1
$\begingroup$

We can classify von Neumann algebras into semi-finite von Neumann algebras and non-semi-finite ones.

I know the fact there are type I,II,III von Neumann algebras.

Do there exist relationships between semi-finiteness and three types?

$\endgroup$
1
  • $\begingroup$ You might want to explain the terms you are using better, if even just with links. $\endgroup$
    – rschwieb
    Oct 9, 2020 at 15:06

1 Answer 1

Reset to default
2
$\begingroup$

Yes. Types I and II are semi-finite, and III are not. Semi-finiteness is equivalent to having a tracial weight, which can never happen in type III.

Never forget that a von Neumann algebra is not necessarily of one of the three types. Rather, it has central summands (or even, central integrands if the algebra is a direct integral) of some of the three types.

$\endgroup$
2
  • $\begingroup$ A VNA is not necessarily of one of the three types. Do there exist concrete examples? Why type III VNAs have no tracial weights? $\endgroup$
    – mathbeginner
    Oct 11, 2020 at 2:41
  • $\begingroup$ Take a type I von Neumann algebra $M$ and a type II von Neumann algebra $N$. Then $M\oplus N$ is neither type I, nor II, nor III. If $A$ is a type III von Neumann algebra, take a non-trivial projection $p$; by halving, there exists a projection $q\leq p$ with $q\simeq p\simeq p-q$. If $\tau$ is a tracial weight with $\tau(p)<\infty$, then $$\tau(q)=\tau(p)=\tau(p-q)=\tau(p)-\tau(q)=0.$$ So the only tracial weights are $\tau=0$ and $\tau=\infty$. $\endgroup$
    – Martin Argerami
    Oct 11, 2020 at 3:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .