Decomposition of an element as a difference of positive elements in the definition subalgebra of a weight (Takesaki)

Question

I originally asked this on MSE, but did not get an answer there.

Let $M$ be a von Neumann algebra. Let $\varphi: M_+ \to [0, \infty]$ be a weight on $M$. Consider \begin{align*}&\mathfrak{p}_\varphi:= \{x\in M_+: \varphi(x) < \infty\}\\ &\mathfrak{n}_\varphi:= \{x \in M: \varphi(x^*x) < \infty\}\\ & \mathfrak{m}_\varphi:=\mathfrak{n}_\varphi^* \mathfrak{n}_\varphi= \left\{\sum_{j=1}^ny_j^*x_j: x_j ,y_j \in \mathfrak{n}_\varphi\right\}\end{align*}

In Takesaki's second volume, chapter VII, (the proof of) lemma 1.9, the following is claimed:

If $x =x^* \in \mathfrak{m}_\varphi$ and $x= u|x|$ is its polar decomposition, then $|x| \in \mathfrak{m}_\varphi$.

Question: Why is this the case?

Attempt: We have $|x|= x^+ + x^-$, so I tried to show that $x^+, x^- \in \mathfrak{m}_\varphi$. Now, we can write $x=p-q$ where $p,q \in \mathfrak{p}_\varphi= \mathfrak{m}_\varphi^+$ so if we would have $p \ge x^+$ and $q \ge x^-$, we would be done by the hereditary property. I think these inequalities are true when $p$ and $q$ commute, but I don't know if we can choose this decomposition with $pq = qp$.

Thanks in advance for any help or suggestions!

Answer 1

I don't know what Takesaki had in mind for the proof, but what you're asking is incorrect. Here is a counterexample where $\phi$ in the counterexample is a normal faithful semifinite (n.f.s.) weight.

Let $H$ be a separable infinite dimensional Hilbert space, and $M = B(H\oplus H)$. Fix a faithful normal state $\psi$ on $B(H)$ and let $\mathrm{Tr}$ be the standard trace on $B(H)$. Define $\phi ((x_{ij})_{i,j=1,2}) = \mathrm{Tr}(x_{11}) + \psi(x_{22})$. This is clearly a n.f.s. weight.

Fix a positive contraction $a\in B(H)$ which is trace-class but so that $a^{1/2}$ is not trace-class. Let $p = \left( \begin{array}{cc} a & (a-a^2)^{1/2} \\ (a-a^2)^{1/2} & 1-a \end{array} \right)$ and $q= \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)$ which are projections that are both in $\mathfrak m_{\phi}$. Hence $x=p-q \in \mathfrak m_\phi$ is self-adjoint. A straightforward computation gives $x^2 = \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right)$ and thus $|x| = \left( \begin{array}{cc} a^{1/2} & 0 \\ 0 & a^{1/2} \end{array} \right)$. As $a^{1/2}$ is not trace-class we have $\phi(|x|) = \infty$ so $|x| \notin \mathfrak m_\phi$.

Answer 2

This is a reply to the comments. I didn't have enough space there.

Assume $M$ acts on the Hilbert space $H$.

Let $ \mathfrak{p},\mathfrak{m}, \mathfrak{n}$ as in lemma 1.2, chapter VII of Takesaki. We claim that $$\mathfrak{m}= \{y^*x: x,y \in \mathfrak{n}\}.$$

Indeed, consider a generic element $\sum_i y_i^* x_i$ of $\mathfrak{m}$, where $x_i, y_i \in \mathfrak{n}$.

Put $a:= \sum_i x_i^* x_i$ and use lemma 1.6 in chapter VII to write $$x_i = s_i a^{1/2}, \quad s_i [aH]^\perp = 0.$$

Then $$\sum_i y_i^* x_i = \sum_i y_i^* s_i a^{1/2} = \left(\sum_i s_i^* y_i\right)^*a^{1/2}.\quad (*)$$ Clearly $a \in \mathfrak{m}\cap M_+ = \mathfrak{p}$ and $a= (a^{1/2})^{*}a^{1/2}$ so that $a^{1/2} \in \mathfrak{n}$. Finally, note that $$\sum_i s_i^* y_i \in \mathfrak{n}$$ since $\mathfrak{n}$ is a left ideal. Thus $(*)$ shows that every element in $\mathfrak{m}$ decomposes as $$t^*s, \quad s,t \in \mathfrak{n}.$$