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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 $III$?

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1 Answer 1

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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 $\| v_n^* \xi \|^2_2 = c^2 n^2 2^{-n}$.

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  • $\begingroup$ Thank you for your answer. Let me just ask for one thing: so in the first part, you really want to use the fact that our construction is with respect to state (namely corresponding to cyclic, separating vector)? I'm asking since one can also obtain GNS construction from weight instead of state and then one can obtain $S$ as isometry. And it can be done even if $M$ is infinite (it is enough that it has semifinite, faithful, normal tracial weight). $\endgroup$
    – truebaran
    Jul 7, 2015 at 0:05
  • $\begingroup$ Yes, my answer only deals with a cyclic and separating unit vector $\xi$, and the corresponding state $x \mapsto \langle x \xi, \xi \rangle$. $\endgroup$ Jul 7, 2015 at 3:46

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