Universal property of tensor products of bounded operators

Question

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.

Answer 1

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.

Answer 2

One can say something positive, though it may not be useful. You say you are willing to "be flexible about what map means". Well, you could define the maps you are interested in to be those which extend! For example, perhaps we want to end up with $\hat F:B(H_1\otimes H_2) \rightarrow B(H)$ which are weak$^*$-continuous and completely bounded. Then just set your class of bilinear maps to be those $F$ which are given by $$ F(a,b) = \hat F(a\otimes b), $$ for such a $\hat F$.

This is obviously a bit tautological. Perhaps you had in mind which $F$ you "really" cared about. You might ask again; or perhaps Yemon's answer has already dealt with the cases you want.

In Banach/Operator Space theory, it is common to study tensor products which don't really come from "universal properties" of this type. For example, the Injective Tensor Product. The maps this linearises would be "bilinear integral" maps, but I don't know if such things have really been studied. Your question is somehow about a weak$^*$-version of this. (In Banach Space theory one is interested in studying the dual space of tensor products: that is, which bilinear maps linearise where the codomain is the scalars. For the injective norm you get the "integral operators".)

If you are interested in linearising the product map (as Yemon's answer discusses) then the relevant tensor product is the (weak$^*$) Haagerup tensor product.

Finally, though off-topic for this question, related is a nice paper of Wiersma which considers ways to tensor two von Neumann algebras: not from the perspective of a universal property, but rather variations on the "spatial" definition.

Answer 3

$\newcommand\range{\operatorname{range}} \newcommand\B{\mathcal B} \newcommand\H{\mathcal H} \newcommand\K{\mathcal K} $ We can achieve the universal property for certain bilinear maps of the specific form $F(x,y) = H(x) K(y)$. More precisely:

Theorem: Let $\mathcal H,\mathcal K,\mathcal L$ be Hilbert spaces. Let $H:\mathcal B(\mathcal H)\to\mathcal B(\mathcal L)$ and $K:\mathcal B(\mathcal K)\to\mathcal B(\mathcal L)$ be normal unital $*$-homomorphisms. Assume that the ranges of $H$ and $K$ commute. Then there is a normal unital *-homomorphism $G:\mathcal B(\mathcal H\otimes\mathcal K)\to \mathcal B(\mathcal L)$ such that $G(h\otimes k)=H(h)K(k)$.

Proof: By this answer, $H(h)=U_H(h\otimes 1_{\mathcal H_0})U_H^*$ for some Hilbert space $\mathcal H_0$ and some unitary $U_H$. Thus $H=\hat H \circ \iota_H$ where $\hat H(x):=U_HxU_H^*$ and $\iota_H(h)=h\otimes 1_{\mathcal H_0}$. $\hat H$ is a unital $*$-isomorphism and therefore normal. $\iota_H$ is a unital $*$-homomorphism. $\iota_H$ is weak*-continuous and therefore normal. Analogously, we get that $K=\hat K\circ\iota_K$ for some normal unital *-isomorphism $\hat K$ and the normal unital *-homomorphism $\iota_K(k):=k\otimes 1_{\mathcal K_0}$. Let $\iota_0(x):=1_{\mathcal H}\otimes x$ for $x\in\mathcal H_0$, also a normal unital *-homomorphism.

The ranges of $H,K$ commute by assumption. Thus $\hat H(\mathcal B(\mathcal H)\otimes \mathbf 1)$ and $\hat K(\mathcal B(\mathcal K)\otimes \mathbf 1)$ commute. Here $\mathbf 1$ denotes the von Neumann algebra in $\mathcal B(\mathcal H_0),\mathcal B(\mathcal K_0)$, respectively, that consists of the multiples of identity. And $\otimes$ is tensor product of von Neumann algebras. Since $\hat H$ is a normal *-isomorphism, $\mathcal B(\mathcal H)\otimes\mathbf 1$ and $S:=\hat H^{-1}\hat K(\mathcal B(\mathcal K)\otimes\mathbf 1)$ commute and $S$ is a von Neumann algebra. Hence $$S \subseteq (\mathcal B(\mathcal H)\otimes\mathbf 1)' \stackrel{(*)}= \mathcal B(\mathcal H)'\otimes \mathbf 1' = \mathbf 1\otimes \mathcal B(\mathcal H_0) = \range \iota_0.$$ Here $X'$ denotes the commutant of $X$ and $(*)$ follows from the commutation theorem for tensor products of von Neumann algebras.

Since $\iota_0$ is an isometry, $\iota_0^{-1}:\mathbf 1\otimes\mathcal B(\mathcal H_0)\to\mathcal B(\mathcal H_0)$ exists and is a normal unital $*$-isomorphism. Thus $$ f : \mathcal B(\mathcal K) \overset{\iota_K}\longrightarrow \mathcal B(\mathcal K) \otimes \mathbf 1 \overset{\hat H^{-1}\hat K}\longrightarrow S \overset{\iota_0^{-1}}\longrightarrow \mathcal B(\mathcal H_0) $$ is a composition of unital normal *-homomorphisms and hence a unital normal *-homomorphism.

Since $1_{\B(\H)}$ and $f$ are normal $*$-homomorphisms and thus completely positive, by [Tak, Prop. 5.13], $1_{\B(\H)}\otimes f:\B(\H\otimes\K)\to\B(\H\otimes\H_0)$ exists and is normal and completely positive. And it satisfies $(1_{\B(\H)}\otimes f)(h\otimes k)=1(h)\otimes f(k)$. Then $1_{\B(\H)}\otimes f$ is also unital and bounded. On the algebraic tensor product $\B(\K)\otimes_{\mathit{alg}}\B(\H)$, $1_{\B(\H)}\otimes f$ is a *-homomorphism, and thus it is a *-homomorpism everywhere by continuity. Let $G := \hat H\circ (1_{\mathcal B(\mathcal H)}\otimes f)$. Since $\hat H$ is also a normal unital $*$-homomorpishm, so is $G$.

We have $$ G(h\otimes 1_{\mathcal K}) = \hat H(h\otimes 1) = H(h) $$ and $$ G(1_{\mathcal H}\otimes k) = \hat H(1_{\mathcal H}\otimes f(k)) = \hat H(\iota_0\circ f(k)) = \hat H(\hat H^{-1}\circ\hat K\circ \iota_K(k)) = \hat K\circ \iota_K(k) = K(k). $$ Since $G$ is a *-homomorphism, this implies $G (h\otimes k) = G(h\otimes 1)G(1\otimes k)=H(h)K(k)$ as desired.

[Tak] Takesaki, M., Theory of operator algebras I., Encyclopaedia of Mathematical Sciences 124. Operator Algebras and Non-Commutative Geometry 5. Berlin: Springer (ISBN 3-540-42248-X/hbk). xix, 415 p. (2002). ZBL0990.46034.