Timeline for Groupoid actions on spaces
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2021 at 15:50 | answer | added | Buschi Sergio | timeline score: 0 | |
| Apr 29, 2016 at 21:50 | answer | added | Rohit D. Holkar | timeline score: 1 | |
| Feb 26, 2014 at 16:57 | comment | added | Ma Ming | @Qfwfq I think this definition agrees with the standard definition of actions for, at least, set-theoretical groupoids. A functor $G\to Sets$ is simply a bifunctor $*^{op}\times G\to Sets$, or a profunctor, or a $*$-$G$ bimodule, or simply a $G$ action. Where $*$ denotes the trivial groupoid with one object and one morphism. | |
| May 18, 2013 at 21:00 | answer | added | Ronnie Brown | timeline score: 7 | |
| Jun 14, 2011 at 21:02 | comment | added | Qfwfq | Your definition of "action of a groupoid on a set" is different from the standard one (and, if I'm not mistaken, non equivalent to it). The standard definition (see Moerdijk) is as follows: let $G=(s,t:G_1\to G_0)$ be a groupoid and $E$ be a space, then an action of $G$ on $E$ is given by an epimorphism $\pi:E\to X$ and a map $\mu : G_1\times_{G_0} E \to E$ that has some properties analogous to the definition of a usual group action. | |
| Apr 30, 2011 at 6:56 | answer | added | Tim Porter | timeline score: 5 | |
| Apr 29, 2011 at 23:15 | comment | added | Dany Majard | @Michael : I believe that the space it would act on is the disjoint union of all the spaces in the image of your functor. In the definition on the action of a groupoid G on a set S, there is a surjective map $p:S\to G_0$ where $G_0$ is the set of objects in your groupoid. Then a morphism $g\in G$ acts on the fiber by p over its taget and sends it to the fiber over its source. This can be viewed as a functor $G\to Set$ that sends objects to their fiber by p. | |
| Apr 29, 2011 at 22:18 | vote | accept | Bill Kronholm | ||
| Apr 29, 2011 at 21:59 | answer | added | Alain Valette | timeline score: 7 | |
| Apr 29, 2011 at 20:23 | answer | added | John Klein | timeline score: 27 | |
| Apr 29, 2011 at 20:09 | comment | added | Niyazi | Once you have a group action you can pass to its pseudogroup and then jump to groupoid generated by the germs of this pseudogroup. Passing from pseudogroups to groupoids have nothing to do with the original group. And , in fact, for every pseudogroup of local homeomorphism of a space you can pass to its groupoid of germs. | |
| Apr 29, 2011 at 19:53 | comment | added | Bill Kronholm | @Hurkyl: Yes, I am. If the groupoid is not connected, then we can think about the action given by each component. | |
| Apr 29, 2011 at 19:03 | comment | added | user13113 | You're assuming the groupoid is connected; that all objects have a morphism between them, and are thus isomorphic. In general, objects of different connected components could map to non-homeomorphic spaces. | |
| Apr 29, 2011 at 17:16 | comment | added | Bill Kronholm | Since all of the morphisms in the groupoid are isomorphisms, they get sent to homeomorphisms by the functor. So, we can probably take the functor to be constant on objects. Otherwise, some objects of G are sent to X and others are sent to spaces which are homeomorphic to X. | |
| Apr 29, 2011 at 17:10 | answer | added | Stefan Waldmann | timeline score: 6 | |
| Apr 29, 2011 at 17:08 | comment | added | Michael Bächtold | On which space is the groupoid acting by this definition? | |
| Apr 29, 2011 at 16:56 | history | asked | Bill Kronholm | CC BY-SA 3.0 |