Tomita Takesaki theory and boundeness of $S$ 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 $III$?
 A: If $S$ is bounded then it's easy to see that the adjoint operation is strong operator continuous on bounded sets. This in turn implies that $M$ is finite: If $p \in M$ were a properly infinite projection then there exists a sequence of partial isometries $v_n \in M$ such that $v_n^* v_n = p$, and $v_n v_n^* \leq p$ are decreasing to $0$. We then have $v_n^* \to 0$ strongly, while $v_n$ is isometric on $p \mathcal H$.   
On the other hand, even in the finite case $S$ may be unbounded. Consider, $M$ any infinite dimensional finite factor which we represent standardly $M \subset \mathcal B(L^2(M, \tau))$. Take a sequence $\{ p_n \}_{n \geq 1}$ of pairwise orthogonal projections such that $\tau(p_n) = 2^{-n}$, and set $\xi = c \sum_{n \geq 1} n p_n$, where $c > 0$ is chosen so that $\| \xi \|_2 = 1$. This is cyclic and separating, and if we take $v_n$ a partial isometry such that $v_n v_n^* = p_n$, and $v_n^* v_n \leq p_1$, then we have $\| v_n \xi \|^2_2 = c^2 2^{-n}$, while $\| v_n^* \xi \|^2_2 = c^2 n^2 2^{-n}$.
