TITLE: Can nuclearity be determined by tensoring with a single C*-algebra?
QUESTION [25 upvotes]: 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?

REPLY [2 votes]: Somewhat related to this, a C*-algebra $A$ is nuclear if and only if $(A^{**}, C^*(\mathbb{F}_\infty))$ is a nuclear pair (that is, has a unique tensor norm), where $A^{**}$ is the enveloping von Neumann algebra of $A$. Not quite what you're after, but related (nuclearity is decided if there is at most one norm not on $A$, but $A^{**}$, tensored with a single other algebra).
This follows first from the fact that $A$ is nuclear if and only if $A^{**}$ is injective, a von Neumann algebra $M$ is injective if and only if it has the WEP, and of course Kirchberg's seminal result [1] that a C*-algebra $A$ has the WEP if and only if $(A, C^*(\mathbb{F}_\infty))$ is a nuclear pair (also mentioned by Prof. Ozawa above). Pisier in his recent book [2] gives a somewhat simpler proof than Kirchberg of this characterization of the WEP, assuming you're familiar with the many interesting nuances of completely positive maps.
Kirchberg's result opened the door for a number of variations on the theme - in fact there are a few other C*-algebras (and operator spaces) you could take in place of $C^*(\mathbb{F}_\infty)$. For instance

$C^*(\mathbb{F}_n)$ for any $n\ge 2$
$C^*(\mathbb{Z}_2 * \mathbb{Z}_3)$ ($*$ denoting the free product)
$C^*(SL_2(\mathbb{Z}))$
the "non-commutative $n$-cube operator system" $NC(n) = \overline{\text{span}}(\{1\}\cup\{u_i\}_{i=1}^n)$, with $u_i$ the generators of $C^*(*_{i=1}^n \mathbb{Z}_2)$
"Farenick's operator space" $J = \mathbb{C}^5/\text{span}((1, 1, -1, -1, -1))$ (the operator space quotient being considered here).

See [3] and [4].
Also on a side note, it appears recent work in quantum complexity theory has refuted the Connes embedding problem (I wish I had a reference, but it's outside my ken), so I take it $B = \mathcal{Q}(\ell_2)$ is definitively insufficient.
[1] - E. Kirchberg. "On non-semisplit extensions, tensor products
and exactness of group C*-algebras". In: Inventiones
mathematicae (1993)
[2] - Gilles Pisier. "Tensor Products of C*-Algebras and Operator Spaces". Cambridge University Press, 2020
[3] - Douglas Farenick, Ali S. Kavruk, Vern I. Paulsen, Ivan G. Todorof, "Characterizations of the Weak Expectation Property". 2013. Retrieved from https://arxiv.org/pdf/1307.1055.pdf.
[4] - Douglas Farenick, Vern I. Paulsen "Operator System Quotients and Tensor Products". 2011. Retrieved from https://arxiv.org/pdf/1010.0380.pdf.