# Completely bounded map and minimal tensor products

Theorem 3.5.2. Let $\phi: A\rightarrow C$ and $\psi: B\rightarrow D$ ($A, B, C, D$ are C*-algebras) be c.p.(completely positive) maps. Then the algebraic tensor product map $$\phi\odot\psi: A\odot B\rightarrow C\odot D$$ extends to a completely positive map. (hence continuous) map on minimal tensor products. Moreover, letting $\phi \otimes \psi: A\otimes B\rightarrow C\otimes D$ we have $||\phi\otimes\psi||=||\phi||~||\psi||$.

Proof. Assume $C\subset B(H)$ and $D\subset B(K)$. Let $\pi_{A}: A\rightarrow B(\bar{H})$, $\pi_{B}: B\rightarrow B(\bar{K})$ be Stinespring dilations of $\phi$ and $\psi$, respectively, and $V_{A}: H\rightarrow \bar{H}$, $V_{B}: K\rightarrow \bar{K}$ the associated bounded linear operators. Then, there is a natural *-homomorphism $\pi_{A}\otimes \pi_{B}: A\otimes B\rightarrow B(\bar{H}\otimes \bar{K})$. Hence we may define $\phi\otimes\psi: A\otimes B\rightarrow C\otimes D$ by the formula $$\phi\otimes \psi(x)=(V_{A}\otimes B_{B})^{*}\pi_{A}\otimes \pi_{B}(x)(V_{A}\otimes V_{B}).$$ Note that $\phi\otimes \psi(a\otimes b)=\phi(a)\otimes\psi(b)$. Hence $\phi \otimes \psi$ is a c.p. which takes values in $C\otimes D$.

I suppose there is nothing difficult in the proof of the Theorem above. But the next Remark confused me.

Remark 3.5.4. It is not hard to extend the previous result to minimal tensor products and c.b. (completely bounded) maps.

My question is how to extend the theorem to c.b. map?

## 1 Answer

You do the same proof but using Wittstock instead of Stinespring.

• But, what is Wittstock? In which text is this Theorem found? – Yan kai May 27 '14 at 15:21
• Brown-Ozawa, page 451. – Martin Argerami May 27 '14 at 15:25
• Yes, thanks, Martin. – Yan kai May 27 '14 at 15:37
• However, the Wittstock' Theorem is for c.c. map not the c.b. map in the Remark. – Yan kai May 27 '14 at 18:30
• But $\phi/\|\phi\|_{\rm cb}$ is completely contractive and you can apply the theorem. So Wittstock works for any completely bounded map, at the cost of allowing one of $V,W$ to not be an isometry. And you don't need isometries in the proof above. – Martin Argerami May 27 '14 at 21:55