# Nuclear $C^*$ algebra and tensor products

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?

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.