Suppose that $A$ and $B$ are unital $C^*$-algebras, and that $A \otimes B$ is a nuclear $C^*$-algebra. I want to see if $B$ is nuclear or not.

Since $A \otimes B$ is nuclear, there exist c.c.p maps $\phi_n: A\otimes B \to M_{k_n}(\mathbb{C})$ and $\psi_n: M_{k_n}(\mathbb{C}) \to A \otimes B$ such that $\psi_n \circ \phi_n \to \text{id}$ in the point norm topology.

So I have the following maps: $$B \xrightarrow[j]{b \to 1_A\otimes b}A \otimes B \xrightarrow{\phi_n}M_{k_n}(\mathbb{C})\xrightarrow{\psi_n}A\otimes B \xrightarrow[\pi]{a \otimes b \to b}B$$

Letting $\tilde{\phi_n}=\phi_n \circ j$ and $\tilde{\psi_n}=\pi \circ\psi_n$, we see that $\tilde{\psi_n}\circ \tilde{\phi_n}=\pi\circ(\psi_n \circ \phi_n)\circ j$ and this converges to $\text{id}$ in the point norm topology.

The same thing can be done for $A$ as well. The only question I have is whether $\pi$ is well defined or not? If not, then is there a way of showing this? Or, is there an example of two non-nuclear $C^*$-algebras whose tensor product is nuclear?

Thanks for the help!!


As stated, your $\pi$ cannot be linear (so, it is not well-defined). If $\pi$ were somehow linear you would have, for any $a\in A$, $$ \pi(a\otimes b)=b=\pi(2a\otimes b). $$ So $b=\pi(a\otimes b)=\pi(2a\otimes b-a\otimes b)=b-b=0$. Not pretty.

What you need to do is take $\pi$ to be a slice map: you fix a state $\phi:A\to\mathbb C$, define $\pi(a\otimes b)=\phi(a)b$, and extend by linearity. The issue is to show that this actually defines a bounded operator. As far as I can tell, this is not obvious. What we can do is identify $B$ with $\mathbb C\otimes B$, so $\phi(a)b$ is identified with $\phi(a)\otimes B$ (this extends properly to an isomorphism of C$^*$-algebras, easy exercise).

So now $\pi$ looks like $\pi(a\otimes b)=\phi(a)\otimes b$, extended by linearity. Then for instance Theorem 3.5.3 in Brown-Ozawa implies that $\pi$ exists and is completely positive, and that $\|\pi\|=\|\phi\|=1$.

With this new $\pi$, your argument works and indeed $A$ and $B$ are nuclear.

  • $\begingroup$ So $A$ and $B$ are both nuclear?? $\endgroup$ – tattwamasi amrutam Apr 12 '18 at 0:23
  • $\begingroup$ Yes, that's what I wrote. $\endgroup$ – Martin Argerami Apr 12 '18 at 0:29
  • $\begingroup$ Thank you. !!!! $\endgroup$ – tattwamasi amrutam Apr 12 '18 at 0:29

A comment on continuity of the slice map: For $\psi \in S(B)$ we define $\eta_\psi : A \odot B \to A : a \otimes b \mapsto a \psi(b)$. Let $\phi \in S(A)$. Then, for $x = \sum_i a_i \otimes b_i \in A \odot B$, we get $$ \phi(\eta_\psi(x)) = \sum_i \phi(a_i)\psi(b_i) = (\phi \otimes \psi)(x). $$ If one knows that $\phi \otimes \psi$ defines a state on $A \otimes B$, it immediately follows that $\lVert \eta_\psi(x) \rVert \leq \lVert x \rVert$.

If not, it is not hard to show that $\phi \otimes \psi \in S(A \otimes B)$. Indeed, working with the universal representation of $A$ and $B$, one can easily show that $$ \lVert x \rVert = \sup_{\phi \in S(A), \psi \in S(B)} \lVert (\pi_\phi \otimes \pi_\psi)(x) \rVert, $$ where $\pi_\phi$ resp. $\pi_\psi$ denote the associated GNS representation. If now $x_\phi, x_\psi$ denote the cyclic vectors for the representations $\pi_\phi$ and $\pi_\psi$, we see that $$ \lvert (\phi \otimes \psi)(x) \rvert = \lvert \langle (\pi_\phi \otimes \pi_\psi)(x)(x_\phi \otimes x_\psi), x_\phi \otimes x_\psi \rangle \lvert \leq \lVert (\pi_\phi \otimes \pi_\psi)(x) \rVert \leq \lVert x \rVert. $$


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