Timeline for Finitely-generated conjugation action on a subgroup that is not normal... what is that?
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Feb 22, 2021 at 16:53 | history | became hot network question | |||
Feb 22, 2021 at 13:13 | answer | added | Benjamin Steinberg | timeline score: 5 | |
Feb 22, 2021 at 12:57 | comment | added | Ville Salo | Glad to hear it's not completely trivial and that generators have come up, I am looking forward to a prospective semianswer. | |
Feb 22, 2021 at 12:50 | comment | added | Benjamin Steinberg | As far as generators go things are complicated. There were some earlier papers that mention the issues but I forget which ones. I think it comes up in Coulbois's papers on profinite topology circa 2000 and maybe in Lawson Kellendonk but I can't quite remember. Or maybe in a later survey by Mark Lawson you might ask him. The I issue is that a partial action is not determined by what it does on generators because it isn't given by a homomorphism into the symmetric inverse monoid but instead by what is called the symmetric inverse monoid. This is getting long so I may give a semianswer | |
Feb 22, 2021 at 12:44 | comment | added | Benjamin Steinberg | You should look for inverse semigroup instead of pseudogroups. You'll see that there is lots of literature on partial group actions and inverse semigroups starting with the work of Exel @YCor mentioned. A good early introduction is the IJAC paper of Lawson and Kellendonk which also discusses discusses globalizing partial actions. They don't speak of your particular partial action so far as I remember. | |
Feb 22, 2021 at 9:32 | comment | added | Ville Salo | So you are saying, I can just say $G$ partially acts on itself by conjugation, and we take the partial subaction on $H$, and then I just say $G$ is finitely-generated. Maybe that is indeed the good setting. But I would like to say that "$G$ partially acts on $H$", and only have the conjugation partial action as a special case; the issue is, what is the big group where $G$ acts? Anyway, maybe I just need to think about this more... | |
Feb 22, 2021 at 9:27 | comment | added | Ville Salo | "it is natural enough to say that a group action $G \curvearrowright H$ is finitely generated when $G$ is" I mean this to be the definition. It's just a property of the group, and I'm ok with just saying that $G$ is finitely-generated, my point was just that that's already non-trivial e.g. in the groupoid approach. | |
Feb 22, 2021 at 9:24 | comment | added | YCor | About finite generation: for a genuine action, what would you mean by "finitely generated"? | |
Feb 22, 2021 at 9:23 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
added 10 characters in body
|
Feb 22, 2021 at 9:23 | comment | added | YCor | There's a good formalism of partial action of a group $G$ on a set $X$ due to Exel (each group element acts as a partial bijection of $X$, that is, a bijection between two given subsets of $X$, with natural axioms), and in particular, if $G$ acts on a set $M$ and $X$ is an arbitrary subset of $M$, then $G$ naturally partially acts on $X$ by restriction. Orbits of partial actions are well-defined (here these are just intersections of orbits on $M$ with $X$). | |
Feb 22, 2021 at 8:52 | history | asked | Ville Salo | CC BY-SA 4.0 |