Proof. (i) It is straightforward.

(ii) Let $(\xi _n)_n$ be a $\|\cdot \|_r$ -bounded sequence in $\operatorname {L}^r(M)$ . There exists $\kappa> 0$ , such that $\sup _{n \in \mathbb {N}} \tau (|\xi _n|^r) < \kappa $ . Set $\xi = (\xi _n)^{\mathcal U} \in \operatorname {L}^2(M)^{\mathcal U}$ . For every $n \in \mathbb {N}$ , write $\xi _n = v_n |\xi _n|$ for the polar decomposition of $\xi _n \in \operatorname {L}^r(M)$ . For every $n \in \mathbb {N}$ and every $t> 0$ , define the spectral projection $p_{n, t} = \mathbf 1_{[0, t]}(|\xi _n|) \in M$ . For every $n \in \mathbb {N}$ and every $t> 0$ , we have

$$ \begin{align*}\|\xi_n \, p_{n, t}^\perp\|_2^2 \leq \| |\xi_n| \, p_{n, t}^\perp \|_2^2 = \tau(|\xi_n|^2 \mathbf 1_{(t, +\infty)}(|\xi_n|)) \leq \frac{1}{t^{r - 2}} \tau(|\xi_n|^r \mathbf 1_{(t, +\infty)}(|\xi_n|)) \leq \frac{\kappa}{t^{r - 2}}.\end{align*} $$

Let $\varepsilon> 0$ , and choose $t> 0$ large enough so that $\frac {\kappa }{t^{r - 2}} \leq \varepsilon ^2$ . For every $n \in \mathbb {N}$ , set $x_n = \xi _n \, p_{n, t}\in M$ and observe that we have $\|\xi _n - x_n\|_2 = \|\xi _n \, p_{n, t}^\perp \|_2 \leq \varepsilon $ . Since $\sup \left \{ \|x_n\|_\infty \mid n \in \mathbb {N} \right \} \leq t$ , Item (i) implies that $\xi = (\xi _n)^{\mathcal U} \in \operatorname {L}^2(M^{\mathcal U})$ .

(iii) Assume that $\xi = (\xi _n)^{\mathcal U} \in \operatorname {L}^2(M^{\mathcal U})$ . Choose $\kappa> 0$ large enough so that $\sup \left \{ \|x_n\|_\infty , \|y_n\|_\infty \mid n \in \mathbb {N} \right \} \leq \kappa $ . Let $\varepsilon> 0$ . By Item (i), there exists a $\|\cdot \|_\infty $ -bounded sequence $(z_n)_n$ in M, such that $\lim _{n \to \mathcal U} \|\xi _n - z_n\|_2 \leq \varepsilon $ . Set $z = (z_n)^{\mathcal U} \in M^{\mathcal U}$ . Then $\|\xi - z\|_2 = \lim _{n \to \mathcal U} \|\xi _n - z_n\|_2 \leq \varepsilon $ . Since $x z y = (x_n z_n y_n)^{\mathcal U} \in M^{\mathcal U}$ , we have

$$ \begin{align*} \|(x_n \xi_n y_n)^{\mathcal U} - x \xi y\|_2 &\leq \|(x_n \xi_n y_n)^{\mathcal U} - (x_n z_n y_n)^{\mathcal U} \|_2 + \| x z y - x \xi y\|_2 \\ &= \lim_{n \to \mathcal U} \|x_n (\xi_n - z_n) y_n\|_2 + \| x (z - \xi) y\|_2 \\ &\leq 2 \kappa^2 \|\xi - z\|_2 \leq 2 \kappa^2 \varepsilon. \end{align*} $$

Since this holds for every $\varepsilon> 0$ , it follows that $(x_n \xi _n y_n)^{\mathcal U} = x \xi y \in \operatorname {L}^2(M^{\mathcal U})$ .

Proof. We may assume that $\sup \left \{\|x_n\|_\infty \mid n\in \mathbb N\right \}\leq 1$ , and thus $\|x\|_\infty \leq 1$ . Note that $|x|^2=x^*x=(x_n^*x_n)^{\mathcal U}=(|x_n|^2)^{\mathcal U}$ . Then, for every $k\in \mathbb N$ , we have $|x|^{2k}=(|x_n|^{2k})^{\mathcal U}$ , and thus $\tau ^{\mathcal U}(|x|^{2k})=\lim _{n\rightarrow \mathcal U}\tau (|x_n|^{2k})$ . Thus, if $P(t)\in \mathbb C[t]$ is a polynomial with complex coefficients and $Q(t)=P(t^2)$ , then $\tau ^{\mathcal U}(Q(|x|))=\lim _{n\rightarrow \mathcal U}\tau (Q(|x_n|))$ . Since by the Stone–Weierstrass theorem $\{P(t^2)\mid P(t)\in \mathbb C[t]\}$ is dense in $\operatorname {C}([0,1])$ in the uniform norm, we get that $\tau ^{\mathcal U}(f(|x|))=\lim _{n\rightarrow \mathcal U}\tau (f(|x_n|))$ , for every $f\in \operatorname {C}([0,1])$ . In particular, $\tau ^{\mathcal U}(|x|^p)=\lim _{n\rightarrow \mathcal U}\tau (|x_n|^p)$ , for every $p\in [1,+\infty )$ , which implies the conclusion.

Proof. We use the notation $H_\varepsilon $ of [Reference Mei and RicardMR16, Section 3].

(i) For every $j \in J$ , set $\varepsilon _j = - 1$ and for every $j \in I\setminus J$ , set $\varepsilon _j = 1$ . Then, with $\varepsilon = (\varepsilon _i)_{i \in I}$ , we have $P_{\mathscr L_J} = \frac {\operatorname {Id} - H_\varepsilon }{2}$ . Therefore, [Reference Mei and RicardMR16, Theorem 3.5] implies that $P_{\mathscr L_J} : \mathscr W \to \mathscr L_J$ extends to a completely bounded operator $P_{\mathscr L_J} : \operatorname {L}^p(M) \to \operatorname {L}^p(\mathscr L_J)$ .

(ii) The proof is completely analogous to Item (i).

(iii) We have $P_{\mathscr W_i} = P_{\mathscr L_{i}} \circ P_{\mathscr R_{i}} = P_{\mathscr R_{i}} \circ P_{\mathscr L_{i}}$ . Therefore, Items (i) and (ii) imply that $P_{\mathscr W_i} : \mathscr W \to \mathscr W_i$ extends to a completely bounded operator $P_{\mathscr W_i} : \operatorname {L}^p(M) \to \operatorname {L}^p(\mathscr W_i)$ .

Proof. To prove the lemma, it suffices to argue that for every finite subset $F\subset M$ , $\varepsilon>0$ and $K\in \mathbb N$ , we can find $u\in \mathscr U(A)$ , such that $\|\operatorname { E}_B(xu^my^*)\|_2<\varepsilon $ , for all $m\in \mathbb Z\setminus \{0\}$ with $|m|\leq K$ . To this end, fix a finite subset $F\subset M$ , $\varepsilon>0$ and $K\in \mathbb N$ . For $u\in \mathscr U(M)$ , set $\psi (u)=\sum _{m\in \mathbb Z\setminus \{0\},|m|\leq K}\sum _{x,y\in F}\|\operatorname { E}_B(xu^my^*)\|_2^2$ . Let $v\in \mathscr U(M)$ with $\{v\}^{\prime \prime } \npreceq _MB$ . For every $N\in \mathbb N$ , set

$$ \begin{align*}\varphi(v,N)=\frac{1}{N}\sum_{k=1}^N\sum_{x,y\in F}\|\operatorname{ E}_B(xv^ky^*)\|_2^2.\end{align*} $$

We claim that $\lim _{N \to \infty }\varphi (v,N)=0$ . Indeed, set $\xi =\sum _{x\in F}xe_Bx^*\in \langle M,B\rangle $ , where $(\langle M, B\rangle , \operatorname {Tr})$ is Jones basic construction of $B\subset M$ . Using that $\|\operatorname { E}_B(z)\|_2^2=\text {Tr}(ze_Bz^*e_B)$ for every $z\in M$ , we obtain that

(2.1) $$ \begin{align} \forall N \in \mathbb{N}, \quad \varphi(v,N)=\operatorname{Tr} \left( \left(\frac{1}{N}\sum_{k=1}^Nv^k\xi {v^{-k}} \right)\xi \right). \end{align} $$

By von Neumann’s ergodic theorem, there exists $\eta \in \operatorname {L}^2(\langle M,B\rangle , \operatorname {Tr})$ , such that $v\eta v^*=\eta $ and $\lim _{N \to \infty }\|\frac {1}{N}\sum _{k=1}^Nv^k\xi {v^{-k}}-\eta \|_{2, \operatorname {Tr}}=0$ . Then $w\eta =\eta w$ , for all $w\in \{v\}^{\prime \prime }$ . Since $\{v\}^{\prime \prime }\npreceq _MB$ , we obtain that $\eta =0$ . In combination with (2.1), this proves our claim that $\lim _{N\to \infty }\varphi (v,N)=0$ .

We are now ready to finish the proof. Since $A\npreceq _MB$ , we can find a diffuse abelian von Neumann subalgebra $A_0\subset A$ , such that $A_0\npreceq _MB$ (see [Reference Brown and OzawaBO08, Corollary F.14]). Let $v\in \mathscr U(A_0)$ be a Haar unitary with $\{v\}^{\prime \prime }=A_0$ . If $m\in \mathbb Z\setminus \{0\}$ , then $\{v^m\}^{\prime \prime }\subset A_0$ has finite index, and thus $\{v^m\}^{\prime \prime }\npreceq _MB$ . The above claim gives that $\lim _{N \to \infty }\varphi (v^m,N)=0$ , for all $m\in \mathbb Z\setminus \{0\}$ . Thus, we can find $N\in \mathbb N$ , such that $\sum _{m\in \mathbb Z\setminus \{0\},|m|\leq K}\varphi (v^m,N)<\varepsilon ^2$ . Since $\sum _{m\in \mathbb Z\setminus \{0\},|m|\leq K}\varphi (v^m,N)=\frac {1}{N}\sum _{k=1}^N\psi (v^k)$ , we find $1\leq k\leq N$ , such that $\psi (v^k)<\varepsilon ^2$ . Thus, $u=v^k$ satisfies the desired conclusion, which finishes the proof.

Proof. Let $x = (x_n)^{\mathcal U} \in \{u\}' \cap M^{\mathcal U}$ , such that $\operatorname { E}_{B^{\mathcal U}}(x) = 0$ . Without loss of generality, we may assume that $\|x_n\|_\infty \leq 1$ for every $n \in \mathbb {N}$ . To prove that $\lim _{n \to \mathcal U} \|x_n - P_{\mathscr W_i}(x_n)\|_2 = 0$ , we show that $\lim _{n \to \mathcal U} \|P_{\mathscr L_{I\setminus \{i\}}}(x_n)\|_2 = \lim _{n \to \mathcal U} \|P_{\mathscr R_{I\setminus \{i\}}}(x_n)\|_2 = 0$ . Since $\mathscr R_{I \setminus \{i\}} = J \mathscr L_{I \setminus \{i\}} J$ , it suffices to prove that $\lim _{n \to \mathcal U} \|P_{\mathscr L_{I\setminus \{i\}}}(x_n)\|_2 = 0$ . To simplify the notation, we set $P_i = P_{\mathscr L_{I\setminus \{i\}}}$ .

By Lemma 2.1(ii) and Theorem 2.3(i), we have $(P_i(x_n))^{\mathcal U} \in \operatorname {L}^2(M^{\mathcal U})$ . Set $\mathscr H_i = \operatorname {L}^2(M^{\mathcal U}) \cap (\mathscr L_{I \setminus \{i\}})^{\mathcal U} \subset \operatorname {L}^2(M^{\mathcal U})$ and denote by $P_{\mathscr H_i} : \operatorname {L}^2(M^{\mathcal U}) \to \mathscr H_i$ the corresponding orthogonal projection. Then we have $P_{\mathscr H_i}(x) = (P_i(x_n))^{\mathcal U} \in \mathscr H_i$ . For every $N \geq 1$ , we have

(3.1) $$ \begin{align} N \cdot \| P_{\mathscr H_i}(x)\|^2_2 &= \sum_{k =1}^N \|u^k P_{\mathscr H_i}(x) u^{-k}\|^2_2 \\ \nonumber &= \sum_{k =1}^N \|P_{u^{k}\mathscr H_i u^{-k}}(u^k x u^{-k})\|^2_2 \\ \nonumber &= \sum_{k =1}^N \|P_{ u^{k}\mathscr H_i u^{-k}}(x)\|^2_2. \end{align} $$

We claim that the Hilbert subspaces $( u^{k}\mathscr H_i u^{-k} )_{k \in \mathbb {Z}}$ are mutually orthogonal in $\operatorname {L}^2(M^{\mathcal U})$ , that is, for every $k \in \mathbb {Z} \setminus \{0\}$ , $ u^{k}\mathscr H_i u^{-k}$ and $ \mathscr H_i $ are orthogonal in $\operatorname {L}^2(M^{\mathcal U})$ . Indeed, for every $k \in \mathbb {Z} \setminus \{0\}$ , since $\operatorname { E}_{B^{\mathcal U}}(u^k) = 0$ , we may write $u^k = (u_{n, k})^{\mathcal U} \in M_i^{\mathcal U}$ , where $(u_{n, k})_{n}$ is a $\|\cdot \|_\infty $ -bounded sequence in $M_i \ominus B$ . Let $\xi = (\xi _n)^{\mathcal U} \in \mathscr H_i$ and $\eta = (\eta _n)^{\mathcal U} \in \mathscr H_i$ , where $(\xi _n)_n$ and $(\eta _n)_n$ are $\|\cdot \|_2$ -bounded sequences in $\mathscr L_{I \setminus \{i\}}$ . By construction, it is plain to see that for all $n \in \mathbb {N}$ and all $k \in \mathbb {Z} \setminus \{0\}$ , the vectors $u_{n, k} \xi _n u_{n, k}^*$ and $\eta _n$ are orthogonal in $\operatorname {L}^2(M)$ . By Lemma 2.1(iii), since $\xi = (\xi _n)^{\mathcal U} \in \operatorname {L}^2(M^{\mathcal U})$ , we have

$$ \begin{align*}\langle u^k \xi u^{-k} , \eta \rangle = \langle (u_{n, k} \xi_n u_{n, k}^*)^{\mathcal U} , (\eta_n)^{\mathcal U} \rangle = \lim_{n \to \mathcal U} \langle u_{n, k} \xi_n u_{n, k}^* , \eta_n \rangle = 0.\end{align*} $$

This finishes the proof of the claim.

From (3.1), since the projections $(P_{u^k \mathscr H_i u^{-k}})_{k \in \mathbb {Z}}$ are mutually orthogonal, we infer that

(3.2) $$ \begin{align} N \cdot \| P_{\mathscr H_i}(x)\|^2_2 = \sum_{k =1}^N \|P_{ u^{k}\mathscr H_i u^{-k}}(x)\|^2_2 \leq \|x\|_2^2. \end{align} $$

Since (3.2) holds for every $N \geq 1$ , we have $\lim _{n \to \mathcal U} \|P_i(x_n)\|_2 = \| P_{\mathscr H_i}(x) \|_2 = 0$ . This finishes the proof of the lemma.

Proof. Set $\mathscr B = \mathbf M_2(B)$ , $\mathscr M_j = \mathbf M_2(M_j)$ for every $j \in I$ and $\mathscr M = \mathbf M_2(M)$ so that we have $\mathscr M = \ast _{\mathscr B, j \in I} \mathscr M_j$ . Let $x = (x_n)^{\mathcal U} \in \{u\}' \cap M^{\mathcal U}$ be such that $\operatorname { E}_{M_i^{\mathcal U}}(x) = 0$ . Since any element of $M_i$ is a linear combination of at most four unitaries of $M_i$ , it suffices to prove that for every $v, w \in \mathscr U(M_i)$ , we have $\lim _{n \to \mathcal U} \|v x_n w - P_{\mathscr W_i}(v x_n w)\|_2 = 0$ .

Let $v, w \in \mathscr U(M_i)$ . Set

$$ \begin{align*}U = \begin{pmatrix} v uv^* & 0 \\ 0 & w^* u w \end{pmatrix} \in \mathscr U(\mathscr M_i^{\mathcal U}) \quad \text{and} \quad X = \begin{pmatrix} 0 & v x w \\ 0 & 0 \end{pmatrix} \in \mathscr M^{\mathcal U} \ominus \mathscr M_i^{\mathcal U}.\end{align*} $$

By construction, we have $U X = X U$ and $\operatorname { E}_{\mathscr B^{\mathcal U}}(U^k) = 0$ for every $k \in \mathbb {Z} \setminus \{0\}$ . We may now apply Lemma 3.1 to $X = (X_n)^{\mathcal U} \in \mathscr M^{\mathcal U} \ominus \mathscr B^{\mathcal U}$ and conclude that

$$ \begin{align*}\lim_{n \to \mathcal U} \|v x_n w - P_{\mathscr W_i}(v x_n w)\|_2 = \lim_{n \to \mathcal U} \| (X_n)_{12} - P_{\mathscr W_i}((X_n)_{12})\|_2 = 0.\end{align*} $$

This finishes the proof of the lemma.

Proof of Theorem A

Keep the same notation as in the statement of Theorem A. Let $k \geq 1$ and $\varepsilon _1, \dots , \varepsilon _k \in I$ be such that $\varepsilon _1 \neq \cdots \neq \varepsilon _k$ . For every $1 \leq j \leq k$ , let $x_j = (x_{j, n})^{\mathcal U} \in \mathbf X_{\varepsilon _j}$ . We may assume that $\sup \left \{\|x_{j, n}\|_\infty \mid 1 \leq j \leq k, n \in \mathbb {N} \right \} \leq 1$ . We show that $\operatorname { E}_{B^{\mathcal U}}(x_1 \cdots x_k) = 0$ .

For every $i \in I$ and every $p \in (1, +\infty )$ , we simply denote by $P_i : \operatorname {L}^p(M) \to \operatorname {L}^p(\mathscr W_i)$ the completely bounded operator (see Theorem 2.3(iii)). For every $p \in (1, +\infty )$ , choose $\kappa _p> 0$ large enough so that

$$ \begin{align*}\sup \left\{\|P_i(x)\|_p \mid i \in \{\varepsilon_1, \dots, \varepsilon_k\}, x \in \operatorname{L}^p(M), \|x\|_p \leq 1 \right\} \leq \kappa_p.\end{align*} $$

Fix $1 < r < 2$ (e.g. $r = \frac 32$ ). Let $p \in (1, +\infty )$ be such that $\frac 1r = \frac 12 + \frac {k - 1}{p}$ . For every $1 \leq j \leq k$ and every $n \in \mathbb {N}$ , write $x_{j, n} = P_{\varepsilon _j}(x_{j, n}) + (x_{j, n} - P_{\varepsilon _j}(x_{j, n}))$ and observe that

• ○ $P_{\varepsilon _j}(x_{j, n}) \in \operatorname {L}^2(\mathscr W_{\varepsilon _j})$ and $\lim _{n \to \mathcal U} \|x_{j, n} - P_{\varepsilon _j}(x_{j, n})\|_2 = 0$ ;

• ○ $P_{\varepsilon _j}(x_{j, n}) \in \operatorname {L}^p(\mathscr W_{\varepsilon _j})$ and $\max \left \{ \|P_{\varepsilon _j}(x_{j, n}) \|_{p}, \|x_{j, n} - P_{\varepsilon _j}(x_{j, n})\|_p \right \} \leq 1 + \kappa _p$ .

For every $n \in \mathbb {N}$ , we may write $x_{1, n} \cdots x_{k, n} - P_{\varepsilon _1}(x_{1, n}) \cdots P_{\varepsilon _k}(x_{k, n})$ as a sum of $2^k - 1$ terms that are products of length k for which at least one of the factors is of the form $x_{j, n} - P_{\varepsilon _j}(x_{j, n})$ for some $1 \leq j \leq k$ . For every $n \in \mathbb {N}$ , using the triangle inequality and the generalized noncommutative Hölder inequality, we obtain

$$ \begin{align*}\|x_{1, n} \cdots x_{k, n} - P_{\varepsilon_1}(x_{1, n}) \cdots P_{\varepsilon_k}(x_{k, n})\|_r \leq (2^{k} - 1) (1 + \kappa_p)^{k - 1} \max_j \|x_{j, n} - P_{\varepsilon_j}(x_{j, n})\|_2. \end{align*} $$

This implies that

(3.3) $$ \begin{align} \lim_{n \to \mathcal U} \|x_{1, n} \cdots x_{k, n} - P_{\varepsilon_1}(x_{1, n}) \cdots P_{\varepsilon_k}(x_{k, n})\|_r = 0. \end{align} $$

Next, set $q = kr $ so that $\frac 1r = \frac {k}{q}$ . For every $1 \leq j \leq k$ and every $n \in \mathbb {N}$ , since $P_{\varepsilon _j}(x_{j, n}) \in \operatorname {L}^q(\mathscr W_{\varepsilon _j})$ , we may choose $w_{j, n} \in \mathscr W_{\varepsilon _j}$ , such that $\|P_{\varepsilon _j}(x_{j, n}) - w_{j, n}\|_q \leq \frac {1}{n + 1}$ . For every $1 \leq j \leq k$ and every $n \in \mathbb {N}$ , write $P_{\varepsilon _j}(x_{j, n}) = w_{j, n} + (P_{\varepsilon _j}(x_{j, n}) - w_{j, n})$ and observe that

• ○ $\max \left \{ \|w_{j, n} \|_{q}, \|P_{\varepsilon _j}(x_{j, n}) - w_{j, n} \|_q \right \} \leq 1 + \kappa _q$ .

We may then write $P_{\varepsilon _1}(x_{1, n}) \cdots P_{\varepsilon _k}(x_{k, n}) - w_{1, n} \cdots w_{k, n}$ as a sum of $2^k - 1$ terms that are products of length k for which at least one of the factors is of the form $P_{\varepsilon _j}(x_{j, n}) - w_{j, n}$ for some $1 \leq j \leq k$ . For every $n\in \mathbb {N}$ , using the triangle inequality and the generalized noncommutative Hölder inequality, we obtain

$$ \begin{align*}\|P_{\varepsilon_1}(x_{1, n}) \cdots P_{\varepsilon_k}(x_{k, n}) - w_{1, n} \cdots w_{k, n}\|_r \leq (2^{k} - 1) (1 + \kappa_q)^{k - 1} \max_j \|P_{\varepsilon_j}(x_{j, n}) - w_{j, n}\|_q. \end{align*} $$

This implies that

(3.4) $$ \begin{align} \lim_{n \to \mathcal U} \| P_{\varepsilon_1}(x_{1, n}) \cdots P_{\varepsilon_k}(x_{k, n}) - w_{1, n} \cdots w_{k, n}\|_r = 0. \end{align} $$

By combining (3.3) and (3.4), it follows that

(3.5) $$ \begin{align} \lim_{n \to \mathcal U} \| x_{1, n} \cdots x_{k, n} - w_{1, n} \cdots w_{k, n}\|_r = 0. \end{align} $$

In particular, since $\operatorname { E}_{B^{\mathcal U}} (x_1 \cdots x_k)=(\operatorname { E}_B (x_{1, n} \cdots x_{k, n}))^{\mathcal U}$ , using Lemma 2.2 and the fact that $\operatorname { E}_B$ is $\|\cdot \|_r$ -contractive, we have

$$ \begin{align*}\|\operatorname{ E}_{B^{\mathcal U}} (x_1 \cdots x_k) \|_r = \lim_{n \to \mathcal U} \|\operatorname{ E}_B (x_{1, n} \cdots x_{k, n}) \|_r = \lim_{n \to \mathcal U} \|\operatorname{ E}_B (w_{1, n} \cdots w_{k, n}) \|_r.\end{align*} $$

Since for every $1 \leq j \leq k$ and every $n \in \mathbb {N}$ , we have $w_{j, n} \in \mathscr W_{\varepsilon _j}$ , and since $\varepsilon _1 \neq \cdots \neq \varepsilon _k$ , it follows that $\operatorname { E}_B(w_{1, n} \cdots w_{k, n}) = 0$ . Thus, we obtain $\operatorname { E}_{B^{\mathcal U}}(x_1 \cdots x_k) =0$ .

Proof. Let $i \in I$ . For every $k \in \mathbb {N}$ , since $A_{i, k}$ is separable, we have $(A_{i, k}' \cap M^{\mathcal U})' \cap M^{\mathcal U} = A_{i, k}$ by [Reference PopaPo13a, Theorem 2.1]. This further implies that

$$ \begin{align*}\mathscr M_i' \cap M^{\mathcal U} = (\bigvee_{k = 1}^\infty A_{i, k}' \cap M^{\mathcal U})' \cap M^{\mathcal U} = \bigcap_{k = 1}^\infty (A_{i, k}' \cap M^{\mathcal U})' \cap M^{\mathcal U} = \bigcap_{k = 1}^\infty A_{i, k} = \mathbb{C} 1.\end{align*} $$

Therefore, $\mathscr M_i \subset M^{\mathcal U}$ is an irreducible subfactor. Since $A_{i, 0}' \cap M^{\mathcal U}$ is nonamenable, $\mathscr M_i$ is nonamenable as well. Let $\mathscr V$ be another nonprincipal ultrafilter on $\mathbb {N}$ . For every $k \in \mathbb {N}$ and every $\lambda \in (0, 1)$ , choose a projection $p_{\lambda , k} \in A_{i, k}$ , such that $\tau ^{\mathcal U}(p_{\lambda , k}) = \lambda $ . Then $p_{\lambda } = (p_{\lambda , k})^{\mathcal V} \in \mathscr M_i' \cap \mathscr M_i^{\mathcal V}$ is a projection such that $(\tau ^{\mathcal U})^{\mathcal V}(p_\lambda ) = \lambda $ . Therefore, $\mathscr M_i' \cap \mathscr M_i^{\mathcal V}$ is a diffuse von Neumann algebra, and so $\mathscr M_i$ has property Gamma.

A combination of Lemma 3.1 and Theorem A implies that all $k, \ell \in \mathbb {N}$ , the family $(A_{i, k}' \cap M^{\mathcal U})_{i \in I}$ is freely independent in $M^{\mathcal U}$ with respect to $\tau ^{\mathcal U}$ . Since for every $i \in I$ , the sequence of von Neumann subalgebras $(A_{i, k}' \cap M^{\mathcal U})_k$ is increasing, Kaplansky’s density theorem further implies that the family $(\mathscr M_i)_{i \in I}$ is freely independent in $M^{\mathcal U}$ with respect to $\tau ^{\mathcal U}$ .

Proof of Theorem C

Keep the same notation as in Theorem A. Let $\mathbf Y_1 \subset \mathbf X_1$ be a subset with the property that $a \mathbf Y_1 b \subset \mathbf X_1$ for all $a, b \in M_1$ . Denote by $M_1 \mathbf Y_1 M_1$ the linear span of all the elements of the form $a Y b$ for $a, b \in M_1$ and $Y \in \mathbf Y_1$ . Then we have $M_1 \mathbf Y_1 M_1 \subset \mathbf X_1$ . Likewise, denote by $M_1 \mathscr W_2 M_1$ the linear span of all the elements of the form $a w b$ for $a, b \in M_1$ and $w \in \mathscr W_2$ . Observe that any word with letters alternating from $\mathbf Y_1$ and $M_1 \mathscr W_2 M_1$ can be written as a linear combination of words with letters alternating from $M_1 \mathbf Y_1 M_1 \cup (M_1 \ominus B)$ and $\mathscr W_2$ . Since $M_1 \mathbf Y_1 M_1 \cup (M_1 \ominus B) \subset \mathbf X_1$ and $\mathscr W_2 \subset \mathbf X_2$ , Theorem A implies that the sets $\mathbf Y_1$ and $M_1 \mathscr W_2 M_1$ are freely independent in $M^{\mathcal U}$ with respect to $\operatorname { E}_{B^{\mathcal U}}$ .

Using Kaplansky’s density theorem, for any element $x \in M \ominus M_1$ , there exists a $\|\cdot \|_\infty $ -bounded sequence $(x_n)_n$ in $M_1 \mathscr W_2 M_1$ , such that $x_n \to x$ for the strong operator topology. Combining this fact with the first paragraph of the proof, we infer that the sets $\mathbf Y_1$ and $M \ominus M_1$ are freely independent in $M^{\mathcal U}$ with respect to $\operatorname { E}_{B^{\mathcal U}}$ .

Proof. Since $\operatorname { E}_{B^{\mathcal U}}(y)=0$ , after replacing $y_n$ with $y_n-\operatorname { E}_B(y_n)$ , we may assume that $\operatorname { E}_B(y_n)=0$ , for all $n\in \mathbb N$ . We may assume that $x_n\in \mathscr U(M)$ and $\|y_n\|_\infty \leq 1$ , for all $n\in \mathbb N$ . For $p\in (1,+\infty )$ , let $\kappa _p=\sup \left \{\|P_{i,j}(x)\|_p\mid i,j\in \{1,2\}, x\in \operatorname {L}^p(M),\|x\|_p\leq 1\right \}.$

For every $n\in \mathbb N$ and $i,j\in \{1,2\}$ , put $y_n^{i,j}=P_{{i,j}}(y_n)$ . Then $y_n=\sum _{i,j=1}^2y_n^{i,j}$ and $\|y_n^{i,j}\|_p\leq \kappa _p$ , for every $n\in \mathbb N,i,j\in \{1,2\}$ and $p\in (1,+\infty )$ . We claim that

(3.6) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|\operatorname{ E}_B(x_ny_n^{2,2}x_n^*{y_n^{i,j}}^*)\|_1=0, \text{for every } (i,j)\not=(1,1). \end{align} $$

Let $(i,j)\not =(1,1)$ . By Lemma 3.1, we have $\lim _{n\rightarrow \mathcal U}\|x_n-P_{1,1}(x_n)\|_2=0$ . By using the noncommutative Hölder inequality and that $1=\frac {1}{2}+3\cdot \frac {1}{6}$ , we get that

$$ \begin{align*}\|x_ny_n^{2,2}x_n^*{y_n^{i,j}}^*-P_{1,1}(x_n)y_n^{2,2}P_{1,1}(x_n)^*{y_n^{i,j}}^*\|_1\leq (\kappa_6^2+\kappa_6^3)\|x_n-P_{1,1}(x_n)\|_2.\end{align*} $$

This implies that

(3.7) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|x_ny_n^{2,2}x_n^*{y_n^{i,j}}^*-P_{1,1}(x_n)y_n^{2,2}P_{1,1}(x_n)^*{y_n^{i,j}}^*\|_1=0.\end{align} $$

Next, let $v_n\in \mathscr W_{1,1}$ and $w_n^{i,j}\in \mathscr W_{i,j}$ , such that $\|P_{1,1}(x_n)-v_n\|_4\leq \frac {1}{n}$ and $\|y_n^{i,j}-w_n^{i,j}\|_4\leq \frac {1}{n}$ , for every $n\in \mathbb N$ and $i,j\in \{1,2\}$ . Then $\|v_n\|_4,\|w_n^{i,j}\|_4\leq \kappa _4+\frac {1}{n}\leq \kappa _4+1$ . Since $1=4\cdot \frac {1}{4}$ , applying the noncommutative Hölder inequality again gives that

$$ \begin{align*}\|P_{1,1}(x_n)y_n^{2,2}P_{1,1}(x_n)^*{y_n^{i,j}}^*-v_nw_n^{2,2}v_n^*{w_n^{i,j}}^*\|_1\leq \frac{15(\kappa_4+1)^3}{n}.\end{align*} $$

This implies that

(3.8) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|P_{1,1}(x_n)y_n^{2,2}P_{1,1}(x_n)^*{y_n^{i,j}}^*-v_nw_n^{2,2}v_n^*{w_n^{i,j}}^*\|_1=0. \end{align} $$

By combining (3.7) and (3.8), it follows that

(3.9) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|x_ny_n^{2,2}x_n^*{y_n^{i,j}}^*-v_nw_n^{2,2}v_n^*{w_n^{i,j}}^*\|_1=0. \end{align} $$

Now, $v_nw_n^{2,2}v_n^*{w_n^{i,j}}^*\in \mathscr W_{1,1}\mathscr W_{2,2}\mathscr W_{1,1}\mathscr W_{i,j}^*\subset \mathscr W_{1,1}\mathscr W_{i,j}^*$ . Since $(i,j)\not =(1,1)$ , $\mathscr W_{1,1}$ and $\mathscr W_{i,j}$ are orthogonal (algebraic) B-bimodules. Thus, $\operatorname { E}_B(\mathscr W_{1,1}\mathscr W_{i,j}^*)=\{0\}$ and therefore $\operatorname { E}_B(v_nw_n^{2,2}v_n^*{w_n^{i,j}}^*)=0$ , for every $n\in \mathbb N$ . In combination with (3.9), this proves (3.6).

Finally, for every $n\in \mathbb N$ , we have that $\|y_n^{2,2}\|_2^2=\langle y_n^{2,2},y_n\rangle =\langle x_ny_n^{2,2},x_ny_n\rangle $ . Since $\lim _{n\rightarrow \mathcal U}\|x_ny_n-y_nx_n\|_2=0$ , we get that

$$ \begin{align*}\lim_{n\rightarrow\mathcal U}\|y_n^{2,2}\|_2^2=\lim_{n\rightarrow \mathcal U}\langle x_ny_n^{2,2},y_nx_n\rangle=\lim_{n\rightarrow\mathcal U}(\sum_{i,j=1}^2\langle x_ny_n^{2,2},y_n^{i,j}x_n\rangle).\end{align*} $$

On the other hand, (3.6) gives that $\lim _{n\rightarrow \mathcal U}\langle x_ny_n^{2,2},y_n^{i,j}x_n\rangle =0$ if $(i,j)\not =(1,1)$ . Thus, we get $\lim _{n\rightarrow \mathcal U}\|y_n^{2,2}\|_2^2=\lim _{n\rightarrow \mathcal U}\langle x_ny_n^{2,2}, y_n^{1,1}x_n\rangle $ , which proves the main assertion. Since $|\langle x_ny_n^{2,2}, y_n^{1,1}x_n\rangle |\leq \|y_n^{2,2}\|_2\|y_n^{1,1}\|_2$ , we get that $\lim _{n\rightarrow \mathcal U}\|y_n^{2,2}\|_2\leq \lim _{n\rightarrow \mathcal U}\|y_n^{1,1}\|_2$ .

Proof. We keep the notation from the proof of Lemma 3.5.

(i) Assume that $\operatorname { E}_{B^{\mathcal U}}(v)=0$ . Write $v=(v_n)^{\mathcal U}$ , where $v_n\in M_2\ominus B$ , for every $n\in \mathbb N$ , and $\sup \left \{\|v_n\|_\infty \mid n\in \mathbb N\right \}<\infty $ . We first claim that

(3.10) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|x_ny_n^{2,2}-y_n^{1,1}x_n\|_2=0. \end{align} $$

Since $vyv^*=y$ , we get that $\lim _{n\rightarrow \mathcal U}\|v_n^*y_nv_n-y_n\|_2=0$ . Thus, we derive that

$$ \begin{align*}\lim_{n\rightarrow\mathcal U}\langle v_ny_n^{1,1}v_n^*,y_n\rangle=\lim_{n\rightarrow\mathcal U}\langle y_n^{1,1},y_n\rangle=\lim_{n\rightarrow\mathcal U}\|y_n^{1,1}\|_2^2.\end{align*} $$

Since $(M_2\ominus B)\mathscr W_{1,1}(M_2\ominus B)\subset \mathscr W_{2,2}$ , we also get that $P_{i,j}(v_ny_n^{1,1}v_n^*)=0$ , for every $(i,j)\not =(2,2)$ and $n\in \mathbb N$ . This implies that $\langle v_ny_n^{1,1}v_n^*,y_n\rangle =\langle v_ny_n^{1,1}v_n^*, y_n^{2,2}\rangle $ , for every $n\in \mathbb N$ . Since $|\langle v_ny_n^{1,1}v_n^*, y_n^{2,2}\rangle \leq \|y_n^{1,1}\|_2\|y_n^{2,2}\|_2$ , for every $n\in \mathbb N$ , we conclude that $\lim _{n\rightarrow \mathcal U}\|y_n^{1,1}\|_2^2\leq \lim _{n\rightarrow \mathcal U}\|y_n^{1,1}\|_2\|y_n^{2,2}\|_2$ , and thus

(3.11) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|y_n^{1,1}\|_2\leq\lim_{n\rightarrow\mathcal U}\|y_n^{2,2}\|_2. \end{align} $$

On the other hand, Lemma 3.5 implies that

(3.12) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|y_n^{2,2}\|_2^2=\lim_{n\rightarrow\mathcal U}\langle x_ny_n^{2,2},y_n^{1,1}x_n\rangle \quad \text{and} \quad \lim_{n\rightarrow\mathcal U}\|y_n^{2,2}\|_2\leq\lim_{n\rightarrow\mathcal U}\|y_n^{1,1}\|_2. \end{align} $$

It is now clear that (3.11) and (3.12) together imply (3.10). Since y also commutes with $x^*=(x_n^*)$ , applying (3.10) to $x^*$ instead of x gives that $\lim _{n\rightarrow \mathcal U}\|x_n^*y_n^{2,2}-y_n^{1,1}x_n^*\|_2=0$ , and thus $\lim _{n\rightarrow \mathcal U}\|y_n^{2,2}x_n-x_ny_n^{1,1}\|_2=0$ . In combination with (3.10), this implies that

(3.13) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|x_n^2y_n^{2,2}-y_n^{2,2}x_n^2\|_2=0. \end{align} $$

Since y commutes with $x^2=(x_n^2)$ and $\operatorname { E}_{B^{\mathcal U}}(x^2)=0$ , (3.6) from the proof of Lemma 3.5 gives that $\lim _{n\rightarrow \mathcal U}\|\operatorname { E}_B(x_n^2y_n^{2,2}{x_n^2}^*{y_n^{2,2}}^*)\|_1=0$ , thus $\lim _{n\rightarrow \mathcal U}\langle x_n^2y_n^{2,2},y_n^{2,2}x_n^2\rangle =0$ . Together with (3.13), we get that $\lim _{n\rightarrow \mathcal U}\|x_n^2y_n^{2,2}\|_2=0$ . Since $x_n\in \mathscr U(M)$ , we get that $\lim _{n\rightarrow \mathcal U}\|y_n^{2,2}\|_2=0$ , and (3.11) gives that $\lim _{n\rightarrow \mathcal U}\|y_n^{1,1}\|_2=0$ , proving part (i).

(ii) Assume that $\operatorname { E}_{B^{\mathcal U}}(v)=\operatorname { E}_{B^{\mathcal U}}(v^2)=0$ . By (i), we have $\lim _{n\rightarrow \mathcal U}\|y_n-(y_{1,2}^n+y_{2,1}^n)\|_2=~0$ . Since $vy=yv$ , we have $\lim _{n\rightarrow \mathcal U}\|v_ny_n-y_n v_n\|_2=0$ , and so

(3.14) $$ \begin{align} \lim_{n\rightarrow\mathcal U}\|v_ny_{1,2}^n+ v_ny_{2,1}^n - y_{1,2}^n v_n - y_{2,1}^n v_n\|_2=0. \end{align} $$

For every $n \in \mathbb {N}$ , we have $v_ny_{1,2}^n = P_{2, 2}(v_ny_{1,2}^n)$ , $y_{2,1}^n v_n = P_{2, 2}(y_{2,1}^n v_n)$ , $v_ny_{2,1}^n = P_{1,1}(v_ny_{2,1}^n) + P_{2,1}(v_ny_{2,1}^n)$ and $y_{1,2}^n v_n = P_{1, 1}(y_{1,2}^n v_n) + P_{1, 2}(y_{1,2}^n v_n)$ . In combination with (3.14), we obtain

(3.15) $$ \begin{align} \lim_{n \to \mathcal U} \|v_ny_{2,1}^n - y_{1,2}^n v_n\|_2 = 0 \quad \text{and} \quad \lim_{n \to \mathcal U} \|v_ny_{2,1}^n - P_{1, 1}(v_n y_{2, 1}^n)\|_2 = 0. \end{align} $$

For every $n \in \mathbb {N}$ , set $\eta _n = P_{1,1}(v_ny_{2,1}^n) \in \operatorname {L}^2(M)$ . Then (3.15), Theorem 2.3 and Lemma 2.1 together imply that $\eta = (\eta _n)^{\mathcal U} = (v_ny_{2,1}^n)^{\mathcal U} = (y_{1,2}^n v_n)^{\mathcal U} \in \operatorname {L}^2(M^{\mathcal U})$ and that $y = v^* \eta + \eta v^*$ . Since $v^* y = y v^*$ , we obtain $(v^*)^2 \eta = \eta (v^*)^2$ , and so $v^2 \eta = \eta v^2$ . Since $\operatorname { E}_{B^{\mathcal U}}(v^2) = 0$ , we may write $v^2 = (w_n)^{\mathcal U}$ , where $w_n \in M_2 \ominus B$ for every $n \in \mathbb {N}$ . For every $n \in \mathbb {N}$ , since $\eta _n = P_{1,1}(\eta _n)$ , we have $w_n \eta _n = P_{2, 1}(w_n \eta _n) \perp P_{1, 2}(\eta _n w_n) = \eta _n w_n$ . Then we obtain $v^2 \eta \perp \eta v^2$ . Since $v^2 \eta = \eta v^2$ , this further implies that $v^2\eta = 0$ , and so $\eta = 0$ . Thus, $y = 0$ .

Proof of Theorem 3.4

Theorem 3.4 follows directly from part $(\mathrm {ii})$ of Lemma 3.6.

Proof. First, we claim that $\xi _1,\dots ,\xi _p$ are linearly independent. Otherwise, we can find $\beta _1,\dots ,\beta _p\in \mathbb R$ , such that $\beta _1\xi _1+\cdots +\beta _p\xi _p=0$ and $\max \left \{|\beta _i|\mid 1\leq i\leq p \right \}>0$ . Let $1\leq j\leq p$ , such that $|\beta _j|=\max \left \{|\beta _i|\mid 1\leq i\leq p \right \}$ . Then $-\beta _j\xi _j=\sum _{i\not =j}\beta _i\xi _i$ , and thus $|\beta _j|\, \|\xi _j\|_2^2 \leq \sum _{i\not =j}|\beta _i|\, |\langle \xi _i,\xi _j\rangle |\leq |\beta _j|\sum _{i\not =j}|\langle \xi _i,\xi _j\rangle |$ . Since $\beta _j\not =0$ , we derive that $\|\xi _j\|_2^2\leq \sum _{i\not =j}|\langle \xi _i,\xi _j\rangle |$ , which implies that $\delta ^2\leq (p-1)\varepsilon $ , contradicting that $\delta ^2> (p-1)\varepsilon $ .

Since $\xi _1,\dots ,\xi _p$ are linearly independent, it follows that we can find $\lambda _1,\dots ,\lambda _p\in \mathbb R$ , such that $h=\sum _{i=1}^p\lambda _i\xi _i$ satisfies $\tau (h\xi _j)=\langle h,\xi _j\rangle =\alpha _j$ , for every $1\leq j\leq p$ . Then $|\alpha _j|=|\sum _{i=1}^p\lambda _i\langle \xi _i,\xi _j\rangle |\geq |\lambda _j|\|\xi _j\|_2^2-\sum _{i\not =j}|\lambda _i||\langle \xi _i,\xi _j\rangle |$ , and thus

(5.1) $$ \begin{align} \forall 1\leq j\leq p, \quad |\alpha_j|\geq \delta^2 |\lambda_j|-\varepsilon\sum_{i\not=j}|\lambda_i|. \end{align} $$

Adding the inequalities in (5.1) for $1\leq j\leq p$ gives $\sum _{j=1}^p|\alpha _j|\geq (\delta ^2 - (p-1)\varepsilon )\sum _{j=1}^p|\lambda _j|$ . Thus, $\|h\|_\infty \leq \sum _{j=1}^p|\lambda _j|\leq \frac {\sum _{j=1}^p|\alpha _j|}{\delta ^2-(p-1)\varepsilon }$ . Since $h=h^*$ , this finishes the proof.

Proof. For every $1\leq i\leq m,1\leq j\leq n$ , set $\xi _{i,j}=-\frac {\mathrm {i}}{2}[x_i,y_j]$ . Then $\xi _{i,j}\in M_{\text {sa},1}$ and $\|\xi _{i,j}\|_2=\frac {\|[x_i,y_j]\|_2}{2}\geq \frac {\delta }{2}$ , for every $1\leq i\leq m,1\leq j\leq n$ . On the other hand, for every $(i,j)\not =(i',j')$ , we have $|\langle \xi _{i,j},\xi _{i',j'}\rangle |=\frac {|\langle [x_i,y_j],[x_{i'},y_{j'}]\rangle |}{4}\leq \frac {\gamma }{4}$ .

By applying Lemma 5.3 to $\xi _{i,j}$ and $\alpha _{i,j}=\frac {\tau (x_iy_j)}{2}$ , we may find $h\in M$ , such that $h=h^*$ ,

(5.2) $$ \begin{align} \forall 1\leq i\leq m,1\leq j\leq m, \quad \tau(h\xi_{i,j})=\frac{\tau(x_iy_j)}{2}, \end{align} $$

and

(5.3) $$ \begin{align} \|h\|_\infty\leq \frac{\sum_{i,j}\frac{|\tau(x_iy_j)|}{2}}{\frac{\delta^2}{4}-(mn-1)\frac{\gamma}{4}}\leq\frac{2mn\varepsilon}{\delta^2 - (mn-1)\gamma}=\lambda. \end{align} $$

Define $u=\exp (\mathrm {i} h)\in \mathscr U(M)$ . We will prove that u satisfies the conclusion. Since for every $x\in \mathbb R$ , $|\exp (\mathrm {i} x)-1|\leq 2|x|$ and $|\exp (\mathrm {i} x)-(1+ \mathrm {i} x)|\leq x^2$ , using (5.3), we get that

(5.4) $$ \begin{align} \|u-1\|_\infty\leq 2\lambda \quad \text{and} \quad \|u-(1+ \mathrm{i} h)\|_\infty \leq \lambda^2. \end{align} $$

Let $1\leq i\leq m$ and $1\leq j\leq n$ . Then, using (5.3) and the second part of (5.4), we get that $\|ux_iu^*y_j-(1+ \mathrm {i} h)x_i(1+ \mathrm {i} h)^*y_j\|_\infty \leq \|u-(1+ \mathrm {i} h)\|_\infty (1+\|1+ \mathrm {i} h\|_\infty )\leq \lambda ^2(2+\lambda )\leq 3\lambda ^2$ and $\|(1+ \mathrm {i} h)x_i(1+ \mathrm {i} h)^*y_j-(x_iy_j+ \mathrm {i} (hx_iy_j-x_ihy_j))\|_\infty =\|hx_ihy_j\|_\infty \leq \lambda ^2.$ Thus, we get

$$ \begin{align*}\|ux_iu^*y_j-(x_iy_j+\mathrm{i} (hx_iy_j-x_ihy_j))\|_\infty \leq 4\lambda^2,\end{align*} $$

and, therefore, $|\tau (ux_iu^*y_j)-\tau (x_iy_j+ \mathrm {i} (hx_iy_j-x_ihy_j))|\leq 4\lambda ^2$ . On the other hand, (5.2) gives $\tau (x_iy_j+ \mathrm {i} (hx_iy_j-x_ihy_j))=\tau (x_iy_j)+\tau (\mathrm {i} h[x_i,y_j])=\tau (x_iy_j)-2\tau (h\xi _{i,j})=0$ . Altogether, we get that $|\tau (ux_iu^*y_j)|\leq 4\lambda ^2$ . Thus, $\varepsilon (uxu^*,y)\leq 4\lambda ^2$ , which proves (iii).

Next, $\|[ux_iu^*,y_j]-[x_i,y_j]\|_2\leq 2\|ux_iu^*-x_i\|_2\leq 4\|u-1\|_2\leq 8\lambda $ , by the first part of (5.4). Hence, $\|[ux_iu^*,y_j]\|_2\geq \|[x_i,y_j]\|_2-8\lambda \geq \delta -8\lambda $ , for every $1\leq i\leq m$ and $1\leq j\leq n$ . This implies that $\delta (uxu^*,y)\geq \delta -8\lambda $ , which proves (ii).

Finally, for every $(i,j),(i',j')$ , we have $\|[ux_{i'}u^*,y_{j'}]\|_2\leq 2$ , $\|[x_i,y_j]\|_2\leq 2$ , and thus

$$ \begin{align*} &|\langle [ux_iu^*,y_j],[ux_{i'}u^*,y_{j'}]\rangle-\langle [x_i,y_j],[x_{i'},y_{j'}]\rangle|\\ & \quad \leq 2\big(\|[ux_iu^*,y_j]-[x_i,y_j]\|_2+\|[ux_{i'}u^*,y_{j'}]-[x_{i'},y_{j'}]\|_2\big)\leq 32\lambda. \end{align*} $$

Thus, $|\langle [ux_iu^*,y_j],[ux_{i'}u^*,y_{j'}]\rangle |\leq |\langle [x_i,y_j],[x_{i'},y_{j'}]\rangle |+32\lambda \leq \gamma +32\lambda $ . This implies that $\gamma (uxu^*,y)\leq \gamma +32\lambda $ , which proves (iv). Since (i) also holds by the first part of (5.4), this finishes the proof.

Proof of Lemma 5.2

We will inductively construct sequences $(u_k)_{k\in \mathbb N}\subset \mathscr U(M)$ and $(\lambda _k)_{k\in \mathbb N}\subset (0,\infty )$ with the following properties: $\lambda _0=1$ , $\lambda _1=\frac {2mn\varepsilon _0}{\delta _0^2-(mn-1)\gamma _0}$ , and if we define $v_0=1$ , $v_k=u_ku_{k-1}\cdots u_1\in \mathscr U(M)$ , $\delta _k=\delta (v_kxv_k^*,y),\varepsilon _k=\varepsilon (v_kxv_k^*,y)$ and $\gamma _k=\gamma (v_kxv_k^*,y)$ , for every $k\geq 0$ , then, for every $k\geq 1$ , we have that

1. 1. $\|u_k-1\|_\infty \leq 2\lambda _k$ .

2. 2. $\delta _k\geq \delta _{k-1}-8\lambda _k$ .

3. 3. $\varepsilon _k\leq 4\lambda _k^2$ .

4. 4. $\gamma _k\leq \gamma _{k-1}+32\lambda _k$ .

5. 5. $\lambda _k\leq \frac {\lambda _{k-1}}{2}.$

Since $\varepsilon _0\leq 1$ , we have that $4mn\varepsilon _0\leq 13mn\sqrt {\varepsilon _0} < \delta _0^2 - (mn-1)\gamma _0$ . Thus, $\lambda _1<\frac {1}{2}$ , and hence condition (v) holds for $k=1$ . By applying Lemma 5.4, we can find $u_1\in \mathscr U(M)$ , such that conditions (i)–(iv) hold for $k=1$ .

Next, assume that we have constructed $u_1,\dots , u_l\in \mathscr U(M)$ and $\lambda _1,\dots ,\lambda _l\in (0,\infty )$ , for some $l\in \mathbb N$ , such that conditions (i)–(v) are satisfied for $k=1,\dots ,l$ . Our goal is to construct $u_{l+1}$ and $\lambda _{l+1}$ . Let $\lambda _{l+1}=\frac {2mn\varepsilon _l}{\delta _l^2-(mn-1)\gamma _l}$ . We continue with the following claim.

Proof of Claim 5.5. First, (ii) implies that $\delta _k^2 \geq (\delta _{k-1}-8\lambda _k)^2 \geq \delta _{k - 1}^2 - 32 \lambda _k$ . Then combining (ii) and (iv) gives that

$$ \begin{align*}\forall 1\leq k\leq l, \quad \delta_k^2-(mn-1)\gamma_k\geq (\delta_{k-1}^2-(mn-1)\gamma_{k - 1}) - 32mn\lambda_k,\end{align*} $$

which implies that $\delta _l^2-(mn-1)\gamma _l\geq (\delta _0^2-(mn-1)\gamma _0)- 32mn(\sum _{k=1}^l\lambda _k).$ By using that (v) holds for $k=1,\dots ,l$ , we also get that $\sum _{k=1}^l\lambda _k\leq 2\lambda _1$ . By combining the last two inequalities, we get that

(5.5) $$ \begin{align} \delta_l^2-(mn-1)\gamma_l\geq (\delta_0^2-(mn-1)\gamma_0) - 64mn\lambda_1. \end{align} $$

Since $13mn\sqrt {\varepsilon _0} < \delta _0^2 - (mn-1)\gamma _0$ , we get that $(\delta _0^2-(mn-1)\gamma _0)^2> 169(mn)^2 \varepsilon _0$ , and thus

(5.6) $$ \begin{align} \delta_0^2-(mn-1)\gamma_0> 80mn\lambda_1. \end{align} $$

By combining (5.5) and (5.6), we derive that

(5.7) $$ \begin{align} \delta_l^2-(mn-1)\gamma_l\geq 16mn\lambda_1. \end{align} $$

Since (v) holds for every $k=1,\dots ,l$ , we get that $\lambda _l\leq \lambda _1$ . Since $\varepsilon _l\leq 4\lambda _l^2$ by (iii), using (5.7), we get that

$$ \begin{align*}\lambda_{l+1}=\frac{2 mn\varepsilon_l}{\delta_l^2 -(mn-1)\gamma_l}\leq \frac{8mn\lambda_l^2}{\delta_l^2-(mn-1)\gamma_l}\leq\frac{16mn\lambda_1}{\delta_l^2-(mn-1)\gamma_l}\cdot\frac{\lambda_l}{2}\leq\frac{\lambda_l}{2}.\end{align*} $$

This finishes the proof of the claim.

By using (v) and Claim 5.5, we get that $\lambda _{l+1}\leq \frac {1}{2^{l+1}}<1$ . Thus, $2 mn\varepsilon _l < \delta _l^2 -(mn-1)\gamma _l$ . We can, therefore, apply Lemma 5.4 to $v_lxv_l^*$ and y to find $u_{l+1}\in \mathscr U(M)$ , such that

1. 1. $\|u_{l+1}-1\|_\infty \leq 2\lambda _{l+1}$ .

2. 2. $\delta _{l+1}=\delta (u_{l+1}(v_lxv_l^*)u_{l+1}^*),y)\geq \delta _{l}-8\lambda _{l+1}$ .

3. 3. $\varepsilon _{l+1}=\varepsilon (u_{l+1}(v_lxv_l^*)u_{l+1}^*,y)\leq 4\lambda _{l+1}^2$ .

4. 4. $\gamma _{l+1}=\gamma (u_{l+1}(v_lxv_l^*)u_{l+1}^*,y)\leq \gamma _l+32\lambda _{l+1}$ .

By induction, this finishes the construction of $(u_k)_{k\in \mathbb N}\subset \mathscr U(M)$ and $(\lambda _k)_{k\in \mathbb N}\subset (0,\infty )$ . Finally, since $\lambda _0=1$ , (v) implies that $\lambda _k\leq \frac {1}{2^k}$ , for every $k\geq 0$ . Using (i), we derive that $\|v_k-v_{k-1}\|_\infty =\|u_k-1\|_\infty \leq \frac {1}{2^{k-1}}$ , for every $k\geq 1$ . Thus, the sequence $(v_k)_{k\in \mathbb N}$ is Cauchy in $\|\cdot \|_\infty $ , and so we can find $v\in \mathscr U(M)$ , such that $\lim _{k\rightarrow \infty }\|v_k-v\|_\infty =0$ . Using (iii), we get that $\varepsilon _k\leq 4\lambda _k^2\leq \frac {1}{4^{k-1}}$ , for every $k\geq 1$ . Thus, $\varepsilon (vxv^*,y)=\lim _{k\rightarrow \infty }\varepsilon _k=0$ . Moreover, using (i) and (v), we get that $\|v_k-1\|_\infty \leq \sum _{l=1}^k\|u_l-1\|_\infty \leq \sum _{l=1}^k2\lambda _l\leq 4\lambda _1.$ Hence, $\|v-1\|_\infty =\lim _{k\rightarrow \infty }\|v_k-1\|_\infty \leq 4\lambda _1= \frac {8mn\varepsilon _0}{\delta _0^2-(mn-1)\gamma _0}$ . This finishes the proof.

Proof of Theorem 5.1

We may clearly assume that $\dim (A)\geq 2$ and $\dim (B)\geq 2$ . Since A and B are separable, we can write $A=(\bigcup _{n\in \mathbb N}A_n)^{\prime \prime }$ , $B=(\bigcup _{n\in \mathbb N}B_n)^{\prime \prime }$ , where $A_n\subset A, B_n\subset B$ are finite dimensional von Neumann subalgebras, such that $A_n\subset A_{n+1}$ , $B_n\subset B_{n+1}$ , $a_n:=\dim (A_n)\geq 2$ and $b_n:=\dim (B_n)\geq 2$ , for every $n\in \mathbb N$ . Fix $n\in \mathbb N$ . Write $A_n=\bigoplus _{i=1}^{a_n}\mathbb Cp_{n,i}$ and $B_n=\bigoplus _{j=1}^{b_n}\mathbb Cq_{n,j}$ , where $(p_{n,i})_{i=1}^{a_n}$ and $(q_{n,j})_{j=1}^{b_n}$ are partitions of unity into projections from A and B, respectively. For every $1\leq i\leq a_n$ and $1\leq j\leq b_n$ , represent $p_{n,i},q_{n,j}\in \prod _{\mathcal U}M_k$ as $p_{n,i}=(p_{n,i}^k)^{\mathcal U}$ and $q_{n,j}=(q_{n,j}^k)^{\mathcal U}$ , where for every $k\in K$ , $(p_{n,i}^k)_{i=1}^{a_n}$ and $(q_{n,j}^k)_{j=1}^{b_n}$ are partitions of unity into projections from $M_k$ . Denote $A_n^k=\bigoplus _{i=1}^{a_n}\mathbb Cp_{n,i}^k$ and $B_n^k=\bigoplus _{j=1}^{b_n}\mathbb Cq_{n,j}^k$ . Moreover, we can arrange that $A_n^k\subset A_{n+1}^k$ and $B_n^k\subset B_{n+1}^k$ , for every $n\in \mathbb N$ and $k\in K$ .

If $(r_l)_{l=1}^m$ is a partition of unity into nonzero projections from a tracial von Neumann algebra $(N,\tau )$ , then $\left \{\tau (r_{l+1}+\cdots +r_m)r_l-\tau (r_l)(r_{l+1}+\cdots +r_m)\mid 1\leq l\leq m-1 \right \}$ is an orthogonal basis for $C\ominus \mathbb C1$ contained in $C_{\text {sa},1}$ , where $C=\bigoplus _{l=1}^m\mathbb Cr_l$ . Using this observation, for every $1\leq i\leq a_n-1,1\leq j\leq b_n-1$ and $k\in K$ , we define

$$ \begin{align*}x_{n,i}=\tau(p_{n,i+1}+\cdots+p_{n,a_n})p_{n,i}-\tau(p_{n,i})(p_{n,i+1}+\cdots+p_{n,a_n}),\end{align*} $$

$$ \begin{align*}y_{n,j}=\tau(q_{n,j+1}+\cdots+q_{n,b_n})q_{n,j}-\tau(q_{n,j})(q_{n,j+1}+\cdots+q_{n,b_n}),\end{align*} $$

$$ \begin{align*}x_{n,i}^k=\tau(p_{n,i+1}^k+\cdots+p_{n,a_n}^k)p_{n,i}^k-\tau(p_{n,i}^k)(p_{n,i+1}^k+\cdots+p_{n,a_n}^k),\end{align*} $$

$$ \begin{align*}y_{n,j}^k=\tau(q_{n,j+1}^k+\cdots+q_{n,b_n}^k)q_{n,j}^k-\tau(q_{n,j}^k)(q_{n,j+1}^k+\cdots+q_{n,b_n}^k).\end{align*} $$

Set $x_n=(x_{n,i})_{i=1}^{a_n-1}\in A_n^{a_n-1}, y_n=(y_{n,j})_{j=1}^{b_n-1}\in B_n^{b_n-1}, x_n^k=(x_{n,i}^k)_{i=1}^{a_n-1}\in M_k^{a_n-1}$ and $y_n^k=(y_{n,j}^k)_{j=1}^{b_n-1}\in M_k^{b_n-1}$ . Let $n\in \mathbb N$ , $1\leq i,i'\leq a_n-1$ and $1\leq j,j'\leq b_n-1$ with $(i,j)\not =(i',j')$ . Since $A_n$ and $B_n$ are $2$ -independent, $x_{n,i}\not =0$ and $y_{n,j}\not =0$ , we have that $\|[x_{n,i},y_{n,j}]\|_2= \sqrt {2}\|x_{n,i}\|_2\|y_{n,j}\|_2>0$ and $\tau (x_{n,i}y_{j,n})=0$ . Moreover, $\langle [x_{n,i},y_{n,j}],[x_{n,i'},y_{n,j'}]\rangle =2\tau (x_{n,i}x_{n,i'})\tau (y_{n,j}y_{n,j'}).$ Since $(x_{n,i})_{i=1}^{a_n-1}$ and $(y_{n,j})_{j=1}^{b_n-1}$ are pairwise orthogonal, we get that $\langle [x_{n,i},y_{n,j}],[x_{n,i'},y_{n,j'}]\rangle =0$ . Altogether, we derive that $\delta (x_n,y_n)>0$ and $\varepsilon (x_n,y_n)=\gamma (x_n,y_n)=0$ .

Thus, we get that $\lim _{k\rightarrow \mathcal U}\delta (x_n^k,y_n^k)=\delta (x_n,y_n)>0$ , $\lim _{k\rightarrow \mathcal U}\varepsilon (x_n^k,y_n^k)=\varepsilon (x_n,y_n)=0$ and $\lim _{k\rightarrow \mathcal U}\gamma (x_n^k,y_n^k)=\gamma (x_n,y_n)=0$ . By applying Lemma 5.2, we find $v_n^k\in \mathscr U(M_k)$ , for every $k\in K$ , such that $\varepsilon (v_n^kx_n^k{v_n^k}^*,y_n^k)=0$ , for every $k\in K$ , and $\lim _{k\rightarrow \mathcal U}\|v_n^k-1\|_\infty =~0$ . Since $x_n^k$ and $y_n^k$ are bases for $A_n^k$ and $B_n^k$ , respectively, we get that $v_n^kA_n^k{v_n^k}^*$ and $B_n^k$ are orthogonal, for every $k\in K$ .

To complete the proof, we consider two cases:

Case 1. $\mathcal U$ is countably cofinal.

In this case, we proceed as in the proof of [Reference Boutonnet, Chifan and IoanaBCI15, Lemma 2.2]. Since $\mathcal U$ is countably cofinal, there exists a decreasing sequence $\{S_n\}_{n\geq 2}$ of sets in $\mathcal U$ , such that $\bigcap _{n\geq 2}S_n=\emptyset $ . For $n\geq 2$ , let $T_n=\{k\in K\mid \|v_m^k-1\|_\infty <\frac {1}{n}, \forall 1\leq m\leq n\}\in \mathcal U$ , and set $K_n=S_n\cap T_n$ . Then $\{K_n\}_{n\geq 2}$ is a decreasing sequence of sets in $\mathcal U$ , such that $\bigcap _{n\geq 2}K_n=\emptyset $ . Let $K_1=K\setminus K_2$ . For every $k\in K$ , let $n(k)$ be the smallest integer $n\geq 1$ , such that $k\in K_n$ . Then $n(k)$ is well-defined and $\lim _{k\rightarrow \mathcal U}n(k)=+\infty $ .

For $k\in K$ , let $C_k=A_{n(k)}^k, D_k^0=B_{n(k)}^k$ and $v_k=v_{n(k)}^k$ . If $n(k)\geq 2$ , then as $k\in K_{n(k)}$ , we have $\|v_k-1\|_\infty < \frac {1}{n(k)}$ . Since $\lim _{k\rightarrow \mathcal U}n(k)=+\infty $ , we get that $\lim _{k\rightarrow \mathcal U}\|v_k-1\|_\infty =0$ .

Let $n\in \mathbb N$ . Since $\{k\in K\mid n(k)\geq n\}\in \mathcal U$ and the sequences $\{A_m^k\}_{m\in \mathbb N}$ and $\{B_m^k\}_{m\in \mathbb N}$ are increasing for every $k\in K$ , we get that $\prod _{\mathcal U}A_n^k\subset \prod _{\mathcal U}C_k$ and $\prod _{\mathcal U}B_n^k\subset \prod _{\mathcal U}D_k^0$ . Since $A_n\subset \prod _{\mathcal U}A_n^k$ and ${B_n\subset \prod _{\mathcal U}B_n^k}$ , we conclude that $A_n\subset \prod _{\mathcal U}C_k$ and $B_n\subset \prod _{\mathcal U}D_k^0$ . As this holds for every $n\in \mathbb N$ , we get that $A\subset \prod _{\mathcal U}C_k$ and $B\subset \prod _{\mathcal U}D_k^0$ . Finally, let $D_k=v_kD_k^0v_k^*$ . Then $C_k=A_{n(k)}^k$ and $D_k=v_{n(k)}^kB_{n(k)}^kv_{n(k)}^*$ are orthogonal, for every $k\in K$ . Since $\lim _{k\rightarrow \mathcal U}\|v_k-1\|_\infty =0$ , we get that $\prod _{\mathcal U}D_k^0=\prod _{\mathcal U}D_k$ and $B\subset \prod _{\mathcal U}D_k$ . This finishes the proof of Case 1.