Timeline for Dirac's Original Operator and the Hodge--Dirac Operator
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 25, 2011 at 14:11 | vote | accept | John McCarthy | ||
| Mar 23, 2011 at 13:47 | comment | added | user1504 | On a smooth function? i's domain is sections of the Clifford bundle. I think maybe you should have a look in Michelson-Lawson. | |
| Mar 23, 2011 at 13:19 | comment | added | John McCarthy | What's the explicit action of $i$ on a smooth function? | |
| Mar 23, 2011 at 12:44 | comment | added | user1504 | The identification -- call it $i$ -- maps sections of one bundle to sections of the other. So the image $i(\psi)$ of a section $\psi$ of the Clifford bundle is a differential form. This form is, in general, of mixed degree, a section of $\Omega^* = \oplus_p \Omega^p$. | |
| Mar 23, 2011 at 12:26 | comment | added | John McCarthy | I see that identifying the Clifford and exterior bundles allows us to view the two operators as operating on the same space. However, I don't see that they are equal: $D$ sends zero forms to zero forms, while $d+d^{\dagger}$ sends zero forms to one forms; how can they be equal? | |
| Mar 22, 2011 at 23:16 | history | answered | Johannes Ebert | CC BY-SA 2.5 |