Timeline for Curvature and Parallel Transport
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2011 at 16:14 | comment | added | Mircea | math.univ-lyon1.fr/~frabetti/GEO-M2/Codogni-Storch.pdf here is a self-contained proof of Ambrose-Singer, taken by students from Kobayashi-Nomizu and Nomizu. (beware: maybe it's too general to really help you!) | |
| Mar 4, 2010 at 8:09 | comment | added | valeri | The area enters because coordinate vectors may be not unit and not normal, but in the formula R(X,Y) they are - after you "normalize" coordinate vectors, say S and T - you have S\wedge T in the integral equals X\wedge Y \times area. After you prove this for small parallelogram - just note that both parallel operator - holonomy around small loop - and integral are additive - if you divide big area in two (many pieces) the formula for small pieces gives the same formula for big film. | |
| Mar 4, 2010 at 1:22 | comment | added | Paul Siegel | Also, I've combed through both Do Carmo and Milnor's Morse theory, as well as Helgason's books and a few others. I can't find anything. | |
| Mar 4, 2010 at 1:21 | comment | added | Paul Siegel | I agree that this should be basically the right idea; this seems to give the correct argument when the curve in question is a small parallelogram, for example. But I can't get all the details to work out, nor can I see where the area enclosed by the curve will come from. Can you elaborate further? | |
| Mar 3, 2010 at 23:54 | history | answered | valeri | CC BY-SA 2.5 |