# 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.

• 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 '12 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 '12 at 6:58
• 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. – user29281 Nov 21 '12 at 7:53
• What do you mean: the operator norm? Do you mean that you are viewing the tensor square as an algebra of operators on the tensor product of Hilbert spaces? Because that is the spatial tensor product as in my answer. – Yemon Choi Nov 21 '12 at 8:15
• @Jesse: in the abelian case the multiplication extends to the spatial tensor square. (Was your comment about normality?) – Yemon Choi Nov 21 '12 at 8:16

## 1 Answer

The answer to your question is No (beyond matrix cases i.e. finite dimensional von Neumann algebras). Even the von Neumann tensor product of one abelian von Neumann algebra, which is indeed $L^\infty(X,\mu)$, does not have this property.

Let $$m: M \odot M \rightarrow M$$ be the multiplication on the algebraic tensor product $\odot$ of a von Neumann algebra, that is $m(v\otimes u) = uv$ for $u,v \in M$.

There is an appropriate tensor product on $M\odot M$ which is suitable for linearizing multiplication map (weak$^*$ continuity) $m$ so-called the normal Haagerup tensor product, usually denoted by $M \otimes^{\sigma h} M$ (introduced initially in ). To see the proof of this fact and learn more about this tensor product look at .

References.

 E. Effros and A. Kishimoto, Module maps and Hochschild-Johnson cohomology, Indiana Univ. Math. J. 36(1987), 257–276.

 E. G. Effros and Z.-J. Ruan, Operator spaces tensor products and Hopf convolution algebras, J. Operator Theory 50 (2003) 131–156.