Timeline for Does the curvature locally determine the connection?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2023 at 11:54 | vote | accept | F.T. | ||
| Apr 22, 2023 at 19:23 | history | became hot network question | |||
| Apr 22, 2023 at 10:42 | answer | added | Robert Bryant | timeline score: 18 | |
| Apr 22, 2023 at 8:53 | comment | added | F.T. | What do you mean by holonomy map? Is then the conclusion of the main question just false? | |
| Apr 21, 2023 at 23:30 | comment | added | Moishe Kohan | The rest of the proof you simply do not have. You cannot recover a connection (up to gauge equivalence) from the holonomy group. However, you can recover it from the holonomy map. | |
| Apr 21, 2023 at 23:10 | comment | added | F.T. | I see. Is the rest of the idea of my proof correct? | |
| Apr 21, 2023 at 22:51 | comment | added | Moishe Kohan | It is really unclear to me what your question really is. Is it that you have a question about the proof or about the statement of their theorem? If you understood the statement, then you know that they describe (in terms of curvature) not just an abstract Lie algebra but a Lie subalgebra of $gl(n,R)$. Given Lie subalgebra determines a unique connected Lie subgroup of $GL(n,R)$ (by exponentiation). Thus, if you have two connections with the same curvature, then the holonomy subgroups of these two connections are the same. | |
| Apr 21, 2023 at 22:30 | comment | added | F.T. | Let us continue this discussion in chat. | |
| Apr 21, 2023 at 22:15 | comment | added | Moishe Kohan | Sorry, I meant Ambrose--Singer Theorem. The second part where you outline a solution of the problem in the 1st part. | |
| Apr 21, 2023 at 22:11 | comment | added | F.T. | Which part of the question are you referencing to? I am not familiar with such a Cartan's theorem. Apologies | |
| Apr 21, 2023 at 20:35 | comment | added | Moishe Kohan | Of course, but this is irrelevant for the purpose of your question. My suggestion is for you to read the statement of Cartan's theorem on the curvature and holonomy Lie algebra and understand exactly what it says. | |
| Apr 21, 2023 at 20:23 | comment | added | F.T. | Yes, but there might be different subgroups (diffeomorphic to the same Lie group) in GL(n,R), right? | |
| Apr 21, 2023 at 20:12 | comment | added | Moishe Kohan | Here the Lie group really means a Lie subgroup of the general linear group. Ditto for the Lie algebra. | |
| Apr 21, 2023 at 20:07 | comment | added | F.T. | I agree with that but there are a couple of doubts: 1) the Lie group is the same but the representation might be different, 2) a connected Lie group is not determined by its Lie algebra unless it is simply connected | |
| Apr 21, 2023 at 19:47 | comment | added | Moishe Kohan | Since the ball is simply connected, the holonomy groups should be connected and, hence, determined by the Lie algebras. | |
| Apr 21, 2023 at 16:23 | history | asked | F.T. | CC BY-SA 4.0 |