Completely positive maps with commuting ranges can be extended to maximal tensor product

Question

I'm trying to do the following exercise from Brown and Ozawa's book.

Exercise $3.5.1.$ Let $\varphi: A \to B(H)$ and $\psi: B \to B(H)$ be c.p. maps with commuting ranges. Show that there exists a c.p. map $\varphi \times_{max} \psi: A \otimes_{max} B \to B(H)$ such that $\varphi \times_{max} \psi(a\otimes b)=\varphi(a)\psi(b)$ for all $a\in A$ and $b\in B$.

Maybe it worth mention that I didn't read the chapters regarding c.p. maps and I know just Stinespring Theorem and the definition of c.p. maps.
So, I'm afraid this is the reason that no solution coming to my mind.
However, I hope that I just need to use continuity of tensor products, Stinespring Theorem + "lifting of commutant", and universality of the maximal tensor product.

However, I don't know how to collect the above to solution.
I prefer a hint than a full-solution, Thank you!

Answer 1

(As far as I can tell, Sabrina's answer is wrong---see my comment there---. Here is a different argument)

$\def\cA{ A}\def\cC{B(H)}\def\cB{B}$ Let $\cA_\varphi =C^*(\varphi(\cA))\subset\cC$ and $\cB_\psi=C^*(\psi(\cB))\subset\cC$. By hypothesis these subalgebras commute with each other. By Theorem 3.5.3 in [Brown-Ozawa 2008, the same section where the exercise belongs] the map $\varphi\otimes\psi:\cA\odot\cB\to\cA_\varphi \odot\cB_\psi$ is completely positive and max-max bounded. Let $\rho:\cA_\varphi \odot\cB_\psi\to\cC$ be the $*$-homomorphism (here is where we use that $\cA_\varphi $ and $\cB_\psi$ commute) induced by $\rho(a\otimes b)=ab$. As $\rho$ is a representation of $\cA\otimes\cB$, we have that $\|\rho(x)\|\leq\|x\|_{\max}$ for all $x\in\cA\odot\cB$ by definition of the max norm. Now $$ \|(\varphi\times\psi)(x)\| =\|(\rho\circ(\varphi\otimes_{\max}\psi))(x)\|\leq\|(\varphi\otimes_{\max}\psi)(x)\|_{\max} \leq\|x\|_{\max}. $$

The complete positivity follows from the fact that we have written $\varphi\times\psi$ as a composition of completely positive maps.

Answer 2

You don't need Stinespring's theorem for this. How this could work (without details):

At first you get a map $\varphi\times \psi:A\odot B\to B(H), a\otimes b\mapsto \varphi(a)\psi (b)$, where $A\odot B$ is the $*$-algebraic tensor product of $A$ and $B$. To extend $\varphi\times \psi$ to a map on $A\otimes_{max}B$ you can do the following: Prove that this map is positive ( for this you need that the ranges of $\varphi$ and $\psi$ commute). Then for every positive linear functional $\eta :B(H)\to \mathbb{C}$ the composition $\eta \circ (\varphi\times \psi):A\odot B\to \mathbb{C}$ extends uniquely to a positive linear functional on $A\otimes_{max}B$. Conclude that there exists a well-defined map $$\Gamma: B(H)^*\to (A\otimes_{max}B)^*,\; \eta \mapsto \eta \circ (\varphi\times \psi),$$where $B(H)^*$ denotes the bounded linear functionals on $B(H)$. This map is bounded (use closed graph theorem for this), thus $\varphi\times \psi$ extends to a bounded map $\varphi\times_{max} \psi:A\otimes_{max}B\to B(H)$. To prove that this map is completely positive, fix an arbitrary $n\in\mathbb{N}$ and prove that $$(\varphi\times_{max} \psi)^{(n)}:M_n(A\otimes_{max}B)\to M_n(B(H)),\; (z_{ij})_{i,j}\mapsto (\varphi\times_{max} \psi (z_{ij}))_{ij}$$ is positive. To do this, note that the product of commuting positive elements is again positive.

Regards