# How to prove the following map is a c.c.p map

Here is a quotation in a book "C*-algebrass and Finite-Dimensional Approximations" by Nate and Taka (P122).

Let $A$ be a C*-algebra, $\Gamma$ be a discrete group and the $\alpha$ is an action of $\Gamma$ on $A$. If $F\subset \Gamma$ is a finite set, define $\psi: A\otimes M_{F}(\mathbb{C})\rightarrow C_{c}(\Gamma, A)$ by $$\psi(a\otimes e_{p, q})=\frac{1}{|F|}\alpha_{p}(a)\lambda_{pq^{-1}}.$$

then $\psi$ is a contractive completely positive map.

In fact, in order to prove completely positive map, it suffices to prove that $\psi$ is positive since Exercise 4.1.3 provides a natural commutative diagram:

$$M_{n}(\mathbb{C})\otimes(A\otimes M_{F}(\mathbb{C}))\cong (M_{n}(\mathbb{C})\otimes A)\otimes M_{F}(\mathbb{C})$$

$$~~~~\downarrow~~~~~~~~~~~~~~~~~~~~~~~~~~~~\downarrow$$

$$M_{n}(\mathbb{C})\otimes(A\rtimes_{\alpha,r}\Gamma)\cong M_{n}(\mathbb{C})\otimes(A\rtimes_{\tau\otimes\alpha,r}\Gamma)$$

Exercise 4.1.3. Let $A$ and $B$ be two C*-algebras and $\Gamma$ be a discrete group. If $\alpha:\Gamma\rightarrow \mathrm{Aut}(A)$ is an action and $\tau\otimes\alpha:\Gamma\rightarrow Aut(B\otimes A)$ is defined by $(\tau\otimes\alpha)_{g}=\mathrm{id}_{B}\otimes\alpha_{g}$, then $$(B\otimes A)\rtimes_{\tau\otimes\alpha, r}\Gamma\cong B\otimes(A\rtimes_{\alpha, r}\Gamma).$$

My question:

1. How to verify $\psi$ is contractive?

2. How to use the commutative diagram to "reduce" the completely positive to positive?

1: On a quick look, I don't immediately see how to show that $\psi$ is contractive in general. But note that when $A$ is unital so is $\psi$, which together with positivity makes it contractive. I think this idea can be extended to the non-unital case.
2: When you prove that $\psi$ is positive, you don't need to use that $A$ is $A$, just that it is a C$^*$-algebra. In particular the same proof works for $M_n(\mathbb C)\otimes A$. This gives you the down arrow on the right of the diagram as a positive map. The commutativity of the diagram then tells you that the left down arrow is a positive map, and this is $\psi^{(n)}$. The same argument applies to contractivity.
• The answer 1 is really a good idea, so in the line 6 of P126 of the book, the author say the map $A\otimes M_{F}(\mathbb{C})\rightarrow A\rtimes_{\alpha, r}\Gamma$ by $b\otimes e_{x,y}\rightarrow \alpha_{x}(b)\lambda_{xy^{-1}}$ is a u.c.p. I think it is wrong ,right? Because, we need to divide by $|F|$. And in your answer 2, do you mean that if we replace $M_{n}(\mathbb{C})\otimes A$ with any other C*-algebra, the map is still a positive? – Yan kai Aug 26 '14 at 18:25
• Yes, it is not ucp. And it doesn't have to be: the map $\psi$ in that proof is ucp by the hypothesis $\sum_gT(g)^2=1_A$. Regarding your question about 2: the proof only uses that $A$ is a C$*$-algebra with an action of $\Gamma$, you don't use anything else. – Martin Argerami Aug 26 '14 at 19:34