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I´m studying the paper of Fredenhagen. There he said that he would prove that the algebra of local observables under certain conditions is of type III, by showing that all modular operators satisfy $spec(\Delta_{\mathcal{O}})=\mathbb{R}_{+}$.

I'm just beginning to study von Neumann algebras and really don't know if it was a sufficient condition.

The definition I have for type III algebras is that there is only projectors of dimension $0$ or $\infty$. My question is, how are this two assertions related?

Many thanks in advance.

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  • $\begingroup$ I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific? $\endgroup$ – Martin Argerami Dec 17 '18 at 4:17
  • $\begingroup$ Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract. $\endgroup$ – Gabriel Palau Dec 17 '18 at 5:27
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To try to make things simple (they are not; modular theory and Connes work on it are very very far from trivial), if $M$ is semifinite then the modular operator is the identity. That is, at least when talking about factors the modular operator is non-trivial only on factors of type III.

That said, the goal of the paper is not that much to show that the factors are to type III, but to show that the factors are of type III_1 (I'm not entirely sure if the algebras in the paper are factors, but at least in talking about factors I'm more sure I'm saying the right thing). This has to do with A. Connes classification.

Among many many other things, Connes proved that the set $$ \Gamma(M)=(0,\infty)\cap\,\bigcap\{\operatorname{sp}\Delta_\phi:\ \phi\ \text{ is a fns weight on }M\} $$ is a closed multiplicative group of $(0,\infty)$. The only possibilities are

  • $\Gamma(M)=\{1\}$; if $M$ is also type III, we say that $M$ is of type III$_0$

  • $\Gamma(M)=\{\lambda^n:\ n\in\mathbb Z\}$ for some $\lambda\in(0,1)$; we say that $M$ is of type III$_\lambda$

  • $\gamma(M)=(0,\infty)$; we say that $M$ is of type III$_1$.

When $M$ is semifinite, you always have $\Gamma(M)=\{1\}$.

So if you can show that the spectrum of $\Delta_\phi$ is $(0,\infty)$, then $M$ is of type III$_1$.

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