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There are two invariants for the type $III$ factor $M$, namely, $S(M)$ and $T(M)$.

When $S(M)=[0, \infty)$, $M$ is a factor of type $III_{1}$.

My question : how to determine whether $M$ is a factor of type $III_{1}$ by using the invariant $T(M)$?

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The invariants $S(M)$ and $T(M)$ of a type III factor $M$ are only partially related.

  • If $M$ is of type III$_1$, then $T(M) = \{0\}$.
  • If $M$ is of type III$_\lambda$ with $\lambda \in (0,1)$, then $T(M) = (2\pi/\log \lambda) \mathbb{Z}$.
  • If $M$ is of type III$_0$, then $T(M)$ ranges over a huge class of subgroups of $\mathbb{R}$, including the two previous cases.

One way to see is, is by considering the flow of weights of $M$. This is an ergodic nonsingular action $\alpha$ of $\mathbb{R}$ on a standard probability space $X$. Any ergodic flow appears as the flow of weights of a type III factor (and of a unique injective type III factor).

Then $T(M)$ is the group of eigenvalues of this flow $\alpha$, namely the set of all $s \in \mathbb{R}$ such that there exists a Borel map $u$ from $X$ to the circle $\mathbb{T}$ such that $u(\alpha_t(x)) = \exp(ist) u(x)$ for all $t \in \mathbb{R}$ and a.e. $x \in X$.

On the other hand, the type is given by:

  • $M$ is of type III$_1$ iff the flow of weights is the trivial action on one point.
  • $M$ is of type III$_\lambda$ with $\lambda \in (0,1)$ iff the flow of weights is periodic with period $|\log \lambda|$.
  • $M$ is of type III$_0$ iff the flow of weights is properly ergodic.
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  • $\begingroup$ In Connes' paper, $T(M)$ is defined as following :$T(M)=\{t\in \Bbb R, \fbox {there exists a weight such that} \sigma_{t}^{\varphi}=1\}$. How to prove that $T(M)$ is the group of eigenvalues of the above flow $\alpha$? Would you mind recommending some papers or books on the the proof of conclusions you mentioned? Thanks so much , Pro Vaes. $\endgroup$ Commented Jun 15, 2021 at 8:44
  • $\begingroup$ You can find this in Definitions 1.3 and 1.5, and Theorem 1.6, of Chapter XII of "Theory of Operator Algebras II" by Masamichi Takesaki. $\endgroup$ Commented Jun 15, 2021 at 18:07

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