Timeline for When do two holonomy maps determine flat bundles that are isomorphic as just bundles (w/o regard to the flat connections)?
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| Mar 7, 2010 at 23:18 | comment | added | Paul | PS: the issue of what path components are in rep spaces $Hom(\pi, G)$ is a dicy one. I assume it is known for surfaces, i.e. the result you attribute to Goldman is not hard, is it? But nothing much general is known. Check out Weinberger's appendix to the Farber-Levine paper (Math Z, 96) for a wierd result along these lines (and the paper for what you can do when the space is connected). A folklore conjecture says that for any 3-manifold $M$ you can join up two $G$ reps of $\pi_1(M)$ by a path after removing a link from $M$. If true, this would help compute Chern-Simons or other invariants. | |
| Mar 7, 2010 at 23:05 | comment | added | Paul | I was thinking of Chern-Weil/Curvature/Chern/Pontryagin classes. Since $BG^d=K(G,1)$ and since every group is a topological group, you can't say anything more general; there are lots of groups $G$ so that $H^*(K(G,1))$ has non-torsion classes. Your second question: $BU(1)=K(Z,2)$ and $K(R/Z,1)=BU(1)^d$ because they have the right homotopy type. Identifying the induced map with the map that takes the $U(1)$ rep to $c_1$ can be seen by working "universally", ie take $X=BU(1)^d$ and look where the identity map goes. Use the fact that $id:BU(1)\to K(Z,2)$ equals $c_1$. | |
| Mar 7, 2010 at 21:10 | comment | added | Ilya Grigoriev | Thank you very much for your answer! It'll take me a while to digest it, but there is a lot of very cool information here. Here are a few more questions: is it clear that the characteristic classes of $BG^d$ are torsion, or do you need to express them via curvature forms (I vaguely know that this is possible, but not how to do it)? How did you calculate $BG^d$, $BG$, and the map in your example (is there an easy way?)? | |
| Mar 4, 2010 at 21:39 | history | answered | Paul | CC BY-SA 2.5 |