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Let $\mathcal A,\mathcal B,\mathcal C$ be von Neumann algebras. Let $F:\mathcal A\otimes\mathcal B\to\mathcal C$ be a bounded linear map. Assume $F(a\otimes b)=0$ for all $a,b\in\mathcal A,\mathcal B$. Does this imply $F=0$?

And does the answer change if we require $\mathcal A,\mathcal B,\mathcal C$ to be sets of all bounded operators?

Here the tensor product $\mathcal A\otimes\mathcal B$ is the tensor product of von Neumann algebras, i.e., the smallest von Neumann algebra containing all $a\otimes b$.

What I know:

  • If we define $\mathcal A\otimes\mathcal B$ as the spatial tensor product (i.e., the completion of the algebraic tensor product w.r.t. the operator norm), then the answer is yes because $F$ is $0$ on the algebraic tensor product and continuous w.r.t. the operator norm.
  • If we restrict $F$ to be WOT-continuous, then it also holds (if I am not mistaken) because the smallest von Neumann algebra is the WOT-closure of the algebraic tensor product. Thus any $x\in \mathcal A\otimes\mathcal B$ is a WOT-limit of $x_i$ in the algebraic tensor product. Thus $F(x)$ is a WOT-limit of $F(x_i)=0$. Thus $F(x)=0$. Hence $F=0$.
  • In the context of von Neumann algebras, normal maps are relevant (i.e., weak$^*$-continuous positive maps). Maybe for weak$^*$-continuous maps the above holds? (Without "positive".) That would be a useful fallback for me, but I don't see the proof.
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    $\begingroup$ Yes if $F$ is weak${}^*$-continuous, no in general. The first because the algebraic tensor product is weak${}^*$-dense in the von Neumann algebra tensor product, the second because the norm closure generally is not the whole thing, so you can find a nonzero bounded linear functional which vanishes on the algebraic tensor product. That's just the Hahn-Banach theorem. $\endgroup$
    – Nik Weaver
    May 18, 2021 at 1:17
  • $\begingroup$ @NikWeaver I see the "no". The fact that the norm closure is in general not the whole space is shown in this answer: mathoverflow.net/a/295762/101775 But when we consider sets of all bounded operators (i.e., $\mathcal A=B(\mathcal H_A)$, etc.) that answer does not apply. Is the spatial tensor product also different from the von Neumann one in that case? $\endgroup$ May 18, 2021 at 9:18
  • $\begingroup$ @NikWeaver The "yes" part I have trouble with: The von Neumann tensor product is the von Neumann algebra generated from the algebraic tensor product (ATP). So it is the WOT-closure and the SOT-closure of the ATP. So the ATP is WOT-dense and SOT-dense. But why weak*-dense? I feel I am overlooking something obvious here. $\endgroup$ May 18, 2021 at 9:23
  • $\begingroup$ @DominiqueUnruh The spatial tensor product of $B(H_A)$ and $B(H_B)$ contains the closed ideals $B(H_A)\otimes K(H_B)$ and $K(H_A)\otimes B(H_B)$ (spatial tensor products), while the von Neumann tensor product is $B(H_A\otimes H_B)$, which only contains $K(H_A\otimes H_B)$ as proper closed ideal. $\endgroup$
    – MaoWao
    May 18, 2021 at 10:05
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    $\begingroup$ Thanks to both of you. Does one of you want to create an answer? Otherwise I will create an answer based on your comments. $\endgroup$ May 18, 2021 at 10:37

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