TITLE: On spatial tensor products of von Neumann algebras
QUESTION [14 upvotes]: Let $H$ be a Hilbert space, and let $A_1,A_2,A_3\subset B(H)$ be three commuting von Neumann algebras.
We write $\odot$ for the algebraic tensor product,
and $\bar\otimes$ for the spatial tensor product of von Neumann algebras.
Suppose that for every $i,j\in\{1,2,3\}$, the map $A_i\odot A_j\to B(H)$ extends to a map $A_i\,\bar\otimes\, A_j\to B(H)$.
Does it follow that the map $A_1\odot A_2\odot A_3\to B(H)$ extends to a map $A_1\,\bar\otimes\, A_2\,\bar\otimes\, A_3\to B(H)$?

REPLY [6 votes]: No. Consider the case where $G$ is an ICC group, $H=\ell_2(G\times G)$, $A_1=(\lambda\otimes\lambda)(G)''$, $A_2=(1\otimes\rho)(G)''$, and $A_3=(\rho\otimes1)(G)''$. By the Fell absorption principle $(g,h)\leftrightarrow (g,gh)$, one has $A_1\vee A_2 \cong (\lambda\otimes1)(G)''\vee(1\otimes\rho)(G)''\cong A_1\mathbin{\bar{\otimes}}A_2$, and similarly for other $A_i \vee A_j$. But since the diagonal $(\lambda\otimes\lambda)(g)\cdot(1\otimes\rho)(g)\cdot(\rho\otimes1)(g)$ fixes $\delta_e\otimes\delta_e$, one has $A_1\vee A_2\vee A_3 \not\cong A_1\mathbin{\bar{\otimes}}A_2\mathbin{\bar{\otimes}}A_3$.