3
$\begingroup$

For $1\leq i\leq n$, let $\psi_i$ be a faithful state on the C$^*$-algebra $A_i$ and $\phi_i$ be a faithful state on the C$^*$-algebra $B_i$. Let $(A,\psi) = *_{i=1}^n (A_i,\psi_i)$ and $(B, \phi) = *_{i=1}^n (B_i,\phi_i)$ be the reduced free products.

Suppose we have unital completely positive maps $\theta_i : A_i \rightarrow B_i$ such that $\phi_i\circ\theta_i = \psi_i$. Choda and Blanchard-Dykema proved that there exists a ucp map $\theta : A \rightarrow B$ such that $\theta|_{A_i} = \theta_i$ and $\phi\circ\theta = \psi$. Note that Mlotkowski proved a similar result to this following the work of Bozekjo-Leinert-Speicher.

Is this $\theta$ unique? Namely, if there exists a ucp map $\varphi : A \rightarrow B$ such that $\varphi|_{A_i} = \theta_i$ and $\phi\circ\varphi = \psi$ do we have that $\varphi = \theta$?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ This isn't true for states. Let $(A_i,\psi_i) = (\mathrm{C}^*{\mathbb Z}/2{\mathbb Z}, \mathrm{tr})$ and $n=2$. Then there are states other than $\psi$ which restrict to $\psi_i$ on $A_i$. $\endgroup$
    – Narutaka OZAWA
    Commented Jan 12, 2016 at 5:37
  • $\begingroup$ @NarutakaOZAWA Wonderful! Can you please spell this out more in an answer as it is still not immediately obvious to me. $\endgroup$
    – Chris Ramsey
    Commented Jan 12, 2016 at 13:08
  • $\begingroup$ Sorry! I overlooked the condition $\phi\circ\varphi=\psi$. $\endgroup$
    – Narutaka OZAWA
    Commented Jan 13, 2016 at 2:44

0

Reset to default

You must log in to answer this question.