Timeline for Flat Principal Connections and Homotopy Groups?
Current License: CC BY-SA 2.5
9 events
when toggle format | what | by | license | comment | |
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Sep 8, 2011 at 21:37 | vote | accept | Abtan Massini | ||
Mar 10, 2011 at 16:40 | answer | added | Micah Miller | timeline score: 1 | |
Mar 9, 2011 at 16:52 | comment | added | André Henriques | @Robot and pAuL: Why don't you guys formulate your answers as answers, as opposed to comments? | |
Mar 8, 2011 at 23:48 | comment | added | Somnath Basu | @ Jos\'{e} - Yes! Thanks for pointing that out. | |
Mar 8, 2011 at 23:34 | comment | added | Paul | To clarify RoBoT's answer a tiny bit. Flat connections on a principal G-bundle P modulo gauge equivalence are in bijection (by taking the holonomy) with a union of some of the components of $\Hom(\pi_1(B),G)/conj$. To get all components you need to consider all (isomorphism classes of) principal G-bunldes over $B$. Also, with some care one can show this identification preserves more structure, i.e. is a homeo, or a (real or complex) analytic isomorphism, etc. | |
Mar 8, 2011 at 22:58 | comment | added | José Figueroa-O'Farrill | @Somnath Basu: You missed the word homotopic in your first sentence. If the two curves are not homotopic, parallel transport can detect it. | |
Mar 8, 2011 at 22:10 | comment | added | Somnath Basu | Given a flat connection for $E\to B$ the parallel transport along two different curves starting and ending at the same points is the same. This is essentially due to Stokes' theorem and the flatness of the connection. Therefore, given a smooth loop $\gamma$ one can parallel transport along $\gamma$ to get an element of $G$. When you quotient out by gauge equivalence on one side, you quotient out by the conjugation action on the right side. | |
Mar 8, 2011 at 19:30 | comment | added | Vít Tuček | To a flat principal connection you can assign its holonomy group. In general, I believe that the answer to your question is an explicit bijective correspondence between moduli space of flat connections modulo gauge equivalence and the set $\mathrm{Hom}(\pi_1(B),G)/G$, where $B$ is the base manifold, $G$ is the principal group and the action of $G$ on homomorphisms is via conjugation on their image. | |
Mar 8, 2011 at 19:18 | history | asked | Abtan Massini | CC BY-SA 2.5 |