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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!

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  • $\begingroup$ I accepted an edit, which makes an important correction to the definition of $\mathfrak{m}_\varphi$ $\endgroup$ Feb 16 at 14:14
  • $\begingroup$ @MatthewDaws My definition was also correct. This follows by polar decomposition (see the proof of lemma 1.8 in takesaki, chapter VII). $\endgroup$
    – Andromeda
    Feb 16 at 15:06
  • $\begingroup$ Interesting... Do you know how the proof works? $\endgroup$ Feb 16 at 16:15
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    $\begingroup$ I believe that Takesaki meant to write in Lemma VII.1.8, that "every element in $\mathfrak m_\phi$ is the linear span of elements $y^\ast x$ with $x,y\in \mathfrak n_\phi$, the fact driven by the polarisation identity." $\endgroup$
    – Jamie Gabe
    Feb 16 at 19:46
  • $\begingroup$ @MatthewDaws I am not 100% sure since I'm new to this stuff, but I posted my own proof in an answer below. $\endgroup$
    – Andromeda
    Feb 16 at 20:09

2 Answers 2

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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 $|x| \notin \mathfrak m_\phi$.

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    $\begingroup$ Very nice. $ \ \ $ $\endgroup$ Feb 16 at 19:26
  • $\begingroup$ @Jamie Gabe Thanks for the answer! If you allow me to ask a follow-up question: this question was inspired by the proof of lemma 1.9, chapter VII, volume II, p46. I was trying to justify why $h^{1/2}\in \mathfrak{n}_\varphi$. Do you see why this would be true? If necessary, I can ask another question. $\endgroup$
    – Andromeda
    Feb 16 at 20:13
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    $\begingroup$ In the language of the proof of Lemma 1.9, $x_0 = uh$ is the polar decomposition, so $h=|x_0|$ and if $h^{1/2}\in\mathfrak{n}$ then by definition $h\in\mathfrak{p}$, which is not so, by this counter-example. So I don't really follow the proof of Lemma 1.9... $\endgroup$ Feb 16 at 20:25
  • $\begingroup$ @MatthewDaws It appears that lemma 1.9 is used in the proof of lemma 1.10, which is used to prove important results later, so it is desirable to find a fix for lemma 1.9. I think I will ask another question. $\endgroup$
    – Andromeda
    Feb 16 at 20:35
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    $\begingroup$ But isn't that part of the proof of Lemma VII.1.9 actually easy? With notation from the proof, write $x_0 = y^\ast z$ with $y,z\in \mathfrak n_\phi$. Then $\rho(x_0) = \psi(x_0) = ( a \eta_\phi(z), \eta_\phi(y) ) = \langle a , \theta_\phi(y^\ast z) \rangle \leq \| \theta_\phi(y^\ast z) \| \| a \| \leq \| \theta_\phi(x_0)\|$. The first inequality follows since $a$ and $y^\ast z = x_0$ are self-adjoint. Or am I missing something? $\endgroup$
    – Jamie Gabe
    Feb 16 at 20:38
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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 $$t^*s, \quad s,t \in \mathfrak{n}.$$

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    $\begingroup$ Thanks! This seems good to me, and a new fact to file away... $\endgroup$ Feb 16 at 20:21
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    $\begingroup$ Yes, I take it back by how I assumed he meant polarisation identity instead of polar decomposition. This is a neat trick. $\endgroup$
    – Jamie Gabe
    Feb 16 at 20:22
  • $\begingroup$ Thanks both for the verification. $\endgroup$
    – Andromeda
    Feb 16 at 20:28
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    $\begingroup$ Neat trick! $\ \ $ $\endgroup$ Feb 16 at 20:28

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