Timeline for Can a sum of commutators of selfadjoint bounded operators be a multiple of the identity?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2021 at 15:30 | comment | added | Fabien Besnard | The reason is the formula $[[a,b],x]=[L_a,L_b]x$ where $L_a$ is the Jordan multiplication by $a$. To see that $[[a,b],[c,d]]$ is in $[A,A]$, you first use Jacobi's identity then the above formula. | |
| Jun 11, 2021 at 4:57 | comment | added | Konstantinos Kanakoglou | @Fabien, i am not sure. How is A+[A,A] a lie algebra ? Is there some reference to see how this works? | |
| Jun 10, 2021 at 8:21 | comment | added | Fabien Besnard | Very interesting ! As it happens, they belong to a Jordan algebra A. In that case A+[A,A] is a Lie algebra. Does your answer still apply ? | |
| Jun 10, 2021 at 4:45 | history | answered | Konstantinos Kanakoglou | CC BY-SA 4.0 |