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Oct 7, 2023 at 12:21 comment added Narutaka OZAWA Delete the part "$K\subset$".
Oct 7, 2023 at 12:12 comment added Narutaka OZAWA @Matthias Ludewig: I wasn't active recently. In any case, it was not $|\xi|$ but $\|\xi|\otimes\delta_0$. Because $\xi$ and $|\xi|\otimes\delta_0$ are unit vectors in $K\subset L^2(A)\otimes\ell_2$, $\xi=v(|\xi|\otimes\delta_0)$ implies $v^*\xi=|\xi|\otimes\delta_0$ (which also follows from the definition). The answer is edited now.
Oct 7, 2023 at 12:06 history edited Narutaka OZAWA CC BY-SA 4.0
typo fixed
Sep 22, 2023 at 15:15 comment added Matthias Ludewig Hi, going back to understanding the proof of the lemma in more detail, I wondered the following: How do you get the existence of these partial isometries $v$ and $w$? And what exactly are the properties that you would like? Should they satisfy $v^*v = w^*w = 1$?
Jan 17, 2022 at 17:59 comment added Matthias Ludewig Thank you for this great answer!
Jan 17, 2022 at 15:57 vote accept Matthias Ludewig
Jan 14, 2022 at 8:55 history edited Matthias Ludewig CC BY-SA 4.0
edited body
Jan 14, 2022 at 1:49 history answered Narutaka OZAWA CC BY-SA 4.0