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 $\hat F(a\otimes b)=F(a,b)$?


  • 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.

3 Answers 3


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.

  • $\begingroup$ Thanks. (Though not what I had hoped.) Do you by any chance know a reference for this impossibility? $\endgroup$ Commented Apr 2, 2021 at 22:47
  • $\begingroup$ Do you know whether this impossibility also holds in the special case where we linearize $F(a,b):=G(a)\cdot H(b)$ where $G,H$ have commuting ranges (and $G,H$ are guaranteed to be bounded unital *-homomorphisms)? $\endgroup$ Commented Apr 2, 2021 at 22:52
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    $\begingroup$ This is also false assuming both G and H have domain B(E) where E is an infinite-dimensional Hilbert space; but it is true if either G or H is defined on a finite-dimensional matrix algebra, and the linearized map has norm 1, so this might be good enough for your purposes. The failure in the infinite-dimensional setting is because the appropriate tensor product for linearizing these special kinds of F is the maximal tensor product of Cstar algebras, and $B(E_1)\otimes_{\max} B(E_2)$ is not $B(E_1\otimes E_2)$ if $E_1$ and $E_2$ are infinite-dimensional $\endgroup$
    – Yemon Choi
    Commented Apr 3, 2021 at 1:02
  • $\begingroup$ I think I understand. Since $\otimes_{\max}$ and $\otimes_\mathrm{spatial}$ both can be written as $F(-)\cdot G(-)$ (using $F(x):=x\otimes1$, $G(x):=1\otimes x$), if both tensor products satisfy the universal property, they can be shown isomorphic. And $\otimes_{\max}$ and $\otimes_\mathrm{spatial}$ are not isomorphic (in the ininite-dimensional case). The only chance to get a tensor product different from $\otimes_{\max}$ to satisfy my universal property would be for a notion of maps where $-\otimes_\max1$, $1\otimes_\max-$ are not maps. $\endgroup$ Commented Apr 3, 2021 at 11:04
  • $\begingroup$ Yemon, I found a proof that the product of normal unital *-homomorphisms can be linearized w.r.t. von Neumann tensor product. Since that stands in contrast to what you said in your comment, you might be interested to have a look (and maybe tell me that I am wrong, though I hope not). $\endgroup$ Commented Apr 16, 2021 at 21:44

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.

  • $\begingroup$ I think Yemon's answer covered my use case. I do currently need to linearize $F(-)\cdot G(-)$ for commuting-range $F,G$. But maybe I have some wiggle-room there that could allow me to use your idea. I will have to think more about it. $\endgroup$ Commented Apr 3, 2021 at 11:25

$\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.


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