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A C*-algebra is nuclear if the algebraic tensor product $A\odot B$ ($B$ is any other C*-algebra) admits a unique C*-norm. This definition requires testing the condition for nuclearity with `all' C*-algebras. But is there a C*-algebra $B$ which is good enough for all separable $A$s? More precisely,

Does there exist a C*-algebra $B$ such that, given a separable C*-algebra $A$, $A$ is nuclear if and only if there is a unique C*-norm on $A\odot B$?

Presumably, $B$ has to be non-separable and highly non-nuclear. Is it known for $B$ being the Calkin algebra?

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    $\begingroup$ I have often wondered about the related question: is there some "natural" construction $B=B_A$ such that A is nuclear iff there is a unique Cstar norm on $A\odot B_A$ ? $\endgroup$ – Yemon Choi Apr 18 '13 at 18:30
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Yes. One can take $B$ to be the direct sum of all separable C*-algebras. A more interesting answer would be $B={\mathcal Q}(\ell_2)\otimes C^{\ast}(F_\infty)$. For an explanation, let me start with another question of you: It is an open problem whether $B = {\mathcal Q}(\ell_2)$, the Calkin algebra, suffices or not. It's written in my textbook with Nate Brown (Problem 10.4.1). If $A\odot{\mathcal Q}(\ell_2)$ has a unique C*-norm, then $A$ is exact and $A\odot{\mathcal B}(\ell_2)$ has also a unique C*-norm. The latter condition is equivalent to that $A$ has the LLP. Kirchberg's QWEP conjecture (which is equivalent to Connes's Embedding Conjecture) asserts that LLP implies WEP. It is known that exact (or local reflexivity) $+$ WEP implies nuclearity. In conclusion, if one finds a non-nuclear $A$ which has a unique C*-norm on $A \odot{\mathcal Q}(\ell_2)$, then one solves the QWEP conjecture in negative. Since WEP is equivalent to that $A \odot C^{\ast}(F_\infty)$ has a unique C*-norm, $B={\mathcal Q}(\ell_2)\otimes C^{\ast}(F_\infty)$, or any C*-algebra $B$ containing both ${\mathcal Q}(\ell_2)$ and $C^{\ast}(F_\infty)$ with conditional expectations, meets the condition. Whether there exists a separable $B$ that meets the condition is not known.

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  • $\begingroup$ In the second sentence, you might want to restrict to one representant from each isomorphism class, so that you get a set. $\endgroup$ – Rasmus Apr 19 '13 at 13:05
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    $\begingroup$ Just out of curiosity, is there a known example of a non-nuclear $C*$-algebra $A$ with a unique C*-norm on $A\odot A$? $\endgroup$ – Tomasz Kania Apr 19 '13 at 14:09
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    $\begingroup$ @Rasmus: Yes. @Kania: Yes. Connes's result in von Neumann theory had suggested that $A$ is nuclear if $A \odot A^{\rm op}$ has a unique C*-norm, but a counterexample was constructed by Kirchberg. One can arrange such $A$ so that $A \cong A^{\rm op}$. $\endgroup$ – Narutaka OZAWA Apr 19 '13 at 15:43
  • $\begingroup$ @Narutaka OZAWA,Pro OZAWA.If $A$ is nuclear,can we deduce that there exists a unique $*$ homomorphism from $A\otimes_{min}A^{op}$ to $M(A)\otimes_{min}M(A)^{op}$? $\endgroup$ – math112358 Oct 8 '18 at 14:27
  • $\begingroup$ @mathrookie: ? Unless trivial, $A$ has many *-automorphisms. $\endgroup$ – Narutaka OZAWA Oct 9 '18 at 7:31

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