Consider the tensor product of bounded operators. Does this tensor product satisfy the universal property of the tensor product, i.e., for any bilinear map $F: B(\mathcal H\_1)\times B(\mathcal H\_2)\to B(\mathcal H)$, there is a unique map $\hat F: B(\mathcal H\_1)\otimes B(\mathcal H\_2)\to B(\mathcal H)$ such that $\hat F(a\otimes b)=F(a,b)$?


**Clarifications:**


* I am flexible about what "map" (and "bilinear map") means here, as long as it is some subset of the linear maps. E.g., bounded linear maps, completely positive maps, etc.
* $B(\mathcal H)$ is the Banach space of bounded operators on the Hilbert space $\mathcal H$.
* The tensor product $B(\mathcal H\_1)\otimes B(\mathcal H\_2)$ is defined as follows: For operators $a,b$, the operator $a\otimes b\in B(\mathcal H\_1\otimes'\mathcal H\_2)$ is the unique operator with $(a\otimes b)(\psi\otimes'\phi)=a\psi\otimes' b\phi$, where $\otimes'$ is the Hilbert space tensor product. Then $B(\mathcal H\_1)\otimes B(\mathcal H\_2)$ is defined as the von-Neumann algebra generated by the $a\otimes b$. We have $B(\mathcal H\_1)\otimes B(\mathcal H\_2) = B(\mathcal H\_1\otimes'\mathcal H\_2)$ according to *Takesaki, Theory of
Operator Algebras I*.



Universal property of tensor products of bounded operators
This is false in general even for very "well-behaved" classes of maps. For instance, the multiplication map on $B(H)$, which is bounded bilinear as a map $B(H)\times B(H) \to B(H)$, does not extend continuously to $B(H\otimes H)\to B(H)$ if $H$ is infinite-dimensional.


In fact, the problem is worse thane one might think, because at the level of matrices the extension of the product map $M\_n \times M\_n \to M\_n$ to $M\_{n^2} \to M\_n$ has a norm which tends to infinity as $n\to \infty$. (I think the growth rate is $O(n)$ but I don't remember the details.)


The relevant universal property is that if $T\_i \in B(H\_i,K\_i)$ for $i=1,2$ then $T\_1\otimes T\_2$ is a well-defined operator $H\_1\otimes H\_2\to K\_1\otimes K\_2$.


If one wants to linearize bilinear maps which are completely bounded (and there are two different notions of complete boundedness for bilinear maps) then you need either the operator-space projective tensor product or the Haagerup tensor product, rather than the spatial von Neumann tensor product that you describe in your question.


