2
$\begingroup$

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!

$\endgroup$
6
  • $\begingroup$ I accepted an edit, which makes an important correction to the definition of $\mathfrak{m}_\varphi$ $\endgroup$ Feb 16, 2022 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, 2022 at 15:06
  • $\begingroup$ Interesting... Do you know how the proof works? $\endgroup$ Feb 16, 2022 at 16:15
  • 1
    $\begingroup$ 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, 2022 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, 2022 at 20:09

2 Answers 2

4
$\begingroup$

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

$\endgroup$
9
  • 1
    $\begingroup$ Very nice. $ \ \ $ $\endgroup$ Feb 16, 2022 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, 2022 at 20:13
  • 2
    $\begingroup$ 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, 2022 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, 2022 at 20:35
  • 2
    $\begingroup$ 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, 2022 at 20:38
2
$\begingroup$

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 $$t^*s, \quad s,t \in \mathfrak{n}.$$

$\endgroup$
4
  • 1
    $\begingroup$ Thanks! This seems good to me, and a new fact to file away... $\endgroup$ Feb 16, 2022 at 20:21
  • 1
    $\begingroup$ 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, 2022 at 20:22
  • $\begingroup$ Thanks both for the verification. $\endgroup$
    – Andromeda
    Feb 16, 2022 at 20:28
  • 1
    $\begingroup$ Neat trick! $\ \ $ $\endgroup$ Feb 16, 2022 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.