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?

  • 3
    $\begingroup$ 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, 2013 at 18:30

2 Answers 2


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.

  • $\begingroup$ 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, 2013 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 $A\odot A$? $\endgroup$ Apr 19, 2013 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 $A \cong A^{\rm op}$. $\endgroup$ Apr 19, 2013 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, 2018 at 14:27
  • $\begingroup$ @mathrookie: ? Unless trivial, $A$ has many *-automorphisms. $\endgroup$ Oct 9, 2018 at 7:31

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.


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