Timeline for Are there principal $G$-bundles whose holonomy group is $G$?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2023 at 5:11 | comment | added | L.F. Cavenaghi | Just adding this in case someone is interested in more "sophisticated examples." Every principal bundle, which is fat, is of such type in the question; see, for instance, the paper of Westein on the subject or the paper "Fatness revisited" due to W. Ziller. | |
| Dec 4, 2016 at 21:46 | comment | added | Robert Bryant | user90041: The connection that you have written down (if I understand you correctly) is the flat connection, and so, of course its holonomy is trivial. However, a generic connection, even on a trivial $G$-bundle (i.e., a product) over a base of dimension $2$ or more and with $G$ connected, will have holonomy $G$, as I show below in my answer. | |
| Dec 4, 2016 at 19:45 | comment | added | user90041 | @RobertBryant Sorry if this is silly, but I do not understand your first comment. If you choose the connection on $B\times \mathbb{R}^n$ define by $(u_b,u_g) \to u_g $ (which is what you perhaps mean by the generic connection on the product bundle), should not the holonomy group be ${e}$ rather than $G$ ? | |
| Jan 10, 2014 at 15:15 | comment | added | Vít Tuček | Thanks. I should have known this, since I read Theorem 11.7 of emis.de/monographs/KSM few months ago. :) | |
| Jan 10, 2014 at 12:28 | comment | added | Robert Bryant | @VítTuček: Sure. For example, any $S^1$-bundle over a compact manifold whose first Chern class is nontorsion has no connection whose holonomy is a proper subgroup of $S^1$. As another example, any $\mathrm{SU}(2)$-bundle $B$ over a compact manifold $M$ whose second Chern class cannot be written as the negative of a square of an element of $H^2(M,\mathbb{Z})$ cannot be given a connection whose holonomy is a proper subgroup of $\mathrm{SU}(2)$. In general, if the structure group of a principal $G$-bundle $B$ cannot be reduced, then the holonomy of any connection on $B$ must be all of $G$. | |
| Jan 10, 2014 at 7:34 | comment | added | Vít Tuček | @RobertBryant: Do you have any nice examples of those $G$-bundles for which all the connections have full holonomy? | |
| Jan 9, 2014 at 21:14 | vote | accept | José Navarro | ||
| Jan 9, 2014 at 19:36 | answer | added | Robert Bryant | timeline score: 16 | |
| Jan 9, 2014 at 18:42 | history | edited | José Navarro | CC BY-SA 3.0 |
corrected the question, to make it more precise
|
| Jan 9, 2014 at 18:39 | comment | added | José Navarro | Yes, that was exactly my question; and that was also the answer I needed, thanks! (I will edit the statement with your correction to make it more precise). | |
| Jan 9, 2014 at 16:44 | comment | added | Robert Bryant | I suppose you mean to ask whether there is a principal $G$-bundle $\pi:P\to B$ for some base $B$ that has a connection $\theta$ whose holonomy is the full group $G$. (You need a connection to define holonomy.) If $G$ is connected, the answer is certainly 'yes' as soon as $B$ has dimension $2$. In fact, in the connected case, you can take $B = \mathbb{R}^2$ and the generic connection on $G\times\mathbb{R^2}$ will have holonomy $G$. If $G$ is not connected, you have to work a little harder. More interesting are those $G$-bundles $P$ for which all the connections on $P$ have holonomy $G$. | |
| Jan 9, 2014 at 16:15 | history | asked | José Navarro | CC BY-SA 3.0 |