. Since $G_v\subseteq \tilde{S}$ and $\tilde{S}$ is a subproduct, we have $F_j\subseteq \tilde{S}$. Since $G_w\nsubseteq N_G(F_j)$, we deduce that $G_w\nsubseteq N_G(\tilde{S})$. Therefore $N\nsubseteq N_G(\tilde{S})$ and in particular $N\nsubseteq N_G(S)$, as desired.

Step 2. We now assume that $(G,P,Q)$ satisfies Properties $\Qadm_1$ and $\Qadm_2$, and prove that $Q$ is contained in a clique-inclusive co-factor of $P$.

Let $B,N$ be as in $\Qadm_2$. Write $P=\mathsf{F}_1\times\dots\times \mathsf{F}_n$, where the $\mathsf{F}_j$ are chosen not to split further (in particular the clique factor of $P$ is decomposed as the product of factors that are conjugate to vertex groups). Let $\tilde{B}$ and $\tilde{N}$ be the respective parabolic supports of $B$ and $N$. Since $N$ normalizes $B$, we have $N\subseteq N_G(\tilde{B})$. For $j\in\{1,\dots,n\}$, let $\tilde{B}_j=\tilde{B}\cap \mathsf{F}_j$.

We claim that there exists $j\in\{1,\dots,n\}$ such that $\tilde{B}_j\neq\{1\}$ and $\tilde{B}_j\neq \mathsf{F}_j$. Indeed, otherwise $\tilde{B}$ would be a subproduct of $P$. Since $N\subseteq N_G(\tilde{B})$ and $N\nsubseteq N_G(P)$ by $\Qadm_2(a)$, we would have $N_G(\tilde{B})\nsubseteq N_G(P)$, in other words $\tilde{B}$ would be a special subproduct. So $(G,P,\tilde{B})$ would satisfy Property $\Qspec$, and the inclusion $N\subseteq N_G(\tilde{B})$ would give a contradiction to $\Qadm_2$(b). Hence our claim.

Let $j\in\{1,\dots,n\}$ be as above. The above claim shows that $\mathsf{F}_j$ contains $\tilde{B}_j$ as a proper and non-trivial parabolic subgroup, in particular $\mathsf{F}_j$ is not conjugate to a vertex group. Let $\tilde{Q}$ be the parabolic support of $Q$ and let us show that $\tilde{Q}\cap \mathsf{F}_j=\{1\}$. Since $Q$ normalizes $B$, we have $Q\subseteq N_G(\tilde{B})$, and therefore $\tilde{Q}\subseteq N_G(\tilde{B})$. In particular $\tilde{Q}\cap \mathsf{F}_j$ is contained in $N_{\mathsf{F}_j}(\tilde{B}_j)$. Since $\tilde{B}_j$ is a proper non-trivial parabolic subgroup of $\mathsf{F}_j$ and $\mathsf{F}_j$ does not split as a product, the normalizer $N_{\mathsf{F}_j}(\tilde{B}_j)$ is a proper parabolic subgroup of $\mathsf{F}_j$. Using Lemma <ref>, lemma:product-parabolic-subgroup we have
\[\tilde{Q}\subseteq \mathsf{F}_1\times\dots\times \mathsf{F}_{j-1}\times N_{\mathsf{F}_j}(\tilde{B}_j)\times \mathsf{F}_{j+1}\times\dots\times \mathsf{F}_n.\]

Let $\tilde{Q}_j:=\tilde{Q}\cap\mathsf{F}_j$, a proper parabolic subgroup of $\mathsf{F}_j$, contained in $N_{\mathsf{F}_j}(\tilde{B}_j)$. Since $Q\unlhd P$, we have $P\subseteq N_G(\tilde Q)$, and we deduce that

\[P\subseteq\mathsf{F}_1\times\dots\times \mathsf{F}_{j-1}\times N_{\mathsf{F}_j}(\tilde Q_j)\times \mathsf{F}_{j+1}\times\dots\times \mathsf{F}_n.\]

Therefore $N_{\mathsf{F}_j}(\tilde Q_j)=\mathsf{F}_j$, which implies that $\tilde Q_j=\{1\}$. We have thus proved that

\[Q\subseteq\tilde{Q}\subseteq \mathsf{F}_1\times\dots\times \hat {\mathsf{F}}_{j}\times\dots\times \mathsf{F}_n.\]

This concludes Step 2.

Step 3. We now assume that $(G,P,Q)$ satisfies Property $\Qadm$, and prove that $Q$ is a clique-inclusive co-factor of $P$.

By Step 2, $Q$ is contained in a clique-inclusive co-factor $Q'$ of $P$. By Step 1, the triple $(G,P,Q')$ satisfies Properties $\Qadm_1$ and $\Qadm_2$. So Property $\Qadm_3$ for $(G,P,Q)$ implies that $Q=Q'$.

Step 4. We are left with showing that if $Q\subseteq P$ is a clique-inclusive co-factor, then $(G,P,Q)$ satisfies Property $\Qadm_3$.

So let $Q'\subseteq P$ be such that $Q\subseteq Q'$ and $(G,P,Q')$ satisfies Properties $\Qadm_1$ and $\Qadm_2$. By Step 2, $Q'$ is contained in a clique-inclusive co-factor $Q''$ of $P$. In particular $Q$ and $Q''$ are two clique-inclusive co-factors of $P$ with $Q\subseteq Q''$, so in fact $Q=Q''$. In particular $Q=Q'$, and Property $\Qadm_3$ holds.

§.§.§ Factors

We now provide the group theoretic analogue of <ref>, de:pfact.

[Property $\Qfact$]Property!$\Qfact$
Let $G$ be a group, and $F\subseteq P$ be subgroups. We say that the triple $(G,P,F)$ satisfies Property $\Qfact$ if the following two assertions hold.

$\Qfact_1$ $F$ is non-trivial, and can be written as an intersection $Q_1\cap\dots\cap Q_\ell$ of subgroups $Q_j$ of $P$ with $\ell \ge 1$, where for every $j\in\{1,\dots,\ell \}$, the triple $(G,P,Q_j)$ satisfies Property $\Qadm$.

$\Qfact_2$ $F$ is minimal among all subgroups of $P$ that satisfy $\Qfact_1$.

The following lemma records the fact that factors arise as minimal non-trivial intersections of co-factors.

Let $G$ be a graph product over a strongly reduced finite simple graph $\Gamma$. Let $P\subseteq G$ be a product parabolic subgroup with clique factor $C_0$. Let $F\subseteq P$ be a subgroup.

* If the type of $P$ is a clique, then $(G,P,F)$ never satisfies Property $\Qfact$.

* If the type of $P$ is not a clique, then

* If $C_0=\{1\}$, then $F$ is a factor of $P$ if and only if $(G,P,F)$ satisfies Property $\Qfact$.

* If $C_0\neq\{1\}$, then $F$ is the clique factor of $P$ if and only if $(G,P,F)$ satisfies Property $\Qfact$.

The first assertion is a consequence of the fact that there does not exist any subgroup $Q\subseteq P$ such that $(G,P,Q)$ satisfies Property $\Qadm$ (Lemma <ref>).

The second assertion follows from Lemma <ref>, together with the fact that

* if $C_0=\{1\}$, then factors of $P$ are exactly the minimal non-trivial intersections of clique-inclusive co-factors;

* if $C_0\neq\{1\}$, then the clique factor of $P$ is the unique minimal non-trivial intersection of clique-inclusive co-factors.

We now aim at a characterization of product parabolic subgroups with trivial clique factor.

[Property $\Qtc$]Property!$\Qtc$
Let $G$ be a group, and $P\subseteq G$ be a subgroup. We say that the pair $(G,P)$ satisfies Property $\Qtc$ if there exist $\ell \geq 1$ and subgroups $Q_1,\dots,Q_\ell\subseteq P$ such that $(G,P,Q_j)$ satisfies Property $\Qadm$ for every $j\in\{1,\dots,\ell\}$, and $Q_1\cap\dots\cap Q_\ell$ is trivial.

The following lemma records the fact that the clique factor of $P$ is trivial if and only if the intersection of all clique-inclusive co-factors is trivial.

Let $G$ be a graph product over a strongly reduced finite simple graph $\Gamma$. Let $P$ be a product parabolic subgroup.

Then the clique factor of $P$ is trivial if and only if $(G,P)$ satisfies Property $\Qtc$.

§.§.§ Proof of Theorem <ref>

The following property is inspired by Lemma <ref>. It can be seen as a group theoretic variation of <ref>, de:pvert. Let us mention that it is closer to the version presented in <ref>, de:pamen.

[Property $\Qvert$]Property!$\Qvert$
Let $G$ be a group, and $V\subseteq G$ be a non-trivial subgroup. We say that the pair $(G,V)$ satisfies Property $\Qvert$ if there exists a chain of subgroups

\[G=F_0\supseteq L_1\supseteq P_1\supseteq F_1\supseteq\dots\supseteq L_n\supseteq P_n\supseteq F_n=V\]

with $n\ge 1$ such that

* for every $j\in\{1,\dots,n\}$, the triple $(G,F_{j-1},L_j)$ satisfies Property $\Qisol$ and $L_j$ is non-abelian;

* for every $j\in\{1,\dots,n\}$, the pair $(L_j,P_j)$ satisfies Property $\Qprod$;

* for every $j\in\{1,\dots,n-1\}$, the pair $(G,P_j)$ satisfies Property $\Qtc$, while $(G,P_n)$ does not;

* for every $j\in\{1,\dots,n\}$, the triple $(G,P_j,F_j)$ satisfies Property $\Qfact$;

* for every non-trivial subgroup $W\subseteq V$, one has $N_G(W)\subseteq N_G(V)$.

Let $G$ be a graph product over a strongly reduced finite simple graph, not reduced to one vertex. A subgroup $V\subseteq G$ is conjugate to an untransvectable vertex group if and only if $(G,V)$ satisfies Property $\Qvert$.

We first assume that $V$ is conjugate to an untransvectable vertex group, and prove that $(G,V)$ satisfies Property $\Qvert$. Let

\[G=F_0\supseteq L_1\supseteq P_1\supseteq F_1\supseteq\dots\supseteq L_n\supseteq P_n\supseteq F_n=V\]

be a chain of parabolic subgroups as in Lemma <ref>, lemma:graph-2. Then

* For every $j\in\{1,\dots,n\}$, the pair $(G,F_{j-1},L_j)$ satisfies Property $\Qisol$ (this uses Assertion 2.(a) of Lemma <ref>, together with Lemma <ref>), and $L_j$ is non-abelian because its type is not a clique.

* For every $j\in\{1,\dots,n\}$, the pair $(L_j,P_j)$ satisfies Property $\Qprod$: this uses Assertion 2.(b) of Lemma <ref>, together with Lemma <ref> (to apply Lemma <ref>, one needs to know that the type of $L_j$ is a connected subgraph of $\Gamma$ which is not reduced to one vertex – this is ensured by Assertion 2.(a) of Lemma <ref>).

* For every $j\in\{1,\dots,n-1\}$, the pair $(G,P_j)$ satisfies Property $\Qtc$, while $(G,P_n)$ does not: this follows from Assertions 2.(c) and 2.(d) of Lemma <ref>, together with Lemma <ref>.

* For every $j\in\{1,\dots,n\}$, the triple $(G,P_j,F_j)$ satisfies Property $\Qfact$. Indeed, we saw that the type of $L_j$ is a connected subgraph which is not a clique; since $P_j$ is a maximal product parabolic subgroup inside $L_j$, its type is not a clique either. Therefore, Assertions 2.(c) and 2.(d) of Lemma <ref>, together with Lemma <ref>, ensure that $(G,P_j,F_j)$ satisfies Property $\Qfact$.

* For every non-trivial subgroup $W\subseteq F_n$, one has $N_G(W)\subseteq N_G(F_n)$: this is exactly Assertion 2.(e) of Lemma <ref>.

So Property $\Qvert$ is satisfied.

Conversely, let us assume that $(G,V)$ satisfies Property $\Qvert$. Let
\[G=F_0\supseteq L_1\supseteq P_1\supseteq F_1\supseteq\dots\supseteq L_n\supseteq P_n\supseteq F_n=V\]
be a chain of subgroups given by Property $\Qvert$. Successive applications of Lemmas <ref>, <ref>, <ref> and <ref> show that

* For every $j\in\{1,\dots,n\}$, $L_j$ is a thick free factor of $F_{j-1}$ (in particular its type is not reduced to one vertex), and by $\Qvert_1$ the subgroup $L_j$ is non-abelian.

* For every $j\in\{1,\dots,n\}$, $P_j$ is a maximal product parabolic subgroup of $L_j$.

* For every $j\in\{1,\dots,n\}$, the type of $P_j$ (and therefore of $L_j$) is not a clique. This follows from $\Qvert_4$ and Assertion 1 of <ref>.

* For every $j\in\{1,\dots,n-1\}$ by $\Qvert_3$ and <ref> the clique factor of $P_j$ is trivial, while the clique factor of $P_n$ is non-trivial.

* For every $j\in\{1,\dots,n\}$, the subgroup $F_j$ is a factor of $P_j$, by $\Qvert_4$ and Assertion 2 of <ref>. Moreover $F_n$ is the clique factor of $P_n$.

Together with Assertion 5 from Property $\Qvert$, this means that all assertions from Lemma <ref> hold. Lemma <ref> therefore implies that $V$ is conjugate to an untransvectable vertex group.

We are now in position to complete the proof of the main theorem of this appendix.

Let $v\in V\Gamma_G$ be an untransvectable vertex. Then $(G,G_v)$ satisfies Property $\Qvert$ (Lemma <ref>), and therefore so does $(H,f(G_v))$ – notice indeed that all properties of the present section are stable under group isomorphisms (Remark <ref>). Applying Lemma <ref> again shows that $f(G_v)$ is conjugate to an untransvectable vertex group.

§.§.§ An extension to products

We finally mention that Theorem <ref> can be extended to include direct products, as follows.

Let $G,H$ be graph products over finite simple graphs $\Gamma_G,\Gamma_H$ that decompose as joins of strongly reduced graphs, none of which is reduced to a vertex or an edge.

Then for every isomorphism $f:G\to H$ and every untransvectable vertex $v\in V\Gamma_G$, there exists an untransvectable vertex $w\in V\Gamma_H$ such that $f(G_v)$ is conjugate to $H_w$.

Write $G=G_1\times\dots\times G_k$ and $H=H_1\times\dots\times H_\ell$, where the $G_j$ and $H_j$ are irreducible graph products over strongly reduced finite simple graphs, not reduced to a vertex. Notice that (by irreducibility) all groups $G_j$ and $H_j$ have trivial center, and by <cit.> they admit a generating set consisting of pairwise non-commuting elements with maximal centralizers. Since $G$ and $H$ are isomorphic, we can therefore apply <cit.> with $A_0=B_0=\{1\}$ and deduce that $k=\ell$, and that there exists a bijection $\sigma:\{1,\dots,k\}\to\{1,\dots,\ell \}$ such that for every $j\in\{1,\dots,k\}$, the image $f(G_j)$ is equal to $H_{\sigma(j)}$. The conclusion then follows from Theorem <ref>, applied to each subgroup $G_j$.

Again, together with <cit.>, we deduce the following consequence, which in particular establishes the version of Theorem <ref>, where “commensurable” and “strongly commensurable” are replaced by “isomorphic”.

Let $G,H$ be graph products over finite simple graphs $\Gamma_G,\Gamma_H$ that decompose as joins of strongly reduced graphs, with $\Gamma_G$ transvection-free.

If $G$ and $H$ are isomorphic, then there exist an isomorphism $\sigma:\Gamma_G\to \Gamma_H$ such that for every $v\in V\Gamma$, the vertex groups $G_v$ and $H_{\sigma(v)}$ are isomorphic.

§.§ Acylindrical hyperbolicity for automorphisms of graph products

A group is acylindrically hyperbolicAcylindrically hyperbolic <cit.> if it admits a non-elementary action on a Gromov-hyperbolic metric space $X$ (i.e. with unbounded orbits and no finite orbit in the Gromov boundary $\partial_\infty X$), which is acylindrical in the sense that for every $R>0$, there exist $L,N\ge 0$ such that given any two points $x,y\in X$ with $d(x,y)\ge L$, one has

\[\left\vert \left\{g\in G \mid d(x,gx)\le R \text{~and~} d(y,gy)\le R \right\} \right\vert\le N. \]

We show the following result, which is essentially due to Genevois <cit.>. See also the remark below for more details.

Let $\Gamma$ be a finite irreducible simple graph, not reduced to a vertex, and let $G$ be a finitely generated graph product over $\Gamma$. Assume that $G$ is not isomorphic to $\mathbb{Z}/2\mathbb{Z}\ast\mathbb{Z}/2\mathbb{Z}$.

Then $\Aut(G)$ is acylindrically hyperbolic.

[Parallel with Genevois’ work]
The above theorem was proved by Genevois <cit.> under the extra assumption that all vertex groups are graphically irreducible (see Remark <ref> for the definition). See also the work of Genevois–Martin <cit.>, where this was proved without any assumption on the vertex groups, but with more restrictions on the defining graph $\Gamma$.

It is unknown whether or not every finitely generated graphically irreducible group admits a decomposition as a graph product over a finite simple graph where all vertex groups are graphically irreducible. Our approach enables us to optimize Genevois' result by bypassing this question. In fact our proof still heavily relies on his work, and we only provide a new argument for Step 1 of Genevois' proof, carried in <cit.>.

In particular all consequences obtained from the acylindrical hyperbolicity of $\Aut(G)$ hold in this generality. Let us mention for instance the result by Fournier-Facio and Wade <cit.>, that now holds for any graph product $G$ as in Theorem <ref>, and says that the space of $\Aut(G)$-invariant homogeneous quasimorphisms on $G$ is infinite-dimensional. Likewise, the theorem of Coulon and Fournier-Facio <cit.>, saying that $G$ has an infinite simple characteristic quotient, also holds in this generality.

If $\Gamma$ is disconnected, then $G$ has infinitely many ends, and the theorem follows from <cit.>. From now on we assume that $\Gamma$ is connected.

By collasping some subgraphs of $\Gamma$ if needed, we can write $G$ as a graph product over a strongly reduced finite simple graph $\bar\Gamma$, not reduced to one vertex, and still irreducible. Therefore, without loss of generality, we can (and shall) assume that $\Gamma$ is strongly reduced to start with.

By <cit.>, there exists an element $g\in G$ such that

* $g$ is not contained in any proper parabolic subgroup of $G$, and

* if $g$ is elliptic in some isometric $G$-action on an $\mathbb{R}$-tree $T$ whose arc stabilizers are contained in proper parabolic subgroups, then $G$ has a global fixed point in $T$.

In particular $g$ is irreducible in the sense of <cit.>, i.e. it is not conjugate in any vertex group or product parabolic subgroup, and up to conjugating $g$ we can further assume that it is graphically cyclically reduced in the sense of <cit.>. By <cit.>, the fixator of $g$ in $\Aut(G)$ has finite image in $\Out(G)$. Now, Theorem <ref> precisely shows that all assumptions from <cit.> are satisfied (our untransvectable vertices correspond to $\prec$-maximal vertices in Genevois' terminology). By applying <cit.>, we deduce that $g$ is a WPD element for some action of $\Aut(G)$ on a hyperbolic space (in the sense of Bestvina–Fujiwara <cit.>). Having a WPD element is enough to ensures that $\Aut(G)$ is either virtually cyclic or acylindrically hyperbolic <cit.>. Finally, $\Aut(G)$ is not virtually cyclic: since $G$ is not virtually cyclic and has trivial center (see <cit.>), the group of inner automorphisms of $G$ is not virtually cyclic.

§ NOTATIONS INDEX

$\preccurlyeq$ See Nota:AsymptoBehaviour;

$[G:H]_\Omega$ Compression constant, equal to ${m(X_H)}/{m(X_G)}$ (see Def:CompressionConstant);

$A_\Gamma$ Right-angled Artin group defined over the graph $\Gamma$;

$\mathfrak{C}_G$ See Nota:FrakC;

$\partial_\infty T_v$ Boundary of the tree $T_v$;

$\left(\partial_\infty T_v\right)^\mathrm{para}$ See Def:Tvpara;

$\left(\partial_\infty T_v\right)^\mathrm{reg}$ See Def:TvReg;

$\bD_G$ Right-angled building of $G$;

$E\Gamma$ Set of edges of the graph $\Gamma$;

$\mathfrak{F}$ The set of all standard clique cosets (see Def:FrakF);

$G \ltimes X$ Groupoid associated to an action of $G$ on a probability space $X$ (see <ref>;

$\calG$, $\calH$ Groupoids;

$\Gamma$, $\Lambda$, $\Upsilon$ Finite simple graphs;

$\Gamma \circ \Lambda$ The graph join of $\Gamma$ and $\Lambda$;

$\Gamma^e_G$ Extension graph of $G$ (see <ref>);

$\mathfrak{I}_G$ See Nota:FrakI;

$\kappa(f)$ Compression constant of $f$, equal to ${\nu(V)}/{\mu(U)}$ (see Def:CompressionConstantf);

$\lk(v)$ The link of the vertex $v$ (see Def:Link)

$\Lambda^\perp$ See notation:lambdaperp;

$\mathcal{P}_{<\infty}(Z)$ Set of all nonempty finite subsets of $Z$;

$\mathcal{P}_{\leq 2}(Z)$ Set of all nonempty subsets of $Z$ of cardinality at most $2$;

$(\Omega,m,X_G,X_H)$ A measure equivalence coupling (see <ref>);

$\mathbb{P}_G$ Set of parabolic subgroups of $G$;

$\mathrm{Prob}(K)$ Space of Borel probability measures on $K$;

$\rho$ A cocycle (groupoid framework);

$\st(v)$ The star of the vertex $v$ (see Def:Star)

$T_v$ Bass-Serre tree associated to the splitting $G=G_{\st(v)}\ast_{G_\lk(v)}G_{\Gamma\setminus\{v\}}$ of a graph product (see <ref>)

$V\Gamma$ Set of vertices of the graph $\Gamma$;

$\underline{\mathsf{w}}$ A word (see <ref>);

$X_G$ A fundamental domain for the action of a group $G$.

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