3
$\begingroup$

Maybe a naive question:

Right now I am reading the paper ``Completely positive maps of order zero'' written by Wilhelm Winter and Joachim Zacharias. I do not quite understand the last corollary when they proved the induced map between Cuntz semigroups of a c.p.c order zero map between algebras preserves addition. Why we need the map is order zero?

To be simple, suppose that $A, B$ are $C^*$-algebras. Let $a,b\in A_+$ and $\varphi: A\rightarrow B$ be a c.p.c order zero map. They claim that if $a,b\in A$ are orthogonal, then so are $\varphi(a), \varphi(b)\in B$, whence $$\varphi(a\oplus b)=\varphi(a)\oplus \varphi(b)$$.

The first $\varphi$ is interpreted to be the amplification of $\varphi$, i.e. $\varphi\otimes id_{M_2}$ I believe. However, I guess that the equation above hold for every c.p.c map. Because $a\oplus b= a\otimes e_{11}+b\otimes e_{22}$ and $\varphi\otimes id_{M_2}(a\otimes e_{11}+b\otimes e_{22})= \varphi(a)\otimes e_{11}+ \varphi(b)\otimes e_{22}=\varphi(a)\oplus \varphi(b)$.

I believe I must miss something. But I cannot find where I was wrong.

Thank you for all helps!

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ You are right. The equation you write holds for every c.p.c. map. At this point, the assumption of "order zero" is not needed. However, at the part of the paper by Winter-Zacharias that you are reading, it is shown that a c.p.c. order zero $\varphi\colon A\to B$ induces a map $\mathrm{Cu}(\varphi)\colon\mathrm{Cu}(A)\to\mathrm{Cu}(B)$ between the respective Cuntz semigroups. A general c.p.c. map need not preserves Cuntz (sub)equivalence, which is why the assumption of order zero is needed. $\endgroup$
    – Hannes Thiel
    Commented Mar 8, 2018 at 13:53
  • $\begingroup$ @HannesThiel Thanks a lot for your clarification! $\endgroup$
    – Targaryen
    Commented Mar 9, 2018 at 16:32

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .