Timeline for Notion of Torsors
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2018 at 16:03 | vote | accept | Praphulla Koushik | ||
| Mar 26, 2018 at 1:04 | comment | added | Praphulla Koushik | @Qfwfq I will try to work out the details.. | |
| Mar 26, 2018 at 1:03 | comment | added | Praphulla Koushik | @ToddTrimble can you take some time and make it as an answer (add more details if you wish) | |
| Mar 25, 2018 at 21:37 | comment | added | Todd Trimble | In other words $\hat{G}$ is a group object in the slice category, and we can speak of actions of groups on other objects in this category. That's what is going on here. | |
| Mar 25, 2018 at 20:58 | comment | added | Todd Trimble | I take exception to the imputation of "sloppiness". Given projection maps $\pi: G \to X$ and $\pi': P \to X$, the notation $G \times_X P$ means a fiber product: $\{(g, p) \in G \times P: \pi(g) = \pi'(p)\}$. There is then an obvious projection $G \times_X P \to X$, whose fiber over $x \in X$ is precisely $G_x \times P_x$. In more categorical language, the fiber product is the cartesian product of the two objects $\pi: G \to X$ and $\pi': P \to X$ in the slice category. So this really does give an action $\hat{G} \times \hat{P} \to \hat{P}$ where $\hat{G}, \hat{P}$ stand for these objects. | |
| Mar 25, 2018 at 20:45 | comment | added | Qfwfq | @cello: if you work out the details, the isomorphism condition should be equivalent to saying that each action $G_x \curvearrowright P_x$ is free. | |
| Mar 25, 2018 at 18:55 | comment | added | Praphulla Koushik | When you have time, can you say few words about the condition for the map $G\times_XP\rightarrow P\times_XP$ is isomorphic. | |
| Mar 25, 2018 at 14:01 | comment | added | Daniel McLaury | Yeah, once people get used to relativizing things it's clear enough to them what things ought to be so they can get a little sloppy with their definitions. I agree it can make things infuriating to read for the uninitiated, though. | |
| Mar 25, 2018 at 13:43 | comment | added | Praphulla Koushik | Even I do not see any condition that says $G_x\times P_x$ maps to $P_x$. | |
| Mar 25, 2018 at 13:30 | comment | added | Daniel McLaury | At the level of sets, $G \times_X P$ is the disjoint unions of the sets $G_x \times P_x$. I don't know if we explicitly said it above but we need to insist that each $G_x \times P_x$ maps into $P_x$. Then you have a bunch of individual actions of $G_x$ on $P_x$. | |
| Mar 25, 2018 at 13:25 | comment | added | Praphulla Koushik | IAs $G$ is not itself a group, it can not act on things. Agree... I do not understand how fiber product gives individual actions $G_x$ on $P_x$. Can you explain a little more on this. I have seen Baez write up. It is useful. | |
| Mar 25, 2018 at 13:22 | comment | added | Praphulla Koushik | “Presumably there's also a condition that the groups "vary continuously" in some sense, just like a vector bundle isn't any disjoint union of arbitrary vector spaces indexed by elements of a space”... I was assuming this as well. I did not just write.. | |
| Mar 25, 2018 at 13:21 | history | edited | Daniel McLaury | CC BY-SA 3.0 |
added 151 characters in body
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| Mar 25, 2018 at 13:14 | history | answered | Daniel McLaury | CC BY-SA 3.0 |