Let $M$ be a von Neumann algebra, $\xi$-separating and cyclic vector for $M$. Let $S$ be antilinear operator acting as $x \xi \mapsto x^* \xi$ where $x \in M$. Then one can show that $S$ is closable and one can use the polar decomposition $S=J\Delta^{\frac12}$ where $\Delta=S^*S$ and $J$ is antilinear isometry. Suppose that $S$ is bounded:
Q1 Does it follow that in this case $M$ can not contain type $III$ part?
Let me say some more about this: it is known that $M$ does not contain type $III$ part (in other words, $M$ is semifinite) iff $M$ admits tracial, faithfull, normal weight). One can perform the construction of Tomita-Takesaki operator with respect to this weight and the weight is tracial iff $S$ is isometry (so iff $\Delta=I$, the identity operator). There is the following invariant $S(M)=\bigcap spec \Delta^{\varphi}$ where $\varphi$ runs over all weights (normal, faithful, semifinte). It is true, that in case of $M$ being a factor, $S(M) \setminus \{0\}$ is (closed) multiplicative subgroup of $\mathbb{R}_+$. Then $S(M)$ could be one of three sets:
- $[0,\infty)$ (type $III_1$
- $\{0\} \cup \{\lambda ^n; n \in \mathbb{Z}\}$ where $\lambda \in (0,1)$ (type $III_{\lambda}$)
- $\{0,1\}$ (type $III_0$)
If $S$ happens to be bounded then two first cases are impossible. However I don't know how to exclude the last case (if it is possible).
On the other side, suppose that for some weight $\varphi$ (normal, semifinite, faithful) $\Delta_{\varphi}$ (or $S$) happens to be unbounded.
Q2 Does it follow that our algebra is of tyle $III$?