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The tensor product of commutative algebras is exactly their coproduct in the category of commutative algebras. In other words, if A and B are two commutative algebras, then the covariant functor that represents A⊗B assigns to an algebra Z the set of all pairs of morphisms f: A→Z and g: B→Z.

Tensor product of noncommutative algebras also admits a categorical characterization. Namely, if A and B are two noncommutative algebras, then the functor that represents A⊗B assigns to an algebra Z the set of all pairs of morphisms f: A→Z and g: B→Z whose images commute in Z, i.e., m(f⊠g)=ms(f⊠g), where m is the multiplication Z⊗Z→Z, s is the symmetry Z⊗Z→Z⊗Z, and ⊠ is the external tensor product: f⊠g: A⊗B→Z⊗Z.

The category of commutative von Neumann algebras also admits a coproduct, which therefore can be thought of as the categorical tensor product of von Neumann algebras. This tensor product can be extended to noncommutative von Neumann algebras in the same way as described above. Apparently this product was first described by Alain Guichardet in his 1966 paper.

The categorical tensor product is much bigger than the spatial tensor product. The difference for commutative algebras is explained in this answer: Is there a category structure one can place on measure spaces so that category-theoretic products exist?

Is there a categorical characterization of the spatial tensor product of von Neumann algebras?

By the universal property of the categorical tensor product for any two von Neumann algebras A and B there is a canonical morphism Q: C→S of von Neumann algebras from the categorical tensor product C to the spatial tensor product S. This morphism is an epimorphism, i.e., it is surjective. However, unless one of the algebras is finite-dimensional, it has non-trivial kernel.

Hence, the algebra S is represented by a subfunctor of the covariant functor that assigns to a von Neumann algebra Z the set of all pairs of morphisms f: A→Z and g: B→Z with commuting images.

Can we characterize categorically the pairs (f,g) that belong to this subfunctor?

Alternatively, the kernel of the morphism Q is a σ-weakly closed two-sided ideal of C, which corresponds to a central projection of C. Can we characterize this central projection categorically?

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  • $\begingroup$ Proposition 8.2 and further of "Sur la categorie des algebres de Von Neumann" by Alain Guichardet (Bull. Sci. Math. 90:41-64, 1966) defines a tensor product of von Neumann algebras that (i) reverts to the coproduct for commutative algebras that (ii) is not quite a coproduct but does satisfy a (categorical) universal property. He also develops some basic results about it. It is not quite the spatial tensor product, but might resemble it enough to cover what you're after? $\endgroup$
    – Chris Heunen
    Commented Jan 8, 2011 at 21:52
  • $\begingroup$ @Chris: Well, Guichardet's article describes the categorical tensor product and actually it is the place where I learned it. I will add a reference to his paper right now. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 8, 2011 at 23:34
  • $\begingroup$ Just an aside, and it comes with the caveat that I know some noncommutative algebra, but absolutely nothing about von Neumann algebra: the description you gave of the tensor product of noncommutative algebras is correct, but isn't entirely "categorical" in the following sense. Namely, a priori your "external tensor product" requires some more data --- at the least, it requires having a "forgetful" functor so that you can talk about underlying vector spaces of the algebras. But, you can give a "purely" categorical description in the Morita framework, which might be all you want. (continued) $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 9, 2011 at 5:13
  • $\begingroup$ (continuation) Namely, let $A,B$ be rings-up-to-Morita, and $\cal A=A\text{-Mod}$ and $\cal B=B\text{-Mod}$ their categories of modules. Then there is a category $\cal A\boxtimes\cal B$, which is the universal cocomplete category receiving a functor from $\cal A \times \cal B$ (the cartesian product in CAT) that is separately cocontinuous in each variable. By a lemma/exercise, $\cal A\boxtimes\cal B\simeq(A\otimes B)\text{-Mod}$, where $A\otimes B$ is the usual tensor product of rings. So this determines $A\otimes B$ up to Morita equivalence. Maybe something similar works for vN algebras? $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 9, 2011 at 5:19
  • $\begingroup$ @Theo: It appears that you construction again gives the categorical tensor product. Therefore your approach gives rise to yet another reformulation of the problem: How can we characterize the full subcategory of A-Mod⊠B-Mod that corresponds to representations of the spatial tensor product (and not just categorical tensor product)? $\endgroup$
    – Dmitri Pavlov
    Commented Jan 9, 2011 at 5:40

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I am not sure whether this helps, but along the lines of the comment of Theo Johnson-Freyd, one can say the following:

If $A,B,C$ are von Neumann algebras and $Mod(A),Mod(B),Mod(C)$ their categories of representations, then there exists a natural isomorphism between morphisms $A \to B \bar{\otimes} C$ and $Mod(B)\times Mod(C) \to Mod(A)$. Here, the representation categories are viewed as categories ${\cal D}$ fibered over (equipped with forgetful functor $U_\cal D$ to) the category of Hilbert spaces, and the product ${\cal D} \times {\cal E}$ of two such fibered categories is the product category, equipped with the forgetful functor $U_{\cal D} \times U_{\cal E}$ followed by the Hilbert space tensor product functor.

This statement depends on the fact that the categories of representations have generators and extends from the spatial tensor product to fiber products of von Neumann algebras relative to some subalgebra. Categorically, this has to be made more precise and may not be quite what you're looking for. Details for the fiber product (which reduces to the spatial tensor product if the subalgebra is just complex numbers) may be found in Section 4.3 of link text

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    $\begingroup$ But does this characterize $B \bar \otimes C$? Is $Mod(B \bar \otimes C)$ determined by your observation? $\endgroup$
    – Andreas Thom
    Commented Jan 11, 2011 at 9:04

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