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 8, 2010 at 0:10 | comment | added | Dan Ramras | I think Ilya is right; the bundle is not flat over the whole product SxI. To find trivializations of the bundle on SxI, just take a neighborhood U of S, over which the universal cover of S trivial, and consider UxI. The bundle will be trivial on this entire product neighborhood. (Remember here that the bundle associated to a representation is formed by mixing the universal cover of S with the representation, i.e. forming $\tilde{S} \times \mathbb{C}$, where S acts by deck transformations on the left and via given representation on the right.) | |
| Mar 8, 2010 at 0:06 | history | edited | Igor Belegradek | CC BY-SA 2.5 |
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| Mar 8, 2010 at 0:05 | comment | added | Igor Belegradek | @Ilya: mentioning "covering homotopy theorem" is a standard way to refer to a basic result of bundle theory that a fiber bundle over the Sx[0,1] is isomorphic to the product of the restricted bundle over Sx{0} and [0,1]. It is proved e.g. in Steenrod's book "Topology of fiber bundles" You are right that the bundle over SxI is not flat, my mistake. I will edit it. | |
| Mar 7, 2010 at 20:59 | comment | added | Ilya Grigoriev | @Joel: Thanks a lot for that example, it's very nice! | |
| Mar 7, 2010 at 20:59 | comment | added | Ilya Grigoriev | @Igor: Oh, and finally, I think your answer has a small mistake in it: the bundle on $S \times I$ can't be flat (otherwise, I think the bundles would have to be isomorphic as flat bundles). Is that right? | |
| Mar 7, 2010 at 20:58 | comment | added | Ilya Grigoriev | @Igor: Thank you for your answer! I did know that lying in the same component of the representation variety should be sufficient, but I didn't know the proof. In fact, I still don't quite understand how the covering homotopy theorem implies isomorphisms of bundles. If I were trying to prove this, I'd look at locally trivial nbhds of the bundle on $S \times I$, but is there a simpler way with covering homotopies? | |
| Mar 5, 2010 at 21:27 | comment | added | Dan Ramras | There are cases in which representations from different path components yield distinct bundles (i.e. cases in which Igor's sufficient condition is necessary). On a non-orientable surface S (other than $RP^2$) there are exactly two principal U(n) bundles, and the space of representations has exactly two components, one component inducing the trivial bundle and the other inducing the non-trivial bundle. In fact, the space of flat connections on these bundles (even before modding out gauge transformations) is highly connected. This was proven by Ho and Liu using Yang-Mills theory (MR2429971). | |
| Mar 4, 2010 at 18:17 | comment | added | Joel Fine | There's a simple example to see how Igor's condition is far from necessary. Consider flat SU(2)-connections over a 3-manifold. In principal this is a discrete set (mod gauge). Indeed the Casson invariant of M is a "count" of the number of elements in this set. On the other hand, all SU(2)-bundles over a 3-manifold are topologically trivial since pi_1 and pi_2 of SU(2) vanish. (More generally if follows from obstruction theory that given a connected Lie group G with pi_j=0 for j=1,...,n-1 then all principle G-bundles over an n-manifold are trivial.) | |
| Mar 4, 2010 at 16:09 | comment | added | Igor Belegradek | @Emerton, continued: the case of a surface is better understood because for one thing there aren't that many GL(n,R)-bundles over a surface, and they can be all classified in terms of characteristic classes. There are extensive studies on components of the representations variety Hom(\pi_1(S), G), and for certain G's components are completely classified. In those cases, the answer is known. | |
| Mar 4, 2010 at 16:07 | comment | added | Igor Belegradek | @Emerton: you have a bundle of geometric origin, how do you determine its isomorphism type? There is no recipe. Of course, the fact that the bundle has a flat structure implies vanishing of appropriate characteristic classes, and sometimes it is known that the principal bundle itself becomes trivial when pulled back to a finitely-sheeted cover. Beyond that little is known. | |
| Mar 4, 2010 at 15:31 | comment | added | Emerton | Dear Igor, Do you know how close this sufficient condition (lying in the same path component) is to being necessary? | |
| Mar 4, 2010 at 15:14 | history | answered | Igor Belegradek | CC BY-SA 2.5 |