Defining the abstract tensor product of W*-algebras via a universal property

Question

I am playing around a bit with $W^*$-algebras, and I'm trying to come up with a definition for the $W^*$-algebraic tensor product. Here is my first attempt:

It is easy to show that such an object exists. Simply represent the $W^*$-algebras as concrete von Neumann algebras on Hilbert spaces and consider the usual von Neumann algebra tensor product (i.e. the weak closure of the algebraic tensor product $M \odot N$).

However, I fail to show that the above definition determines the $W^*$-algebra $M \overline{\otimes} N$ uniquely (up to (normal) $*$-isomorphism). Is this true? If not, can we still modify the above universal property to get a well-working definition?

Basically, I want my universal property to convey the intuition that the $W^*$-tensor product is simply the usual von Neumann tensor product when we appropriately represent the spaces involved.

Thanks in advance for any help!

Answer 1

You can see this is false by taking $N = \mathbb{C}$. Then, given a von Neumann algebra $M$, you are asking for a von Neumann algebra $\widetilde{M}$ and a weak* dense embedding $\iota: M \to \widetilde{M}$, such that there is a weak* continuous homomorphism $\pi: \widetilde{M} \to M$ with $\pi\circ\iota = {\rm id}_M$. Your proposed solution is $\widetilde{M} = M$, but you could also take $\widetilde{M} = M^{**}$, for instance. Here $\iota: M \to M^{**}$ would be the natural embedding and $\pi: M^{**} \to M$ would be the adjoint of the natural embedding from $M_*$ into $M^*$.

To bring this a little more down to earth, consider $M = l^\infty \cong C(\beta \mathbb{N})$ and take $\widetilde{M} = l^\infty(\beta\mathbb{N})$ (which is not all of $M^{**}$, only part of it). Here $\iota$ is the identity map from $C(\beta\mathbb{N})$ into $l^\infty(\beta\mathbb{N})$ and $\pi$ is the restriction map from $l^\infty(\beta\mathbb{N})$ to $l^\infty(\mathbb{N})$.

Answer 2

Let me add that the basic flaw in your attempt is that you are working with a l.c. topology which is not complete. This is a problem common to the array of standard topologies which are used in von Neumann algebras--weak, strong, strong $\ast$, ultraweak, ultrastrong (not, of course for the norm but it is useless here). However, this is not a "remedy is none" situation. There is a standard ploy (the mixed or strict topology) which saves the day in each case. One replaces each of the first three by the so-called $\beta$ version--the finest l.c. topology which agrees with it on the unit ball. These are complete. They are described in the chapter on von Neumann algebras of the monograph "Saks Spaces and Applications to Functional Analysis". Carrying out an analogous construction with one of these topologies validates your argument.

Answer 3

Something similar was studied by Wiersma in arXiv:1506.01671 [math.OA]. However, that paper takes a different definition of the universal property: it wants the individual maps $\sigma$ and $\tau$ to be weak$^*$-homeomorphisms onto their ranges, and the map $\iota$ needs to have this property as well.

However, even here you do not recover the usual von Neumann tensor product, but rather something larger.