Let $X\subseteq B(H)$ be an operator system and let $M\subseteq B(K)$ be a von Neumann algebra. We form the Fubini-tensor product $$X \otimes_\mathcal{F} M := \{z \in B(H\otimes K): (\sigma\otimes \iota)(z) \in M \text{ and } (\iota \otimes \tau)(z)\in X \text{ for all }\sigma\in B(H)_*, \tau \in B(K)_*\}.$$ We have a natural injective linear map $$X \otimes_\mathcal{F}M \to B(M_*,X): z \mapsto (\omega \mapsto (\iota \otimes \omega)(z)).$$

Is it true that the image of this map consists of all completely bounded maps $M_*\to X$?

Note that this is at least true if $X$ itself is a von Neumann algebra (However, this result is not very easy to find in the literature).

  • 1
    $\begingroup$ I'm not sure what is meant in the last paragraph although I agree that finding a properly self-contained account is not so easy (at least I never quite succeeded). If X and M are vN algebras then their Fubini tensor product equals their normal spatial tensor product by Tomita's commutant theorem, and this in turn equals $CB(M_*,X)$ by a result of Effros and Ruan (Int JMath 1990 Theorem 3.2). $\endgroup$
    – Yemon Choi
    Sep 7 at 23:13
  • $\begingroup$ In your definition of the Fubini tensor product you are only considering slices by $\sigma$-weakly continuous functionals. Does this mean you are assuming $X$ to be $\sigma$-weakly closed? $\endgroup$
    – Yemon Choi
    Sep 7 at 23:17
  • $\begingroup$ @YemonChoi I'm only assuming X to be norm-closed, but if you know a solution when X is sigma weakly closed that's also welcome! $\endgroup$ Sep 8 at 5:57
  • $\begingroup$ I deleted some comments which no longer make sense given the edits to the question. In recompense, I have added an answer. $\endgroup$ Sep 8 at 6:43

1 Answer 1


I follow the book of Effros+Ruan (which is a book, so not viewable online, but really is the nicest source I think). For any operator spaces $X,Y$ we can consider the operator space projective tensor product $\newcommand{\proten}{\widehat\otimes}X\proten Y$ whose dual satisfies $$ (X\proten Y)^* = CB(X,Y^*). $$ (To be precise, there is some completely isometric isomorphism here.) This is in Chapter 7, Corollary 7.1.5 to be precise.

The Fubini tensor product is also considered in Chapter 7, Theorem 7.2.3, which shows that $(X\proten Y)^* \cong X^* \bar\otimes_{\mathcal F} Y^*$ in general, so that $X^* \bar\otimes_{\mathcal F} Y^* \cong CB(X,Y^*)$.

So now apply this to a von Neumann algebra $M$ with predual $M_*$, and a dual operator space $X$ with predual $X_*$. Then $$ CB(M_*, X) = (M_* \proten X_*)^* = M \bar\otimes_{\mathcal F} X. $$ Again there are (canonical) isomorphisms involved here, but chasing them down will show that they match the isomorphism given in the original question. In particular, this includes the isomorphism $M \bar\otimes_{\mathcal F} X \cong X \bar\otimes_{\mathcal F} M$.

Here I used that $X$ is a dual space, and in Chapter 7 of Effros and Ruan, we need this, and a "dual realisation" of $X$ as a weak$^*$-closed subspace of $B(H)$. That is, the Original Question has a positive answer when $X \subseteq B(H)$ is weak$^*$-closed.

When $X$ is only assumed norm closed, we can use the definition of the Fubini tensor product given in the original question, though I am not aware of much study of this. However, let $\overline{X}^{w^*}$ be the weak$^*$-closure of $X$ in $B(H)$. Then by definition(s), $$ X \bar\otimes_{\mathcal F} M \subseteq \overline{X}^{w^*} \bar\otimes_{\mathcal F} M \cong CB(M_*, \overline{X}^{w^*}). $$ The isomorphism clearly identifies $X \bar\otimes_{\mathcal F} M$ with those CB maps $T:M_*\rightarrow \overline{X}^{w^*}$ which map into $X$, and by definition of what a CB map is, this is just the space $CB(M_*,X)$. Thus the original question is answered in the affirmative.

  • $\begingroup$ So if I understand correctly: your last paragraph actually shows that we have $X\overline{\otimes}_\mathcal{F} M \cong CB(M_*,X): z \mapsto (\omega \mapsto (\iota \otimes \omega)(z))$ even if $X$ is norm-closed? $\endgroup$ Sep 8 at 7:43
  • $\begingroup$ Yes; I added an edit to make this clear. $\endgroup$ Sep 8 at 9:37
  • $\begingroup$ Thanks! This is splendid! This result is very useful to me :) $\endgroup$ Sep 8 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.