Here is a proof of (continuity of tensor product maps) theorem:

Theorem 3.5.2. Let $\phi: A\rightarrow C$ and $\psi: B\rightarrow D$ ($A, B, C, D$ are C*-algebras) be c.p. 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 maximal tensor products. Moreover, letting $\phi \otimes_{max} \psi: A\otimes_{max} B\rightarrow C\otimes_{max} D$ we have $||\phi\otimes_{max}\psi||=||\phi||~||\psi||$.

Proof. let's tackle the case that $B=D$ and $\psi=id_{B}$. Fix a faithful representation $C\otimes_{max} B\subset B(H)$. Then, taking the restrictions, we assume that $C\subset B(H)$ and $B\subset B(H)$ commute (and generate $C\otimes_{max} B$) thus allowing us to regard $\phi$ as a c.p. map into $B(H)$ with $B\subset \phi(A)'$.

Applying Stinespring to $\phi$ - also lifting $B$ with the commutant $\phi(A)'$ (Proposition 1.5.6) - we get a *-representation of $A\otimes_{max} B$ which we can cut to recover the original map $\phi\odot id_{B}: A\odot B\rightarrow C\otimes_{max} B\subset B(H)$.

Since an arbitrary map $\phi\odot \psi: A\odot B\rightarrow C\odot D$ can be decomposed as $(\phi\odot id_{D})\circ (id_{A}\odot \psi)$, the proof is complete.

Proposition 1.5.6. Let ($\pi, \bar{H}$, V) be the minimal Stinespring dilation of a contractive completely map $\phi: A\rightarrow B(H)$. Then, there exists a *-homomorphism $$\rho: \phi(A)' \rightarrow \pi(A)'\subset B(\bar{H})$$.

I have several questions on the proof:

  1. How apply Proposition 1.5.6 to get a *-representation of $A\otimes_{max} B?$ and how to comprehend this *-representation can cut to recover the original map $\phi\odot id_{B}$?

  2. In the end of the proof, how to use the decomposition to get the result?

  3. How to verify the norm-equation?


1) Let $(\pi,K,V)$ be a minimal Stinespring decomposition for $\phi$. Prop. 1.5.6 gives you a representation $\rho:\phi(A)'\to\pi(A)'\subset B(K)$. Then $a\otimes b\mapsto \pi(a)\rho(b)$ defines a representation $\alpha:A\odot B\to B(K)$, which is contractive by definition of the max norm, and so it extends to $A\otimes_\max B$. Then $$ \beta:a\otimes b\longmapsto V^*\alpha(a\otimes b)V $$ extends to a completely positive map $A\otimes_\max B\to B(H)$. And by the Proposition, $$ \beta(a\otimes b)=V^*\pi(a)\rho(b)V=\phi(a)b=\phi\odot\text{id}_B(a\otimes b)\in C\odot B, $$ and $\beta(A\otimes_\max B)\subset C\otimes_\max B$ by continuity of $\beta$ (being completely positive).

2) You write $\phi\otimes\psi=(\phi\odot\text{id}_D)(\text{id}_A\odot \psi)$. Applying 1) to both maps in the brackets, we get $$ A\otimes_\max B\xrightarrow{\text{id}_A\otimes_\max\psi} A\otimes_\max D\xrightarrow{\phi\otimes_\max\text{id}_D} C\otimes_\max D. $$

3) Since on elementary tensors $\|\phi\otimes_\max\psi(a\otimes b)\|=\|\phi(a)\otimes\psi(b)\|=\|\phi(a)\|\,\|\psi(b)\|$, we get that $\|\phi\otimes_\max\psi\|\geq\|\phi\|\,\|\psi\|$. On the other hand, $$ \|\phi\otimes_\max\psi\|=\|(\phi\odot\text{id}_D)(\text{id}_A\odot \psi)\|\leq\|\phi\odot\text{id}_D\|\,\|\text{id}_A\odot \psi\|=\|\phi\|\,\|\psi\|. $$

  • $\begingroup$ In line 7 of your first answer, you say $\phi(a)b=\phi \odot id_{B}(a\otimes b)$. But I think it should be $\phi(a)\otimes b=\phi \odot id_{B}(a\otimes b)$. So there may be a little gap in your answer. $\endgroup$ – Yan kai May 27 '14 at 18:17
  • $\begingroup$ No. As the proof says, you are assuming that both $C\subset B(H)$ and $B\subset B(H)$. So no tensor product, it is just $\phi(a)b$. $\endgroup$ – Martin Argerami May 27 '14 at 21:51
  • $\begingroup$ Should $\phi(a)\otimes b$ contain in $B(H)\odot B(H)$? What do you mean by "there is no tensor product"? $\endgroup$ – Yan kai May 28 '14 at 0:30
  • $\begingroup$ Just read the proof: "Then, taking the restrictions, we assume that $C\subset B(H)$ and $B\subset B(H)$ commute (and generate $C\otimes_\max B$) thus allowing us to regard $\phi$ as a c.p. map into $B(H)$ with $B\subset \phi(A)'$". Since $B\subset \phi(A)'$, it means that $\phi(a)b=b\phi(a)$ for all $a\in A$, $b\in B$. $\endgroup$ – Martin Argerami May 28 '14 at 1:25
  • $\begingroup$ Yes, but I think $\phi(a) b= \phi \times 1_{B}(a\otimes b)$, not $\phi\odot 1_{B}(a\otimes b)$. $\endgroup$ – Yan kai May 28 '14 at 4:18

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