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A C*-algebra $A$ is called nuclear iff there is a unique cross C*-norm on the algebraic tensor product $A\otimes B$ for any other C*-algebra $B$. It is equivalent that the minimal (spatial) and maximal norms on $A\otimes B$ coincide.

I know (cf. this MathOverflow question) that an example of a non-nuclear C*-algebra ist $C_r^\ast(\mathbb F_2)$ where $\mathbb F_2$ is the free group on two generators and $C_r^\ast(\mathbb F_2)$ denotes its reduced group C*-algebra.

Now my question is: Is there an explicit example of a C*-algebra $B$ such that the minimal and maximal norms on $C_r^\ast(\mathbb F_2)\otimes B$ do not coincide? With explicit I mean that one can write down both norms and see that they are different, e.g. by evaluating them on some specific element. Are there easier examples for other C*-algebras instead of $C_r^\ast(\mathbb F_2)$?

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Takesaki's original proof that $C_r^*(\mathbb F_2)\otimes C_r^*(\mathbb F_2)$ admits at least two different tensor norms is as "explicit" as it can be, although it cannot explicitly calculate elements of the norms.

The problem here is with the word "explicit": it is a relative notion. You want a specific element of the tensor product. And you want an explicit calculation of its norm. As far as I can tell, with the exception of very carefully crafted trivial examples, there is no way to explicitly calculate the norm of $\sum_ja_j\otimes b_j$.

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  • $\begingroup$ Well, this is actually very much what I thought of. Mainly, I need a simple counterexample to give at the end of a introductory talk about C*-algebras, just to show that tensor products are no trivial construction with C*-algebras. $\endgroup$ – Benedikt Hunger Oct 8 '16 at 18:51
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An answer to your second question (it depends on what you mean by "easier", however), the following $C^*$-algebra is well understood: $B(H)$ for an infinite-dimensional Hilbert space is not nuclear [S. Wassermann, "on Tensor products of certain group $C^*$-algebras"].

Regards

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