Timeline for Are there always flat connections?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2023 at 6:57 | vote | accept | CommunityBot | ||
| Sep 28, 2023 at 0:26 | comment | added | Igor Khavkine | @RobertBryant I may be out of my depth here, but it seems to me that I.B.'s answer also works if you just replace "tangent bundle" by "vector bundle" (which is how I had read it). Anyway, this type of question seems to come up often enough that it might be worth recording an answer of sufficient generality to be useful in the future. I've opened a question to give someone an opportunity to do that. | |
| Sep 28, 2023 at 0:13 | answer | added | Robert Bryant | timeline score: 4 | |
| Sep 27, 2023 at 11:09 | comment | added | Robert Bryant | @IgorKhavkine: However, that answer has something wrong with it. The statement 'If a vector bundle admits a flat connection, then the rational Pontryagin classes of the tangent bundle vanish..." is just false. I suspect that I.B. meant to start with "If a tangent bundle admits a flat connection...", which would have been a true statement. Also, the final statement that 'most vector bundles don't admit a flat connection' is true, but that says nothing about the vector bundles over particular manifolds, such as $G/\Gamma$. | |
| Sep 27, 2023 at 9:11 | comment | added | Igor Khavkine | @RobertBryant That older question was different, of course. But the linked answer was phrased to apply to any vector bundle (along the same lines as the comments by Ben McKay and yourself, with a couple of different references). | |
| Sep 27, 2023 at 8:59 | comment | added | Robert Bryant | @IgorKhavkine: The OP has not asked that $E$ be the tangent bundle of $G/\Gamma$. Indeed, the tangent bundle of $G/\Gamma$ is trivial, so it does have a flat connection. | |
| Sep 26, 2023 at 23:58 | comment | added | Igor Khavkine | I think this old answer is general enough to addresses also this question. Is it not? | |
| Sep 26, 2023 at 14:40 | comment | added | Ben McKay | Chern classes are computed in terms of curvature of any connection. A flat connection has zero Chern classes. See Chern, Vector bundles with a connection, Global Differential Geometry. | |
| Sep 26, 2023 at 14:35 | comment | added | user473423 | Can you give me a reference as to why non-vanishing of the chern class implies non-flatness? | |
| Sep 26, 2023 at 14:05 | comment | added | Robert Bryant | You may be missing some hypotheses. Otherwise, there are very simple examples provided by smooth complex line bundles with nontrivial first Chern class over $\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$. | |
| Sep 26, 2023 at 11:58 | history | asked | user473423 | CC BY-SA 4.0 |