Timeline for Various definitions of Connections on bundles
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2010 at 22:41 | comment | added | David MJC | In other words, the omega's in the second calculation are minus the omega's in the first. Hence the sign difference in the transition formulae. | |
| Dec 9, 2010 at 22:36 | comment | added | Deane Yang | I encourage you to read the second paragraph of my answer (which I just added). | |
| Dec 9, 2010 at 22:31 | vote | accept | Rex | ||
| Dec 9, 2010 at 22:31 | |||||
| Dec 9, 2010 at 22:29 | comment | added | Rex | If both formulas are correct, then we would have the following: Start with a vector bundle $E\to X$. Then using trivialisations, we get connection forms $\omega^U$ which satisfy $\omega^V=g\omega^Ug^{-1}-dgg^{-1}$. Then $-\omega^U$ satisfy $\omega^V=g\omega^Ug^{-1}+dgg^{-1}$ which is exactly what is needed to define a connection on the principal $GL_n$ bundle. Conversely, we can go backwards also. This seems to suggest that for a principle bundle we should define curvature as $d\omega-\omega\wedge\omega$, as opposed to $d\omega+\omega\wedge\omega$ in the case of a vector bundle. | |
| Dec 9, 2010 at 22:02 | history | edited | Deane Yang | CC BY-SA 2.5 |
added 502 characters in body
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| Dec 9, 2010 at 21:58 | comment | added | Deane Yang | Both formulas appear correct to me. | |
| Dec 9, 2010 at 21:51 | comment | added | Rex | @t3suji: I am not sure I understand your question, but I think the answer is no. I am getting $\-\omega^U$. | |
| Dec 9, 2010 at 21:36 | comment | added | Rex | @Deane Yang : The first calculation is correct. It's the second one I am worried about, the one which involves the principal bundle. Are you saying that the second one is fine too? | |
| Dec 9, 2010 at 21:27 | comment | added | Deane Yang | t3suji (and anyone else), you're more than welcome to expand on my answer or even rewrite it differently as a separate answer. I'm being a little coy, because I prefer students to struggle and figure these things out on their own. But if someone else wants to explain what's going on, they're welcome to do so. | |
| Dec 9, 2010 at 21:25 | comment | added | Deane Yang | Your calculation is correct. Please focus on an actual section of $E$ and expand it using two different frames. You'll see why everything is actually OK. | |
| Dec 9, 2010 at 20:08 | comment | added | t3suji | Let me try to extend Deane Yang's answer (hope it is OK): Suppose you are looking at the same connection and the same trivialization in these two different ways. Do you get the same form $\omega^U$? | |
| Dec 9, 2010 at 19:44 | comment | added | Rex | My problem is on the one hand I get $\omega^V=g\omega^Ug^{-1}-dgg^{-1}$ and on the other hand I get $\omega^V=g\omega^Ug^{-1}+dgg^{-1}$. I have checked the calculation several times and so I think I am goofing up with some definition?? | |
| Dec 9, 2010 at 19:06 | history | answered | Deane Yang | CC BY-SA 2.5 |