1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Yan kai Aug 26 '14 at 18:25
  • 1
    $\begingroup$ 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. $\endgroup$ – Martin Argerami Aug 26 '14 at 19:34
  • $\begingroup$ Clarification: "it is not ucp, it is cp". $\endgroup$ – Martin Argerami Aug 26 '14 at 21:22
  • $\begingroup$ Yes, thanks Martin. :P $\endgroup$ – Yan kai Aug 27 '14 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.