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I am somewhat a beginner in the field of operator algebras and was wondering about the following:

Let $T$ be a linear map between the space of bounded operators $B(H)$ on some Hilbert space and $S$ a map between the space of trace-class operators that we denote by $N(H)$ in the sequel.

Then, one defines maps $T_n: M_n(B(H)) \rightarrow M_n(B(H))$ by $$T_n((a_{ij})):=(T(a_{ij}))$$

Then, $T$ is completely bounded if and only if $\sup_n \left\lVert T_n \right\rVert< \infty.$

And analogously for $S.$

I would like to know whether a map $T$ is completely bounded if and only if $$\left\lVert T \otimes \operatorname{id}_{X} \right\rVert< \infty$$ or in case of $S$ whether complete boundedness is equivalent to $$\left\lVert S \otimes \operatorname{id}_{X} \right\rVert< \infty$$

for some $X$?

For $T$ I think I found a reference on math.stackexchange saying that $X=B(K)$ works for any infinite-dimensional Hilbert space $K$ (nothing on whether $K$ needs to be separable discussed in this thread.) This should imply that $X=N(H)$ works for $S$ as well.

Assuming this to be true, I am wondering about two things now:

1.)Assume I can take $X=B(H)$ for $T$ and $X=N(H)$ for $S$, then $T \otimes \operatorname{id}_{B(H)}$ is a map from $B(H) \otimes B(H)$ into itself. Can the space $B(H) \otimes B(H)$ be identified with $B(H \otimes H)$?- I assume that by a duality argument this would be equivalent to asking whether $N(H) \otimes N(H)$ is isomorphic to $N(H\otimes H)$.

2.) Are there any other choices of $X$ allowed in the above two examples?

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Okay, I think the most important thing to start with is that there are different ways of defining tensor products of C*-algebras. There's a brief introduction here. So when you talk about "the space $B(H) \otimes B(H)$" this could mean different things. The von Neumann tensor product --- which is the weak* completion of the spatial C*-algebra tensor product --- does satisfy $B(H)\otimes B(H) \cong B(H\otimes H)$, where $H \otimes H$ is the usual Hilbert space tensor product.

Moving to question 2, the most obvious choice is $X = K(H)$, the space of compact operators. For any C*-algebra $A$ there is only one way to define $A\otimes K(H)$, and the fact that $\|T\|_{cb} = \|T\otimes I_{K(H)}\|$ is practically immediate. As you suggest, the same is true with $X = B(H)$, but now you need to specify that you're talking about the spatial tensor product.

The corresponding questions for the space of trace class operators might be most easily answered by dualization, since $TC(H)^* \cong B(H)$ and the norm of any Banach space operator equals the norm of its dual. In this dual setting you'd be using the projective tensor product.

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  • $\begingroup$ thank you. Just to understand this fully: The projective tensor product has the property that: $TC(H) \otimes_{\pi}TC(H) \simeq TC(H\otimes K)$? Such that by duality $TC(H \otimes K)^* \simeq B(H \otimes K) \simeq B(H) \otimes_{\varepsilon} B(K).$ $\endgroup$ – Hörmander123 Jul 15 '18 at 3:09
  • $\begingroup$ Yes, I think that's right. $\endgroup$ – Nik Weaver Jul 15 '18 at 4:14
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    $\begingroup$ Not exactly, the injective tensor product needs to be completed in the weak operator topology for this to hold. $\endgroup$ – Mateusz Wasilewski Jul 15 '18 at 6:02

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