Timeline for Norm of a multiplier of a right-ideal in C*-algebras
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 9, 2020 at 0:12 | comment | added | Narutaka OZAWA | @mathbeginner: It's a standard fact that any SOT-closed right ideal $J$ in a von Neumann algebra is of that form; Consider the unit $p$ of the von Neumann algebra $J\cap J^*$. | |
| Jul 8, 2020 at 18:34 | comment | added | mathbeginner | How to show that the $\sigma(A^{**},A^*)$-closure of $I$ is of the form $pA^{**}$ for some projection $p$? | |
| Jun 8, 2020 at 2:11 | comment | added | Narutaka OZAWA | By construction, $qa$ is near to a partial isomery and so any nonzero "segment" of $qa$ has norm at least $\|a\|-\varepsilon$. To be precise, since $aa^*\geq(\|a\|-\varepsilon)^2q$, one gets $\|va\|^2=\|vaa^*v^*\|\geq(\|a\|-\varepsilon)^2\|vv^*\|$. | |
| Jun 7, 2020 at 12:59 | comment | added | AlexE | Sorry to bother you again, but I'm confused by the last step. We have $\|pva\| = \|vqa\|$ and I see that $\|qa\| \ge \|a\|-\varepsilon$. But how do you let the $v$ disappear for this estimate? | |
| Jun 3, 2020 at 13:07 | vote | accept | AlexE | ||
| Jun 3, 2020 at 10:36 | comment | added | Narutaka OZAWA | The existence of $v$ follows from comparison theory of projections, but here's a proof. Since $q$ is dominated by the central cover $z$ of $p$, one has $pA^{**}q\neq\{0\}$. Pick any nonzero $x\in pA^{**}q$ and consider the polar decomposition $x=v|x|$. The nonzero partial isometry $v$ belongs to $pA^{**}q$. | |
| Jun 3, 2020 at 7:36 | comment | added | AlexE | I'm trying to understand your argument. Why do we have $v=pv$? | |
| Jun 3, 2020 at 1:37 | history | edited | Narutaka OZAWA | CC BY-SA 4.0 |
added 253 characters in body
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| Jun 3, 2020 at 1:13 | history | answered | Narutaka OZAWA | CC BY-SA 4.0 |