Timeline for Stabilizer groups of Yang-Mills connections
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2022 at 15:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 22, 2021 at 15:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 25, 2021 at 12:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 27, 2021 at 12:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 24, 2021 at 21:45 | comment | added | Tobias Diez | Why do you believe that they are equal? I outlined a proof of my initial conjecture below, but maybe I've overlooked something. I would appreciate if you could have a look, thanks! | |
| Feb 24, 2021 at 21:44 | answer | added | Tobias Diez | timeline score: 1 | |
| Feb 19, 2021 at 15:23 | comment | added | Nicolast | I am not sure I understand your notations but I think $\mathrm{Gauge}(P^{\mathbb C})_A = (\mathrm{Gauge}(P)_A)^{\mathbb C}$. The case of $G=\mathrm{U}(n)$ can be worked out by decomposing your vector bundle into a direct sum of stable bundles, but probably also follows from some abstract nonsense in representation theory. | |
| Feb 18, 2021 at 14:21 | history | edited | Tobias Diez | CC BY-SA 4.0 |
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| Feb 18, 2021 at 14:21 | comment | added | Tobias Diez | Thanks for pointing this out. The bijection is of course between connections on $P$ and holomorphic structures on $P^c$. | |
| Feb 16, 2021 at 10:30 | comment | added | Nicolast | It is not true that a connection on $P$ defines a holomorphic connection on $P_{\mathbb C}$. Take for instance $P= \mathrm{U}(n)$ and $P_{\mathbb C} =\mathrm{GL}(n,\mathbb C)$. Then you are looking at a complex Hermitian vector bundle. A Hermitian connection defines a $\bar \partial$-operator, hence a holomorphic structure, but not a holomorphic connection (which would be flat on a Riemann surface). | |
| Feb 8, 2021 at 12:22 | history | asked | Tobias Diez | CC BY-SA 4.0 |