Timeline for Can a sum of commutators of selfadjoint bounded operators be a multiple of the identity?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2021 at 7:05 | answer | added | Mikael de la Salle | timeline score: 2 | |
| Jun 10, 2021 at 4:45 | answer | added | Konstantinos Kanakoglou | timeline score: 1 | |
| Jun 9, 2021 at 14:55 | comment | added | Liviu Nicolaescu | Oops! You're right. | |
| Jun 9, 2021 at 11:59 | comment | added | Fabien Besnard | @LiviuNicolaescu Yes, the sum will be anti-selfadjoint, so it could be equal to i times the identity. | |
| Jun 9, 2021 at 9:38 | comment | added | Liviu Nicolaescu | No. $[a,b]^*=-[a,b]$ for $a,b$ bounded selfadjoint. | |
| Jun 9, 2021 at 9:17 | history | asked | Fabien Besnard | CC BY-SA 4.0 |