Suppose $A,B$ are $C^*$ algebras, $\alpha$ is some $C^*$-norm on the algebraic tensor product of $A$ and $B$. If $A\otimes_{\alpha}B$ is nuclear, can we conclude that $A$ and $B$ are nuclear? What about the converse?
2026-03-26 03:00:54.1774494054
Nuclear $C^*$ algebra and tensor products
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Yes.
Let us assume that $A$ and $B$ are unital.
Fix a $C^*$-norm $\alpha$ on $A \odot B$, by which I denote the algebraic tensor product. Next, we note that for any state $\varphi \in S(A)$ and $\psi \in S(B)$ there exists, by [Prop. 3.4.7., $C^*$-algebras and Finite-dimensional Approximation; Brown and Ozawa], a state $$ \varphi \otimes \psi \colon A \otimes_\alpha B \to \mathbb C: a \otimes b \mapsto \varphi(a)\psi(b). $$
In particular, the slice map $$ \eta_\psi \colon A \odot B \to A : a \otimes b \mapsto a\psi(b) $$ is $\alpha$-continuous. To see this, let $x \in A \odot B$ and $\varphi \in S(A) $ such that $$ \varphi(\eta_\psi(x)) = \lVert \eta_\psi(x) \rVert_A. $$ Now, by the previous discussion, we get $$ \lVert \eta_\psi(x) \rVert_A = \varphi(\eta_\psi(x)) = \lvert (\varphi \otimes \psi)(x) \rvert \leq \lVert x \rVert_{\alpha}. $$
In particular, $\lVert \eta_\psi \rVert \leq 1$, whenever $\psi \in S(B)$.
Now, assume that $A \otimes_{\alpha} B$ is nuclear. We show that $A$ is nuclear (and similarly you can show that $B$ is nuclear).
Let $\mathscr F \subset A$ be finite and $\varepsilon > 0$. Denote by $\iota_A \colon A \to A \otimes_{\alpha} B$ the obvious inclusion. Since $A \otimes_{\alpha} B$ is nuclear, there are a finite dimensional $C^*$-algebra $F$ and c.p.c. maps $\alpha \colon A \otimes_{\alpha} B \to F$ and $\beta \colon F \to A \otimes_\alpha B$ such that for all $a \in \mathscr F$, $$ \lVert (\beta \circ \alpha)(\iota_A(a)) - \iota_A(a) \rVert < \varepsilon . $$ Now, let $\psi$ be any state on $B$. Then
\begin{align*} \lVert \eta_\psi((\beta \circ \alpha)(\iota_A(a))) - a \rVert \\ = \lVert \eta_\psi((\beta \circ \alpha)(\iota_A(a)) - \iota_A(a)) \rVert \\ \leq \lVert (\beta \circ \alpha)(\iota_A(a)) - \iota_A(a) \rVert \\ < \varepsilon \end{align*}
It follows that $A$ is nuclear.
If $A$ or $B$ are not unital, you have to look at $\iota_A \colon A \to A \otimes_\alpha B$ given by $\iota_A(a) = A \otimes h$, where $h$ is a positive element of norm one.
The converse direction can be easily proved for the minimal and maximal tensor prodcut. However, if $\alpha$ is just some $C^*$-norm I am not sure if the tensor product of c.p.c. maps with finite dimensional domain or codomain might be $\alpha$-continuous.