Timeline for Why does a principal G-bundle with a discrete structure group G have a unique flat connection?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2021 at 8:56 | comment | added | NDewolf | So if I understand correctly, a general $G$-bundle can admit many nonequivalent flat connections, but by passing to a specific $G^\delta$-reduction (where $G^\delta$ is just $G$ with the discrete topology) a single flat connection is chosen? In other words the equivalence classes of $G^\delta$-reductions are in bijection with the equivalence classes of flat connections? | |
| Nov 21, 2019 at 10:34 | comment | added | Ben McKay | In physicists' notation, a connection expressed in local coordinates is represented by some field $A$, called a gauge field, valued in the Lie algebra. If the Lie algebra is zero, then $A=0$, unique. | |
| Nov 21, 2019 at 6:39 | vote | accept | pyroscepter | ||
| Nov 21, 2019 at 6:39 | vote | accept | pyroscepter | ||
| Nov 21, 2019 at 6:39 | |||||
| Nov 21, 2019 at 6:39 | vote | accept | pyroscepter | ||
| Nov 21, 2019 at 6:39 | |||||
| Nov 21, 2019 at 5:26 | comment | added | user17945 | Within any local trivialisation of the bundle, a curve in the base manifold will lift to a curve in the trivialisation with a constant $g$ value. However patching together local trivialisations along a closed curve in the base that is homotopically non-trivial can produce a non-trivial holonomy - think for example of the bundle $p:U(1)\to U(1)$ given by $z\mapsto z^2$, considered as a $\mathbb{Z}_2$-bundle over $U(1)$. | |
| Nov 21, 2019 at 5:19 | answer | added | user17945 | timeline score: 4 | |
| Nov 21, 2019 at 5:13 | comment | added | pyroscepter | @WillSawin I'm stuck at the difficulty arising from discreteness, let alone how to prove the statement itself! How does one even define a connection on a discrete G-bundle, if the underlying manifold is smooth? If a change in the position on the manifold results in a discrete change in g, then what about if I change the position by half as much? there's no notion of smoothness on the fiber. Perhaps the manifold being smooth is a faulty assumption? | |
| Nov 21, 2019 at 4:55 | review | Close votes | |||
| Dec 3, 2019 at 3:05 | |||||
| Nov 21, 2019 at 3:34 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
added 47 characters in body; edited tags; edited title
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| Nov 21, 2019 at 1:06 | answer | added | Tsemo Aristide | timeline score: 4 | |
| Nov 21, 2019 at 0:47 | comment | added | Will Sawin | What have you thought about so far? Did you pick a discrete group and draw a picture? | |
| Nov 21, 2019 at 0:34 | history | asked | pyroscepter | CC BY-SA 4.0 |