Timeline for Lie groupoids in practice
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2019 at 13:42 | comment | added | Praphulla Koushik | oh.. you are taking pairs $(g,m)$ with the condition that $g.n\in N$... I was misreading... Is it obvious that it's a Lie groupoid.?? I am checking | |
| Nov 25, 2019 at 13:21 | comment | added | DamienC | For a given $m\in N$, there may be a $g\in G$ such that $g.m\notin N$. So, the $G$-action on $M$ doesn't necessarily restrict to a $G$-action on $N$. | |
| Nov 25, 2019 at 12:12 | comment | added | Praphulla Koushik | I am really missing something. I will take a break and respond to your answer... You are writing $(g,m)\in G\times N$ with $g.m\in N$... I am seeing this as action of $G$ on $N$... May be I need some fresh air... | |
| Nov 25, 2019 at 12:02 | comment | added | Bugs Bunny | No, I am not. In general, $G$ does not act on $N$, only on $M$. So there is no such groupoid. | |
| Nov 25, 2019 at 11:59 | comment | added | Praphulla Koushik | you are mentioning the Lie groupoid $[G\times N\rightrightarrows N]$.. Isn't it? I do not understand the point of considering a submanifold that is invariant under action of G and then considering the associated action Lie groupoid.. This is already covered in example 3. Am I missing something? | |
| Nov 25, 2019 at 11:58 | comment | added | Bugs Bunny | What do you mean? If $N$ is not closed under $G$-action, it is not an action groupoid of $G$. In general, it could be quite complicated groupoid. | |
| Nov 25, 2019 at 11:50 | comment | added | Praphulla Koushik | I am having difficulty in understanding how this is different from example $3$? :O | |
| Nov 25, 2019 at 11:31 | history | answered | Bugs Bunny | CC BY-SA 4.0 |