Proof. Throughout the proof we use the terminology that a set $F\subset \Lambda$ is said to be small relative to $\{\Lambda _0\}$ if it is contained into a finite union of $s\Lambda _0 t$, where $s,t\in \Lambda$. For any $F\subset \Lambda$, let $\mathcal {H}_F\subset L^2(M)$ be the $\|\cdot \|_2$-closed linear span of $\{B v_{\lambda }\mid \lambda \in F\}$ and denote by $P_F:L^2(M)\to \mathcal {H}_F$ the orthogonal projection onto $\mathcal {H}_F$. Let $\epsilon >0$ and denote $T=E_{A'\cap pMp}(e)$. Note that $T\in (A\cup \mathcal {G})'\cap pMp\subset A$ and that $T$ belongs to the $\|\cdot \|_2$-closed convex hull of $\{aea^* \mid a\in \mathcal {U}(A)\}$. Thus, we can take $a_1,\ldots,a_n\in \mathcal {U}(A)$ and $\alpha _1,\ldots,\alpha _n\in [0,1]$ such that if we denote $T_0=\sum _{i=1}^n \alpha _i a_i ea_i^*$, then $\|T-T_0\|_2\leq \epsilon$.

Since $\mathcal {G}''e\prec _M^s B\times \Lambda _0$, it follows from [Reference VaesVae13, Lemma 2.5] that there exists $F\subset \Lambda$ that is small relative to $\{\Lambda _0\}$ such that $\|we-P_{F}(we)\|_2\leq \epsilon /n$, for all $w\in \mathcal {G}$. Hence, for all $a\in \mathcal {U}(A), w\in \mathcal {G}$, we have

(2.1)\begin{equation} \biggl\|awT - \alpha_i\sum_{i=1}^n a (wa_iw^*)P_F(we)a_i^* \biggr\|_2\leq \epsilon + \sum_{i=1}^n \alpha_i\| aw(a_iea_i^*) - a(wa_iw^*)P_F(we) a_i^* \|_2\leq 2\epsilon. \end{equation}

Since $A\subset pBp$, we have $a (wa_iw^*)P_F(we)a_i^* \in \mathcal {H}_F$ and, hence, $\|awT-P_F(awT)\|_2\leq 2\epsilon$, for all $a\in \mathcal {U}(A),w\in \mathcal {G}$. Therefore, there exists a sequence $\{F_n\}_{n\ge 1}$ of subsets of $\Lambda$ that are small relative to $\{\Lambda _0\}$ such that $\|awT-P_{F_n}(awT)\|_2\to 0$ uniformly in $a\in \mathcal {U}(A),w\in \mathcal {G}$.

For every $\delta >0$ define the spectral projection $q_\delta =\chi _{(\delta,\infty )}(T)\in A$ and let $T_\delta \in A$ satisfying $TT_\delta =q_\delta.$ If we denote by $q_0$ the support projection of $T$, then $\|q_\delta -q_0\|_2\to 0$ as $\delta \to 0$. These altogether imply that $\|awq_\delta -P_{F_n}(awq_\delta )\|_2\to 0$ uniformly in $a\in \mathcal {U}(A),w\in \mathcal {G}$ and, hence, $\|awq_0-P_{F_n}(awq_0)\|_2\to 0$ uniformly in $a\in \mathcal {U}(A),\ w\in \mathcal {G}$. Finally, note that $q_0\in (A\cup \mathcal {G})'\cap pMp$ and $q_0\ge e$. This concludes the proof.

Proof. (1) Let $\mathcal {M}=M_0\bar \otimes M$ and $\tilde {\mathcal {M}}=M_0\bar \otimes \tilde M.$ The assumption implies that the bimodule $_{Q\tilde z} L^2(\tilde z\tilde {\mathcal {M}})_{\tilde {\mathcal {M}}}$ is weakly contained in $_{Q\tilde z} L^2(\tilde z \langle \tilde {\mathcal {M}},e_{M_0}\rangle )_{\tilde {\mathcal {M}}}$. If we denote by $z\in Q'\cap q(M_0\bar \otimes M)q)$ the support projection of $E_{\mathcal {M}}(\tilde z),$ we obtain that

(2.2)\begin{equation} {}_{Qz} L^2(z{\mathcal{M}})_{\mathcal{M}} \text{ is weakly contained in }_{Qz} L^2(z \langle \tilde{\mathcal{M}},e_{M_0}\rangle)_{{\mathcal{M}}}. \end{equation}

Note that $_{\mathcal {M}} L^2(\langle \tilde {\mathcal {M}},e_{M_0}\rangle )_\mathcal {M}\cong _{\mathcal {M}} L^2(\tilde {\mathcal {M}})\otimes L^2(\tilde M) _{\mathcal {M}}$. Note also that $_{\mathcal {M}} L^2(\tilde {\mathcal {M}})_{M_0}$ is weakly contained in $_{\mathcal {M}} L^2({\mathcal {M}})_{M_0}$ and $_{\mathbb {C}} L^2(\tilde M)_M$ is weakly contained in $_{\mathbb {C}}L^2(M)_M.$

These altogether imply that $_{Qz} L^2(\langle \tilde {\mathcal {M}},e_{M_0}\rangle )_\mathcal {M}$ is weakly contained in $_{Qz} L^2(\langle {\mathcal {M}},e_{M_0}\rangle )_\mathcal {M}$. Using (2.2) we deduce that $_{Qz} L^2(z{\mathcal {M}})_{{\mathcal {M}}}$ is weakly contained in $_{Qz} L^2(z \langle {\mathcal {M}},e_{M_0}\rangle )_{{\mathcal {M}}}$, which shows that $Qz$ is amenable relative to $M_0$ inside $\mathcal {M}.$

(2) By assuming the contrary, there exists a sequence $u_n\in \mathcal {U} (Q)$ such that

(2.3)\begin{equation} \|E_{M_0}(xu_ny)\|_2\to 0, \quad \text{for all } x,y\in M_0\bar\otimes M. \end{equation}

We want to show that (2.3) holds for all $x,y\in M_0\bar \otimes \tilde M,$ which will contradict the assumption. Note that it is enough to consider $x=1$ and $y\in \tilde M$. In this case, by using (2.3) we obtain $E_{M_0}(u_ny)=E_{M_0}(E_{M_0\bar \otimes M}(u_ny))=E_{M_0}(u_nE_{M_0\bar \otimes M}(y))$, which goes to $0$ in the $\|\cdot \|_2$-norm. This finishes the proof.

Proof. Denote $M=L(\Gamma )$. By applying [Reference DrimbeDri20a, Lemma 2.10], we get that $\Psi (P)$ is strongly non-amenable relative to $\mathcal {M}\bar \otimes 1$ inside $\mathcal {M}\bar \otimes M$. Then Remark 3.4 implies that $\Psi (P'\cap p\mathcal {M} p)$ is ${\rm (id}\otimes \alpha )$-rigid. For proving the second part, assume in addition that $P'\cap p\mathcal {M} p\nprec _{\mathcal {M}} B$ and let $A\subset M$ be an amenable subalgebra as given by the assumption that $M$ belongs to $\boldsymbol {\mathscr M}$. By using [Reference IoanaIoa12, Lemma 9.2(1)] we get $\Psi (P'\cap p\mathcal {M} p)\nprec _{\mathcal {M}\bar \otimes M} \mathcal {M}\otimes 1$, and by assumption we must have $\Psi (P'\cap p\mathcal {M} p)\nprec _{\mathcal {M}\bar \otimes M} \mathcal {M}\otimes A$. Hence, in combination with $\Psi (P'\cap p\mathcal {M} p)$ being ${\rm (id}\otimes \alpha )$-rigid and the fact that $L^2(\mathcal {M}\bar \otimes \tilde M)\ominus L^2(\mathcal {M}\bar \otimes M)$ is a mixing $\mathcal {M}\bar \otimes M$-bimodule relative to $\mathcal {M}\bar \otimes A$, we get from [Reference de Santiago, Hayes, Hoff and SinclairdSHHS20, Corollary 6.7] that $\Psi (P\vee (P'\cap p\mathcal {M} p))$ is ${\rm (id}\otimes \alpha )$-rigid.

Notation 3.7 Let $\Lambda$ be a countable icc group that is measure equivalent to a product $\Gamma =\Gamma _1\times \cdots \times \Gamma _n$ of $n\ge 1$ groups. By using [Reference FurmanFur99, Lemma 3.2], there exist $d\ge 1$, free ergodic pmp actions $\Gamma \curvearrowright (X,\mu )$ and $\Lambda \curvearrowright (Y,\nu )$ such that

\[ \mathcal{R}(\Lambda\curvearrowright Y)=\mathcal{R}(\Gamma\times \mathbb{Z}/d\mathbb{Z} \curvearrowright X \times \mathbb{Z}/d\mathbb{Z})\cap (Y\times Y). \]

Here, we considered that $\mathbb {Z}/d\mathbb {Z} \curvearrowright (\mathbb {Z}/d\mathbb {Z},c)$ acts by addition and $c$ is the counting measure. We also identified $Y$ as a measurable subset of $X\times \mathbb {Z}/d\mathbb {Z}$ and denote $p=1_Y \in L^\infty (X\times \mathbb {Z}/d\mathbb {Z})$. Note that $L^\infty (\mathbb {Z}/d\mathbb {Z})\rtimes \mathbb {Z}/d\mathbb {Z}=\mathbb {M}_d(\mathbb {C})$. Hence, by letting $B=L^\infty (Y)$, $A=L^\infty (X)\otimes M_d(\mathbb {C})$, and $M=A\rtimes \Gamma$, we have $pMp=B\rtimes \Lambda$ and $B\subset pAp$. Denote by $\{u_g\}_{g\in \Gamma }$ and $\{v_\lambda \}_{\lambda \in \Lambda }$ the canonical unitaries implementing the actions $\Gamma \curvearrowright A$ and $\Lambda \curvearrowright B$, respectively.

Following [Reference Popa and VaesPV10] we define the $*$-homomorphism $\Delta : pMp\to pMp \bar \otimes L(\Lambda )$ by $\Delta (bv_\lambda )=bv_\lambda \otimes v_\lambda$, for all $b\in B,\lambda \in \Lambda$. One can extend $\Delta$ to a $*$-homomorphism $\Delta :M\to M\bar \otimes L(\Lambda )$ and verify that $\Delta (M)'\cap M\bar \otimes L(\Lambda )=\mathbb {C} 1$ since $\Lambda$ is icc (see the first part of [Reference Drimbe, Hoff and IoanaDHI16, § 5] for more details).

For any $i\in \overline {1,n}$, let $\Psi ^i: M\to M\bar \otimes L(\Gamma _i)$ be the $*$-homomorphism given by $\Psi ^i(xu_g)=xu_g\otimes u_g$, for all $x\in A\rtimes \Gamma _{\widehat i} ,g\in \Gamma _i$.

Proof. Let $(\tilde M_i, (\alpha ^i_t)_{t\in \mathbb {R}})$ be the associated s-malleable deformation of $L(\Gamma _i) \in {\rm {\boldsymbol {\mathscr {M}}}}$. By assuming that $L(\Lambda )$ has property Gamma, we can use [Reference Houdayer and UedaHU16, Theorem 3.1] to obtain a decreasing sequence of diffuse abelian von Neumann subalgebras $Q_n\subset L(\Lambda )$ with $n\ge 1$ such that $L(\Lambda )=\bigvee _{n\ge 1} (Q_n'\cap L(\Lambda )$). Since $Q_1$ is abelian, it follows that $(Q_1'\cap pMp)'\cap pMp\subset Q_1'\cap pMp$. Hence, for all $k, n\ge 1$, we have

(3.1)\begin{equation} \mathcal{Z}(Q_n'\cap pMp)\subset Q_1'\cap pMp\subset Q_k'\cap pMp. \end{equation}

Using Zorn's lemma and a maximality argument, one can show that for any $m\ge 1$, there exist maximal projections $r_m^1,\ldots, r_m^n\in Q_m'\cap pMp$ satisfying $Q_mr_m^i\nprec _{M}A\rtimes \Gamma _{\widehat i}$, for any $i\in \overline {1,n}$. One can check that $r_m^i\in \mathcal {Z}(Q_m'\cap pMp)$ and $Q_m (p-r_m^i)\prec _M^s A\rtimes \Gamma _{\widehat i}$, for any $i\in \overline {1,n}$ (see the proof of [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.4]).

Since $Q_{m}\nprec _{M} A$, [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.8(2)] implies that $\bigwedge _{i=1}^n (p-r_m^i)=0$, which proves that $\bigvee _{i=1}^n r_m^i =p$. Hence, for any $m\ge 1$ there is $i_m\in \overline {1,n}$ such that $\tau (r_m^{i_m})\ge {\tau (p)}/{n}$. Up to passing to a subsequence, we can assume that there is $j\in \overline {1,n}$ such that $i_m=j$, for all $m\ge 1$. Next, note that (3.1) gives that $r_{m}^j\in \mathcal {Z}(Q_m'\cap pMp)\subset Q_{m-1}'\cap pMp$. Since $Q_{m-1} r_m^j \nprec _M A\rtimes \Gamma _{\widehat j}$, it follows from the choice of all the $r_m^j$'s that $\{r_{m}^j\}_{m\ge 1}$ is a decreasing sequence of projections. If we let $r^j=\bigwedge _{m\ge 1} r_{m}^j$, we deduce that $r^j$ is a non-zero projection since $\tau (r^j)\ge {\tau (p)}/{n}$. For all $m\ge k\ge 1$, since $Q_m\subset Q_k$ we have that $r_m^j \in (Q_k'\cap pMp)'\cap pMp$. Consequently, by letting $m\to \infty$, we deduce that $r^j \in (Q_k'\cap pMp)'\cap pMp$, for all $k\ge 1$, which implies that $r^j\in L(\Lambda )'\cap pMp=\mathbb {C} p$. Since $r^j\neq 0$, we derive that $r^j=p$ and, therefore, we must have $r_m^j=p$, for any $m\ge 1$. This implies that $Q_m\nprec _M A\rtimes \Gamma _{\widehat j}$, for any $m\ge 1$.

Since $\Gamma _j$ is non-amenable, it follows from [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.9] that $L(\Lambda )$ is non-amenable relative to $A\rtimes \Gamma _{\widehat j}$ inside $M$. Since relative amenability is closed under inductive limits (see [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.7]), there exists $k\ge 1$ such that $Q_k'\cap pMp$ is non-amenable relative to $A\rtimes \Gamma _{\widehat j}$ inside $M$. Using [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.6] there is a non-zero projection $z^j\in \mathcal {Z}(Q_k'\cap pMp)$ such that

(3.2)\begin{equation} (Q_k'\cap pMp)z^j \text{ is strongly non-amenable relative to } A\rtimes \Gamma_{\widehat j} \text{ inside }M. \end{equation}

This implies by Lemma 3.6 that $\Psi ^j (Q_k z^j) \text { is }{\rm (id}\otimes \alpha ^j) \text {-rigid}$. Fix an arbitrary $m\ge k$. Since $Q_m\subset Q_k$, we have $z^j\in \mathcal {Z}(Q_k'\cap pMp)\subset Q_m'\cap pMp$ and

(3.3)\begin{equation} \Psi^j (Q_m z^j) \text{ is }{\rm (id}\otimes \alpha^j) \text{-rigid} \text{ and }Q_mz^j\nprec_M A\rtimes\Gamma_{\widehat j}. \end{equation}

Equation (3.2) also implies that

(3.4)\begin{equation} z_j(Q_m'\cap pMp)z^j \text{ is strongly non-amenable relative to } A\rtimes \Gamma_{\widehat j} \text{ inside }M. \end{equation}

By combining (3.3) and (3.4), it follows from the second part of Lemma 3.6 that $\Psi ^j (z^j(Q_m'\cap pMp) z^j) \text { is }{\rm (id}\otimes \alpha ^j) \text {-rigid}$, for any $m\ge k$. Note that (3.3) gives, in particular, that $z^j(Q_m'\cap pMp) z^j\nprec _M A\rtimes \Gamma _{\widehat j}$, for any $m\ge k$. Therefore, we may apply [Reference de Santiago, Hayes, Hoff and SinclairdSHHS20, Theorem 3.5] (see also [Reference DrimbeDri21, Theorem 3.2]) to deduce that $\Psi ^j (z^ j \bigvee _{m\ge k}(Q_m'\cap pMp) z^j) \text { is }{\rm (id}\otimes \alpha ^j) \text {-rigid}$. Using [Reference de Santiago, Hayes, Hoff and SinclairdSHHS20, Proposition 5.6] there exists a non-zero projection $\tilde z^j\in \mathcal {Z}(\bigvee _{m\ge n}(Q_m'\cap pMp))$ such that $\Psi ^j ( \bigvee _{m\ge k}(Q_m'\cap pMp) \tilde z^j) \text { is }{\rm (id}\otimes \alpha ^j) \text {-rigid}$. Note, however, that $\bigvee _{m\ge k}(Q_m'\cap pMp)$ is a factor since $\bigvee _{m\ge k}(Q_k'\cap L(\Lambda ))=L(\Lambda )$ and $\Lambda$ is icc. In particular, $\Psi ^j (L(\Lambda ) )$ is ${({\rm id}}\otimes \alpha ^j)$-rigid. Since $\Psi ^j(B)\subset M\otimes 1$, it follows that $\Psi ^j (M)$ is ${({\rm id}}\otimes \alpha ^j)$-rigid, which gives that $L(\Gamma _j)$ is $\alpha ^j$-rigid, contradiction. This ends the proof of the proposition.

Proof. Let $i\in \{1,\ldots,n\}$. Since $\Gamma _i$ is non-amenable, by [Reference Krogager and VaesKV15, Proposition 2.4] we get that $\Delta (L(\Gamma _i))$ is strongly non-amenable relative to $M\otimes 1$ inside $\mathcal {M}$. It follows that for every non-zero projection $z\in \Delta (L(\Gamma ))'\cap \mathcal {M}$, there exist $f(i,z)\in \{1,2\}$ and a non-zero projection $p(i,z)\in \Delta (L(\Gamma ))'\cap \mathcal {M}$ with $p(i,z)\leq z$ such that

(4.1)\begin{equation} \Delta(L(\Gamma_i))p(i,z) \text{ is strongly non-amenable relative to } M\bar\otimes P_{f(i,z)} \text{ inside } \mathcal{M}. \end{equation}

Indeed, otherwise there exists a non-zero projection $z\in \Delta (L(\Gamma ))'\cap \mathcal {M}$ such that for any $k\in \{1,2\}$ and non-zero projection $z_0\in \Delta (L(\Gamma ))'\cap \mathcal {M}$ with $z_0\leq z$, there exists a non-zero projection $\tilde z_0\in \Delta (L(\Gamma ))'\cap \mathcal {M}$ with $\tilde z_0\leq z_0$ for which $\Delta (L(\Gamma _i))\tilde z_0$ is amenable relative to $M\bar \otimes P_k$ inside $\mathcal {M}$. By using [Reference Popa and VaesPV14, Proposition 2.7] we derive that there exists a non-zero projection $\tilde z_1\in \Delta (L(\Gamma ))'\cap \mathcal {M}$ with $\tilde z_1\leq z$ for which $\Delta (L(\Gamma _i))\tilde z_1$ is amenable relative to $M\bar \otimes 1$ inside $\mathcal {M}$, contradiction.

By applying (4.1) finitely many times, the proof will be obtained as follows. Define $z_1=p(1,1)$ and $f(1)=f(1,1)$. For any $i\in \{2,\ldots,n\}$ we recursively define $z_i=p(i,z_{i-1})$ and $f(i)=f(i,z_{i-1})$. Note that $z_1\ge z_2\ge \cdots \ge z_n$ are non-zero projections in $\Delta (L(\Gamma ))'\cap \mathcal {M}$. Hence, the lemma follows by letting $S_1=f^{-1}(2)$, $S_2=f^{-1}(1)$, and $z=z_n$.

Proof. By assuming the contrary, there exist $j\in \{1,\ldots,n\}$ and a projection $z$ as in the statement such that $\Psi ^{j,4}(\sigma (\Delta (M)\bar \otimes \Delta (M))z$ is $\alpha ^{j,5}$-rigid. Hence, for any $\epsilon >0$, there is $t_0>0$ such that

\[ \| \Psi^{j,4}(\sigma( v_g\otimes v_g\otimes v_h\otimes v_h ))z - \alpha_t^{j,5}( \Psi^{j,4}(\sigma( v_g\otimes v_g\otimes v_h\otimes v_h ))z) \|_2\leq \epsilon, \]

for all $g,h\in \Lambda$ and $|t|\leq t_0.$ Since $\sigma$ acts trivially on $( M\otimes 1)\bar \otimes ( M \otimes 1)$, we obtain that

\[ \| \Psi^{j,4}(\sigma( 1\otimes v_g\otimes 1\otimes v_h ))z - \alpha_t^{j,5}( \Psi^{j,4}(\sigma( 1\otimes v_g\otimes 1\otimes v_h ))z) \|_2\leq \epsilon, \]

for all $g,h\in \Lambda$ and $|t|\leq t_0.$ If we let $\mathcal {G}=\{\Psi ^{j,4}(\sigma ( 1\otimes v_g\otimes 1\otimes v_h ))|\; g,h\in \Lambda \}$, we get that $\mathcal {G}''=\Psi ^{j,4}(1\otimes L(\Lambda )\otimes 1\bar \otimes L(\Lambda ))=1\otimes L(\Lambda )\otimes 1\bar \otimes \Psi ^j(L(\Lambda ))$ since $(1\otimes L(\Lambda ))\bar \otimes (1\otimes L(\Lambda ))$ is $\sigma$-invariant. Note that $\mathcal {N}_{M\bar \otimes L(\Lambda )\bar \otimes M\bar \otimes pMp\bar \otimes L(\Gamma _j)}(\mathcal {G}'')\subset 1\otimes 1\otimes 1\otimes (\Psi ^{j}(L(\Lambda ))'\cap (pMp\bar \otimes L(\Gamma _j)))$.

By applying Proposition 3.2 we obtain a non-zero projection $z_0\in \Psi ^{j}(L(\Lambda ))'\cap (pMp\bar \otimes L(\Gamma _j))$ such that $\Psi ^{j}(L(\Lambda ))z_0$ is $({\rm id}\otimes \alpha ^j)$-rigid. Since $\Psi ^j(B)=B\otimes 1$ and $\Psi ^{j}(pMp)'\cap (pMp\bar \otimes L(\Gamma _j))=\mathbb {C} (p\otimes 1)$, it follows from Proposition 3.2 that $\Psi ^{j}(pMp)$ is $({\rm id}\otimes \alpha ^j)$-rigid, and hence, $\Psi ^{j}(M)$ is $({\rm id}\otimes \alpha ^j)$-rigid. This shows that $L(\Gamma _j)$ is $\alpha ^j$-rigid: contradiction.

Proof of Theorem 4.2 Denote $\mathcal {M}=M\bar \otimes L(\Lambda )$ and $\tilde {\mathcal {M}}=M\bar \otimes M$. For proving this theorem, we use the following variation of the flip automorphism method from [Reference Isono and MarrakchiIM19]. Namely, since $L(\Lambda )=P_1\bar \otimes P_2$, we define $\sigma _{P_1}\in {\rm Aut}(\mathcal {M}\bar \otimes \mathcal {M})$ by letting $\sigma _{P_1}(m\otimes p_1\otimes p_2 \otimes m'\otimes p_1'\otimes p_2')=m\otimes p_1'\otimes p_2 \otimes m'\otimes p_1\otimes p_2'$, for all $m,m'\in M,p_1,p_1'\in P_1, p_2,p_2'\in P_2$.

By applying Lemmas 4.4 and 2.4 we obtain a partition $S_1\sqcup S_2=\{1,\ldots,n\}$ and a non-zero projection $z\in \Delta (L(\Gamma ))'\cap \mathcal {M}$ such that

(4.2)\begin{equation} \begin{aligned} & \Delta(L(\Gamma_i))z\otimes 1 \text{ is strongly non-amenable relative to }(M\bar\otimes P_2)\bar\otimes (M\bar\otimes P_1) \text{ inside } \mathcal{M}\bar\otimes\mathcal{M},\\ & 1\otimes \Delta(L(\Gamma_j))z \text{ is strongly non-amenable relative to }(M\bar\otimes P_2)\bar\otimes (M\bar\otimes P_1) \text{ inside } \mathcal{M}\bar\otimes\mathcal{M}, \end{aligned} \end{equation}

for all $i\in S_1$ and $j\in S_2$. By applying the flip automorphism $\sigma _{P_1}$ to (4.2), we derive that

\begin{align*} \sigma_{P_1}(\Delta(L(\Gamma_i))z\otimes 1) \text{ is strongly non-amenable relative to }\mathcal{M}\bar\otimes (M\otimes 1) \text{ inside } \mathcal{M}\bar\otimes\mathcal{M},\\ \sigma_{P_1}(1\otimes \Delta(L(\Gamma_j))z) \text{ is strongly non-amenable relative to }\mathcal{M}\bar\otimes (M\otimes 1) \text{ inside } \mathcal{M}\bar\otimes\mathcal{M}, \end{align*}

for all $i\in S_1$ and $j\in S_2$. By using Lemma 2.4, we further deduce that

(4.3)\begin{equation} \begin{aligned} & \sigma_{P_1}(\Delta(L(\Gamma_i))z\otimes 1) \text{ is strongly non-amenable relative to }\mathcal{M}\bar\otimes (M\otimes 1) \text{ inside } \mathcal{M}\bar\otimes\tilde{\mathcal{M}},\\ & \sigma_{P_1}(1\otimes \Delta(L(\Gamma_j))z) \text{ is strongly non-amenable relative to }\mathcal{M}\bar\otimes (M\otimes 1) \text{ inside } \mathcal{M}\bar\otimes\tilde{\mathcal{M}}, \end{aligned} \end{equation}

for all $i\in S_1$ and $j\in S_2$. Denote $\widehat z=\sigma _{P_1}(z\otimes z)\in \sigma _{P_1}(\Delta (L(\Gamma ))\bar \otimes \Delta (L(\Gamma )))'\cap (\mathcal {M}\bar \otimes \mathcal {M})$ and note that $\widehat z\leq \sigma _{P_1}(z\otimes 1)$, $\widehat z\leq \sigma _{P_1}(1\otimes z)$. For ease of notation, we denote $Q_i=\Delta (L(\Gamma _i))\otimes 1$ and $R_i=\Delta (L(\Gamma _{\widehat i})\bar \otimes \Delta (L(\Gamma ))$, for all $i\in S_1$. Similarly, denote $Q_j=1\otimes \Delta (L(\Gamma _j))$ and $R_j=\Delta (L(\Gamma ))\bar \otimes \Delta (L(\Gamma _{\widehat j})$, for all $j\in S_2$. Note that $Q_i \vee R_i=\Delta (L(\Gamma ))\bar \otimes \Delta (L(\Gamma ))$, for any $i\in \{1,\ldots,n\}$.

By applying a similar argument to that used in the proof of Lemma 4.4, we deduce from (4.3) that there exist a non-zero projection $\tilde z\in \sigma _{P_1}(\Delta (L(\Gamma ))\bar \otimes \Delta (L(\Gamma )))'\cap (\mathcal {M}\bar \otimes \tilde {\mathcal {M}})$ and a function $\varphi :\{1,\ldots,n\}\to \{1,\ldots,n\}$ such that

(4.4)\begin{equation} \sigma_{P_1}(Q_i)\tilde z \text{ is strongly non-amenable relative to } \mathcal{M}\bar\otimes M\bar\otimes (A\rtimes \Gamma_{\widehat{\varphi(i)}}), \end{equation}

for all $i\in \{1,\ldots,n\}$. By Lemma 3.6, we get that for any $i\in \{1,\ldots,n\}$,

(4.5)\begin{equation} \Psi^{\varphi(i),4}(\sigma_{P_1}(R_i)\tilde{z}) \text{ is } \alpha^{\varphi (i),5} \text{ -rigid}. \end{equation}

Next, we claim that the map $\varphi$ is bijective. If this does not hold, it is easy to see that we can deduce from (4.4) that there exists $j\in \{1,\ldots,n\}$ such that

(4.6)\begin{equation} \Psi^{j,4}(\sigma_{P_1}(\Delta(L(\Gamma))\bar\otimes \Delta(L(\Gamma)))\tilde{z}) \text{ is }\alpha^{j,5}\text{-rigid}. \end{equation}

Using the position of $B\subset A$, and $\sigma _{P_1}(\Delta (B)\bar \otimes \Delta (B))\subset \mathcal {M}\bar \otimes (M\otimes 1)$, we obtain that

(4.7)\begin{equation} \Psi^{j,4}(\sigma_{P_1}(\Delta(A)\bar\otimes \Delta(A)) \text{ is } \alpha^{j,5} \text{-rigid}. \end{equation}

Relations (4.6) and (4.7) in combination with Proposition 3.2 gives a contradiction to Lemma 4.5. This shows that $\varphi$ is indeed bijective. Next, we claim that for all $i\in \{1,\ldots,n\}$,

(4.8)\begin{equation} \sigma_{P_1}(R_i)\tilde{z}\prec^s_{\mathcal{M}\bar\otimes \tilde{\mathcal{M}}} \mathcal{M}\bar\otimes M\bar\otimes (A\rtimes \Gamma_{\widehat{g(i)}}), \end{equation}

Assume by contradiction that there is $i\in \{1,\ldots, n\}$ for which (4.8) does not hold. Then by using [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.4(2)], it follows that, up to replacing $\tilde z$ by a smaller non-zero projection, we have $\sigma _{P_1}(R_i)\tilde {z}\nprec _{\mathcal {M}\bar \otimes \tilde {\mathcal {M}}} \mathcal {M}\bar \otimes M\bar \otimes (A\rtimes \Gamma _{\widehat {g(i)}})$. Using (4.4) and (4.5) we may apply Lemma 3.6 to deduce that $\Psi ^{j,4}(\sigma _{P_1}(\Delta (L(\Gamma ))\bar \otimes \Delta (L(\Gamma )))\tilde {z}) \text { is }\alpha ^{j,5}\text {-rigid}.$ As before, Proposition 3.2 leads to a contradiction.

Finally, by applying [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.8(2)] finitely many times, we deduce from (4.8) that

\[ \sigma_{P_1}(\Delta(L(\Gamma_{\widehat{S_1}}))\bar\otimes \Delta(L(\Gamma_{\widehat{S_2}}))) \prec_{\mathcal{M}\bar\otimes \tilde{\mathcal{M}}} \mathcal{M}\bar\otimes (M \otimes 1). \]

By applying Lemma 2.4 we further obtain that $\sigma _{P_1}(\Delta (L(\Gamma _{\widehat {S_1}}))\bar \otimes \Delta (L(\Gamma _{\widehat {S_2}}))) \prec _{\mathcal {M}\bar \otimes {\mathcal {M}}} \mathcal {M}\bar \otimes (M \otimes 1)$. By applying the flip automorphism $\sigma _{P_1}$ to the previous intertwining relation, we deduce that $\Delta (L(\Gamma _{S_i}))\prec _{\mathcal {M}} M\bar \otimes P_i$, for all $i\in \{1,2\}$. Since $\Delta (A)\prec ^s_{\mathcal {M}}B\otimes 1$, we may use [Reference Berbec and VaesBV14, Lemma 2.3] to get that $\Delta (A\rtimes \Gamma _{S_i})\prec _{\mathcal {M}} M\bar \otimes P_i$, for all $i\in \{1,2\}$. Since $\mathcal {N}_{\mathcal {M}}(\Delta (A\rtimes \Gamma _{S_i}))'\cap \mathcal {M}\subset \Delta (M)'\cap \mathcal {M}=\mathbb {C}1$, we obtain $\Delta (A\rtimes \Gamma _{S_i})\prec ^s_{\mathcal {M}} M\bar \otimes P_i$, for all $i\in \{1,2\}$.

For showing that $S_1$ and $S_2$ are non-empty sets, we suppose the contrary. Hence, without loss of generality, assume that $S_2$ is empty. This shows that $\Delta (M)\prec _{\mathcal {M}} M\bar \otimes P_1$, which implies from [Reference IoanaIoa11, Lemma 9.2] that $L(\Lambda )\prec _{L(\Lambda )} P_1$. This shows that $P_2$ is not diffuse: a contradiction.

Proof of Theorem 4.6 Theorem 4.2 implies that there exist projections $r_1\in \Delta (L(\Gamma _{S_1}))$, $q_1\in M\bar \otimes P_1$, a non-zero partial isometry $w_1\in q_1 (M\bar \otimes M) r_1$ and a $*$-homomorphism $\varphi _1: r_1\Delta (L(\Gamma _{S_1}))r_1\to q_1(M\bar \otimes P_1)q_1$ such that $\varphi _1(x)w_1=w_1x$, for all $x\in r_1\Delta (L(\Gamma _{S_1}))r_1$. Fix an arbitrary $j_0\in S_1$. Since $L(\Gamma _{j_0})$ is a II$_1$ factor we can apply [Reference Chifan, de Santiago and SucpikarnonCdSS18, Lemma 4.5] and therefore assume without loss of generality that $r_1\in L(\Gamma _{j_0})$. In addition, we can assume that the support projection of $E_{M\bar \otimes P_1}(w_1w_1^*)$ equals $q_1$. For any $j\in S_1$, denote $Q_1^j=\varphi _1 (r_1\Delta (L(\Gamma _j))r_1)\subset q_1(M\bar \otimes P_1)q_1$ and let $Q_1=\bigvee _{j\in S_1} Q_1^j$. Note that for any subset $S\subset S_1$, we have $\Delta (L(\Gamma _S))\prec ^{s'}_{M\bar \otimes P_1} \bigvee _{j\in S}Q_1^j$. Indeed, let $S\subset S_1$ and consider a non-zero projection $z\in Q_1'\cap q_1 (M\bar \otimes P_1) q_1$. Note that $\tilde w_1:=zw_1\neq 0$ since otherwise $zE_{M\bar \otimes P_1}(w_1w_1^*)=0$, which implies that $z=0$, false. This shows that the $*$-homomorphism $\tilde \varphi _1:r_1\Delta (L(\Gamma _{S}))r_1\to \bigvee _{j\in S} Q_1^jz$ satisfies $\tilde \varphi _1 (x)\tilde w_1=\tilde w_1 x$, for all $x\in r_1\Delta (L(\Gamma _{S}))r_1.$ By replacing $\tilde w_1$ by the partial isometry from its polar decomposition, we derive that $\Delta (L(\Gamma _S))\prec _{M\bar \otimes P_1} \bigvee _{j\in S}Q_1^jz$. By using [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.4] it follows that $\Delta (L(\Gamma _S))\prec ^{s'}_{M\bar \otimes P_1} \bigvee _{j\in S}Q_1^j$. By applying [Reference DrimbeDri20b, Lemma 2.3], we derive that for any subset $S\subset S_1$,

(4.9)\begin{equation} \Delta(L(\Gamma_S))\prec^{s'}_{M\bar\otimes M} \bigvee_{j\in S} Q_1^j. \end{equation}

The rest of the proof is divided between three claims.

Proof of Claim 1 We assume by contradiction that there exist $j\in S_1$ and a non-zero projection $z\in Q_1'\cap q_1 (M\bar \otimes P_1) q_1$ such that $Q_1^{j} z$ is amenable relative to $M\bar \otimes (A\rtimes \Gamma _{\widehat {k}}),$ for all $k\in \{1,\ldots,n\}$. By applying [Reference Popa and VaesPV14, Proposition 2.7] we get that $Q_1^{j}z$ is amenable relative to $M\otimes 1$ inside $M\bar \otimes M$. Relation (4.9) implies that $\Delta (L(\Gamma _j))\prec ^{}_{M\bar \otimes M} Q_1^j z.$ We can apply [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.4(3) and Lemma 2.6(3)] and derive that there exists a non-zero projection $r'\in \Delta (L(\Gamma _{j}))'\cap M\bar \otimes M$ such that $\Delta (L(\Gamma _{j}))r'$ is amenable relative to $Q_1^{j}z\oplus \mathbb {C}(1-z)$. Using [Reference Ozawa and PopaOP10, Proposition 2.4(3)] we derive that $\Delta (L(\Gamma _{j}))r'$ is amenable relative to $M\otimes 1$. By using [Reference Ioana, Popa and VaesIPV13, Lemma 10.2(5)] we deduce that $\Gamma _{j}$ is amenable: a contradiction. Thus, there exist $k\in \{1,\ldots, n\}$ such that $Q_1^{j} z$ is non-amenable relative to $M\bar \otimes (A\rtimes \Gamma _{\widehat {k}}).$ By [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.6], there exists a non-zero projection $z_0 \in \mathcal {N}_{q_1(M\bar \otimes M)q_1}(Q_1^j)'\cap q_1 (M\bar \otimes M_1) q_1\subset Q_1'\cap q_1(M\bar \otimes P_1)q_1$ with $z_0\leq z$ such that $Q_1^{j} z_0$ is strongly non-amenable relative to $M\bar \otimes (A\rtimes \Gamma _{\widehat {k}}).$

Proof of Claim 2 Fix an arbitrary $j\in S_1$. We are in one of the following situations. First, if we assume that $(1\otimes P_2)z\prec _{M\bar \otimes M}M\bar \otimes (A\rtimes \Gamma _{\widehat {k_j}})$, we get $P_2\prec _M A\rtimes \Gamma _{\widehat {k_j}}$. Since $\mathcal {N}_{pMp}(P_2)'\cap pMp\subset L(\Gamma )'\cap pMp=\mathbb {C}p$, the claim follows from [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.4(3)]. Second, assume that $(1\otimes P_2)z\nprec _{M\bar \otimes M}M\bar \otimes (A\rtimes \Gamma _{\widehat {k_j}})$. Since $Q_1^j z\subset ((1\otimes P_2)z)'\cap z (M\bar \otimes M) z$, (4.10) implies that $((1\otimes P_2)z)'\cap z (M\bar \otimes M) z$ is strongly non-amenable relative to $M\bar \otimes (A\rtimes \Gamma _{\widehat {k_j}})$. Altogether, we can apply Lemma 3.6 to deduce that $(1\otimes \Psi ^{k_j}) (z(M\bar \otimes L(\Lambda ))z)$ is $({\rm id}\otimes {\rm id}\otimes \alpha ^{k_j})$-rigid. Since $M\bar \otimes L(\Lambda )$ is a II$_1$ factor and $(1\otimes \Psi ^{k_j})(1\otimes B)\subset 1\otimes B\otimes 1$, it follows that $(1\otimes \Psi ^{k_j}) (M\bar \otimes M)$ is $({\rm id}\otimes {\rm id}\otimes \alpha ^{k_j})$-rigid, which implies that $L(\Gamma _{k_j})$ is $\alpha ^{k_j}$-rigid: a contradiction. This completes the proof of the claim.

Proof of Claim 3 Assume by contradiction that there exist two distinct elements $j_1,j_2\in S_1$ such that $k:=k_{j_1}=k_{j_2}$. Thus, $(S_1\setminus \{j_1\})\cup (S_1\setminus \{j_2\})=S_1$. Since the algebras $Q_1^jz, j\in S_1$, are commuting, we deduce from (4.11) that $(1\otimes \Psi ^k)(Q_1z) \text { is } ({\rm id}\otimes {\rm id}\otimes \alpha ^{k})\text {-rigid.}$ As in the proof of Claim 1, we note that $zw_1\neq 0$. Note also that $Q_1zw_1=zw_1r_1\Delta (L(\Gamma _{S_1}))r_1$. By applying Proposition 3.2 we obtain a non-zero projection $e_1\in (1\otimes \Psi ^k)(\Delta (L(\Gamma )))'\cap M\bar \otimes M\bar \otimes L(\Gamma )$ such that

(4.12)\begin{equation} (1\otimes\Psi^k)(\Delta(L(\Gamma_{S_1})))e_1 \text{ is } ({\rm id}\otimes{\rm id}\otimes \alpha^{k})\text{-rigid.} \end{equation}

Since $z\in q_1(M\bar \otimes P_1)q_1$ and $M\bar \otimes P_1$ is a II$_1$ factor, one can check that (4.11) implies

(4.13)\begin{equation} \Psi^k(P_2) \text{ is } ({\rm id}\otimes \alpha^{k}) \text{ -rigid.} \end{equation}

Next, since $\Delta (L(\Gamma _{S_2}))\prec ^s_{M\bar \otimes L(\Lambda )} M\bar \otimes P_2$, we obtain from [Reference Drimbe, Hoff and IoanaDHI16, Remark 2.2] that

\[ (1\otimes \Psi^k)(\Delta(L(\Gamma_{S_2})))\prec^s_{M\bar\otimes M\bar\otimes L(\Gamma_k)} M\bar\otimes \Psi^k(P_2). \]

Therefore, $(1\otimes \Psi ^k)(\Delta (L(\Gamma _{S_2})))e_1\prec _{M\bar \otimes M\bar \otimes L(\Gamma _k)} M\bar \otimes \Psi (P_2)$, which implies by (4.13) that there is a projection $0\neq e_2\in (1\otimes \Psi ^k)(\Delta (L(\Gamma _{S_2})))'\cap (M\bar \otimes M\bar \otimes L(\Gamma _k))$ with $e_2\leq e_1$ such that

(4.14)\begin{equation} (1\otimes\Psi^k)(\Delta(L(\Gamma_{S_2})))e_2 \text{ is } ({\rm id}\otimes{\rm id}\otimes \alpha^{k})\text{-rigid.} \end{equation}

Note that (4.12) implies that $e_2(1\otimes \Psi ^k)(\Delta (L(\Gamma _{S_1})))e_2 \text { is } ({\rm id}\otimes {\rm id}\otimes \alpha ^{k})\text {-rigid.}$ Together with (4.14) and the fact that $\Psi ^k(A)\subset A\otimes 1$, we deduce from Proposition 3.2 that there exists a non-zero projection $e_3\in (1\otimes \Psi ^k)(\Delta (M))'\cap M\bar \otimes M\bar \otimes L(\Gamma _k)$ such that

(4.15)\begin{equation} (1\otimes\Psi^k)(\Delta(M))e_3 \text{ is } ({\rm id}\otimes{\rm id}\otimes \alpha^{k})\text{-rigid.} \end{equation}

This implies that for any $\epsilon >0$, there exists $t_0>0$ such that for all $|t|\leq t_0$ and $g\in \Lambda$,

\[ \|(1\otimes \Psi^k)(v_g\otimes v_g)e_3 -({\rm id}\otimes{\rm id}\otimes \alpha_t^{k}) ((1\otimes \Psi^k)(v_g\otimes v_g)e_3) \|_2\leq \epsilon, \]

and, therefore,

\[ \|(1\otimes \Psi^k)(1\otimes v_g)e_3 -({\rm id}\otimes{\rm id}\otimes \alpha_t^{k}) ((1\otimes \Psi)(1\otimes v_g)e_3) \|_2\leq \epsilon. \]

Note that $\Psi ^k(B)\subset B\otimes 1$. By applying Proposition 3.2 we get that $\Psi ^k(M)e_0$ is $({\rm id}\otimes \alpha ^k)$-rigid for a projection $0\neq e_0\in \Psi ^k(M)'\cap ( M\bar \otimes L(\Gamma _k))$. Since $\Gamma _k$ is icc, we get $\Psi ^k(M)'\cap ( M\bar \otimes L(\Gamma _k))=\mathbb {C}1$. Thus, we obtain that $L(\Gamma _k)$ is $\alpha ^k$-rigid, contradiction.

Denote $R_1=\{k_j\mid j\in S_1\}\subset \{1,\ldots,n\}$. Claim 3 implies that $|S_1|=|R_1|$ while Claim 2 together with [Reference Drimbe, Hoff and IoanaDHI16, Lemma 2.8(2)] gives that $P_1\prec _M^s B\rtimes \Lambda _{\widehat {R_1}}$. In a similar way, there exists a subset $R_2\subset \{1,\ldots,n\}$ with $|S_2|=|R_2|$ such that $P_2\prec _M^s A\rtimes \Gamma _{\widehat {R_2}}$. By using [Reference Chifan, Diaz-Arias and DrimbeCDD21, Proposition 4.4] we deduce that $L(\Gamma )\prec _M^s A\rtimes \Gamma _{\widehat {R_1}\cup \widehat {R_2}}$. Using [Reference Berbec and VaesBV14, Lemma 2.3] we get that $M\prec _{M} A\rtimes \Gamma _{\widehat {R_1}\cup \widehat {R_2}}$, which implies that $\widehat {R_1}\cup \widehat {R_2}=\{1,\ldots,n\}$.

Finally, we let $T_1=\widehat {R_1}$ and $T_2=\widehat {R_2}$. Since $S_1\sqcup S_2=\{1,\ldots,n\}$ is a partition, it follows that $T_1\sqcup T_2=\{1,\ldots,n\}$ is a partition as well. This ends the proof.

Proof. For the proof, we assume Notation 3.7. Using Proposition 3.8, we get that $L(\Gamma )$ does not have property Gamma. Next, by applying Theorems 4.2 and 4.6 and Lemma 6.1, we obtain a partition $T_1\sqcup T_2=\{1,\ldots,n\}$ and some subgroups $\Sigma _1,\Sigma _2<\Lambda$ such that for all $i\in \{1,2\}$ we have:

1. (1) $B\rtimes \Sigma _i\prec ^s_M A\rtimes \Gamma _{T_i}$ and $A\rtimes \Gamma _{S_i}\prec ^s_M B\rtimes \Sigma _i$;

2. (2) $P_i\prec ^s_{L(\Lambda )} L(\Sigma _i)$ and $L(\Sigma _i)\prec _{L(\Lambda )}^s P_i$.

Part (2) together with [Reference Drimbe, Hoff and IoanaDHI16, Theorem 6.1] give a product decomposition $\Lambda =\Lambda _1\times \Lambda _2$, a decomposition $L(\Lambda )=P_1^{t_1}\bar \otimes P_2^{t_2}$, for some $t_1,t_2>0$ with $t_1t_2=1$, and a unitary $u\in L(\Lambda )$ such that $P_1^{t_1}=uL(\Lambda _1)u^*$ and $P_2^{t_2}=uL(\Lambda _2)u^*$. In addition, we have that $\Lambda _i$ is measure equivalent to $\Sigma _i$, for any $i\in \{1,2\}$.

Part (1) together with [Reference Drimbe, Hoff and IoanaDHI16, Proposition 3.1] implies that for any $i\in \{1,2\}$, $\Gamma _{T_i}$ is measure equivalent to $\Sigma _i$ and, hence, to $\Lambda _i$.