Tensor product of von Neumann algebras

Hi everyone,

I currently encountered some difficulties on my research related to operator algebras and I wondered if by any chance someone could find my question quite trivial. Here is the context: given a von Neumann algebra $A$, I know that we can construct the so-called von Neumann tensor product $A\otimes A$ of $A$, providing us with a natural completion of the algebraic tensor product of $A$. My question is then the following: can we continuously extend the multiplication map $u\otimes v \to uv$ (defined on the algebraic tensor product) to a linear map defined on the whole product $A\otimes A$? In other words, is this multiplication map continuous with respect to the topology of $A\otimes A$? If not, what if $A$ is endowed with a faithful trace in addition? These questions may be obvious for some operator algebras specialists, but I can’t find any paper/book on this specific issue…

Thank you for any help.

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 I'm afraid I don't recall the proof, but I think that if you take $A= M_n$ then the linear extension of the multiplication map to $M_{n^2}\to M_n$ has norm $O(n)$. So if you now take $A=\prod_n M_n$ (a perfectly nice Type I finite von Neumann algebra) then you will see that the multiplication map does not extend continuously to the spatial tensor product of $A$ with itself – Yemon Choi Nov 20 at 21:56 Which topology on $A \otimes A$ are you considering? In any case, try $A = L^\infty[0, 1]$, if continuity fails here then it also fails for any von Neumann algebra which contains a copy of $L^\infty[0, 1]$. – Jesse Peterson Nov 21 at 6:58 @Jesse: in the abelian case the multiplication extends to the spatial tensor square. (Was your comment about normality?) – Yemon Choi Nov 21 at 8:16 @Yemon: Yes, my comment was about normality. If $A = L^\infty[0, 1]$ then the map will not extend to a normal map on $A \overline \otimes A \cong L^\infty([0, 1] \times [0, 1])$. – Jesse Peterson Nov 21 at 8:38

As a Von Neumann algebra, $A\otimes A$ (recall my notation) is a set of bounded operators, and so I use the operator norm on this space. Anyway, thanks a lot to both of you for your attention and your enlightening examples, I will think about it and will come back with (hopefully) some final answer. Thanks again.