2
$\begingroup$

I have actually a basic quastion about maps between von Neumann algebras.

If I have a map $f:N \to M$ being $N$ and $M$ von Neumann algebras. when this map is considered: completely positive, normal and unital?

I suppose that $f$ is unital if $f(1)=1$, and suppose that $f$ is completely positive if $f$ maps any operator with positive spectrum to other with positive spectrum too. But I really don't know.

Many thanks in advance. And apologise for this basic question.

$\endgroup$
2
$\begingroup$

You are right about unital.

Normal means that $f$ respect suprema of bounded nets of selfadjoints. That is, if $\{a_j\}\subset M$ and $a=\sup a_j$, then $f(a)=\sup f(a_j)$. This is the same as saying that $f$ is sot-sot continuous.

Positive means that $f(x)\geq0$ if $x\geq0$. Note that $x\geq0$ not only means that $\sigma(x)\subset[0,\infty)$ but also that $x$ is selfadjoint.

Completely positive means that $f^{(n)}=f\otimes I_n:M\otimes M_n(\mathbb C)\to N\otimes M_n(\mathbb C)$ is positive for all $n\in\mathbb N$. That is if $X\in M_n(M)$ is positive, then $[f(X_{kj})]_{k,j}\in M_n(N)$ is positive.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks @Martin Argerami, Your answer was very usfull, but I have a new question. The point is that I have never heard the consept of supremum in the context of operators, so I was a bit shocked when I read the definition of normal, but then you say that that is equivalent to sot-sot continuity, and I can understand that. However, I stayed with the doubt about the sup of operators. $\endgroup$ – Gabriel Palau Dec 29 '18 at 3:52
  • $\begingroup$ Supremum="least upper bound". If you have an order, you have notion of supremum. $\endgroup$ – Martin Argerami Dec 29 '18 at 3:53
  • $\begingroup$ yes, you are right. Thanks again $\endgroup$ – Gabriel Palau Dec 29 '18 at 4:01

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.