Can we characterize the spatial tensor product of von Neumann algebras categorically? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:49:46Z http://mathoverflow.net/feeds/question/51490 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51490/can-we-characterize-the-spatial-tensor-product-of-von-neumann-algebras-categorica Can we characterize the spatial tensor product of von Neumann algebras categorically? Dmitri Pavlov 2011-01-08T19:26:34Z 2011-01-10T09:32:09Z <p>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.</p> <p>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.</p> <p>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 <a href="http://math.berkeley.edu/~pavlov/scans/guichardet.pdf" rel="nofollow">1966 paper</a>.</p> <p>The categorical tensor product is much bigger than the spatial tensor product. The difference for commutative algebras is explained in this answer: <a href="http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th/49542#49542" rel="nofollow">http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th/49542#49542</a></p> <p>Is there a categorical characterization of the spatial tensor product of von Neumann algebras?</p> <p>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.</p> <p>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.</p> <p>Can we characterize categorically the pairs (f,g) that belong to this subfunctor?</p> <p>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?</p> http://mathoverflow.net/questions/51490/can-we-characterize-the-spatial-tensor-product-of-von-neumann-algebras-categorica/51636#51636 Answer by Thomas Timmermann for Can we characterize the spatial tensor product of von Neumann algebras categorically? Thomas Timmermann 2011-01-10T09:32:09Z 2011-01-10T09:32:09Z <p>I am not sure whether this helps, but along the lines of the comment of Theo Johnson-Freyd, one can say the following:</p> <p>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. </p> <p>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 <a href="http://arxiv.org/abs/0907.4846" rel="nofollow">link text</a></p>