Timeline for topological actions
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2011 at 4:28 | comment | added | Kevin Wray | Ok, this makes more sense to me. Plus, I can kinda see why/how they came to picking this term when you note that the choice should obey factorization (as they mention). I would just a logical line of reasoning from A to B, but guess and check also works too. Thanks for your update! | |
| Apr 28, 2011 at 3:24 | history | edited | Aaron Bergman | CC BY-SA 3.0 |
added 709 characters in body
|
| Apr 28, 2011 at 2:15 | comment | added | Kevin Wray | Yes, I meant $S_0 = \frac{1}{n}\int_B \langle F\wedge F\rangle$, which is not correct (what they are saying in the paper). | |
| Apr 27, 2011 at 21:45 | comment | added | Aaron Bergman | The action they wrote down is completely unambiguous. You could have written down a different one like S = 1/n \int F /\ F, but that wouldn't have the correct properties. | |
| Apr 27, 2011 at 19:43 | comment | added | Kevin Wray | Do you mean that adding $\langle \gamma^\ast(\omega), B\rangle$ will give you an action that doesn't depend on bounding $B$, homotopy, etc.? If so, then how does this immediately get rid of the ambiguity? Thanks. | |
| Apr 27, 2011 at 19:22 | comment | added | Aaron Bergman | If it would make you feel better, you can think of this as a definition of the Chern-Simons action. It's sort of the only thing you can really write down that makes sense. | |
| Apr 27, 2011 at 17:31 | comment | added | Kevin Wray | So, you think it was more of a guess and check, rather than starting at point A and arrive at point B? | |
| Apr 27, 2011 at 16:47 | comment | added | Aaron Bergman | As I recall, it's justified post facto by the fact that it works. I wouldn't be surprised if they knew about the differential character and worked backwards, but it's also possible they just worked it out themselves. | |
| Apr 27, 2011 at 14:13 | comment | added | Kevin Wray | Thanks for the explanation! I think I'm fine with everything except for how they arrive at the correction they use to remove the ambiguity (see the update at the top). Yes, I've also been looking at this paper by Freed and Quinn, but they assume finite $G$ which implies the general action reduces to $\langle \gamma_M^\ast(\omega),[M]\rangle$ and they just state this to be the action without explaining why (unless I'm missing something). | |
| Apr 27, 2011 at 11:46 | history | answered | Aaron Bergman | CC BY-SA 3.0 |