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 $A\_1\,\bar\otimes\, A\_2\,\bar\otimes\, A\_3\to B(H)$?



On spatial tensor products of von Neumann algebras
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 $A\_1\vee A\_2\vee A\_3 \not\cong A\_1\mathbin{\bar{\otimes}}A\_2\mathbin{\bar{\otimes}}A\_3$. 


