# On spatial tensor products of von Neumann algebras

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 $A_1\,\bar\otimes\, A_2\,\bar\otimes\, A_3\to B(H)$?

• Can you show that this implies that $(A_1\vee A_2) \odot A_3$ extends to a map $(A_1\vee A_2) \mathbin{\bar\otimes} A_3$? Jul 4, 2016 at 7:06
• If I could show that, then I could solve my problem. Indeed, $A_1\vee A_2=A_1\,\bar\otimes\, A_2$ by assumption, and $(A_1\,\bar\otimes\, A_2)\,\bar\otimes\, A_3=A_1\,\bar\otimes\, A_2\,\bar\otimes\, A_3$ by definition. So, to answer your question: unfortunately no, I don't know. Jul 4, 2016 at 9:27
• Exactly. Why is naively intersecting the type I factors not working? Jul 4, 2016 at 16:23
• Because I don't know that the intersection still type I. Jul 4, 2016 at 20:26

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 $A_1\vee A_2\vee A_3 \not\cong A_1\mathbin{\bar{\otimes}}A_2\mathbin{\bar{\otimes}}A_3$.
• The construction can even be generalised to yield $n$ commuting actions such that any $n-1$ of them extend to the spatial tensor product, but this does not happen for all actions at once. Take $H=\ell_2(G^{n-1})$ and the $n$ actions to be given by $\lambda^{\otimes n-1}$ and $1\otimes...\otimes \rho \otimes...\otimes 1$. Jun 22, 2017 at 10:11