Timeline for Finitely-generated conjugation action on a subgroup that is not normal... what is that?
Current License: CC BY-SA 4.0
20 events
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| Feb 23, 2021 at 12:35 | comment | added | Benjamin Steinberg | I don't remember well which examples are given, but I think in either that paper or a different one, they show that if you globalize a partial action of a group on a meet semilattice (by order isomorphisms) you just get an action on a poset by order preserving maps and you lose meet operation. | |
| Feb 23, 2021 at 4:57 | comment | added | Ville Salo | Maybe you can globalize "inside partial actions on groups" also, and when you globalize the action on a subgroup you get the normal closure. (I have no idea whether that makes sense.) | |
| Feb 23, 2021 at 4:53 | comment | added | Ville Salo | Heh, I realize you said that in your answer, but I didn't get it. And you already went further, globalization probably wouldn't preserve the extra structure of acting by partial group homomorphisms. I think that article does not discuss preserved extra structure at all, unfortunately. | |
| Feb 23, 2021 at 4:49 | comment | added | Ville Salo | Yes I find it illuminating. I also learned that apparently you can globalize partial actions (w.r.t. a symmetric generating set). (Maybe) related to the uniqueness discussion, I wonder what you get from globalizing the conjugation actions on a subgroup. | |
| Feb 22, 2021 at 20:28 | comment | added | Benjamin Steinberg | I really find the Kellendonk-Lawson article is a good place to start. | |
| Feb 22, 2021 at 19:29 | comment | added | Ville Salo | It sounds important to me, but it does not sound so dangerous to me :) Honestly it is not entirely clear to me what I want, and I'll probably.need to give this another go now that I have more trust in inverse semigroups. | |
| Feb 22, 2021 at 16:05 | comment | added | Diego Martinez | @VilleSalo why doesn't the reconstruction issue sound important to you? From where I'm at, it seems that the difference between $\theta(g)\theta(h)$ and $\theta(gh)$ is precisely the problematic point. I guess you could do something if, for instance, that difference was allowed to only happen in a small number of points (a la Følner). Otherwise, no matter how large your generating set is, you'll probably be able to find sufficiently large cycles in the group whose domain is the whole thing, but that $\theta(g_1) \dots \theta(g_k) = 0$. | |
| Feb 22, 2021 at 14:43 | comment | added | Ville Salo | So my first instinct was to give the definitions for a partial action of $G$ on $H$ by partial automorphisms, and then specialize to the case $G$ acts on the subgroup $H$ by conjugation. But ranging over generating sets made no sense, which lead me formulate this question. | |
| Feb 22, 2021 at 14:40 | comment | added | Ville Salo | In my application, I am actually saying something like "suppose $G$ admits a finite generating set $T$ such that the partial action of each $t \in T$ by conjugation on $H$ has good properties [related to paradoxical decompositions]". (Hard to explain without going in more detail, and I don't want to bore you to death.) The reconstruction issue does not sound so dangerous to me, I'm really looking for the correct general statement because I was hoping it would clarify my thoughts about what's really going on. | |
| Feb 22, 2021 at 14:30 | comment | added | Benjamin Steinberg | action on arbitrary elements. However, I suspect that for free groups, the conjugation action in some sense does come from the free inverse monoid and so you probably would recover the original partial action. In general I am not sure there is a good sense of how to build a partial action of $G$ from just the actions of the generators. | |
| Feb 22, 2021 at 14:29 | comment | added | Benjamin Steinberg | Rereading your post more carefully, the problem is for a general partial action (and it wouldn't surprise me if this is the case in your setup of conjugation), if $G=\langle S\rangle$ and you just allow yourself the actions coming from productions of partial bijections coming from elements of $S$, you will certainly in most cases miss the actions of some of your original elements of $G$. If your group $G$ is free on $S$, there is a partial action of $G$ on $S$ which agrees with your original partial action on the generators, but I suppose it need not agree with original partial (part 1) | |
| Feb 22, 2021 at 14:25 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 22, 2021 at 14:20 | comment | added | Benjamin Steinberg | Sorry, I also forgot to stipulate that both 1,g have to belong to the subset in the Birget-Rhodes expansion. I hadn't thought about this stuff in over 15 years and was working from faulty memory. | |
| Feb 22, 2021 at 14:18 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Feb 22, 2021 at 14:15 | comment | added | Benjamin Steinberg | If you try to generated an inverse monoid by the the partial bijections coming from your group, then you a quotient inverse monoid of that $F$-inverse cover which may no longer be $F$-inverse, that is, each element of your inverse monoid is a restriction of group elements but it may be a restriction of more than one of them. You will certainly get many elements which are restrictions of a given group element but the case that elements are restrictions of more than one element is in some sense worse. | |
| Feb 22, 2021 at 14:12 | comment | added | Benjamin Steinberg | The Kellendonk-Lawson paper is here worldscientific.com/doi/10.1142/S0218196704001657 | |
| Feb 22, 2021 at 13:40 | comment | added | Ville Salo | The issues I had in mind were more about the formula $\theta(g)\theta(h) \leq \theta(gh)$; I see why you need this, but the issue is that these domain restrictions seem to necessarily lead to having the same element with various domains, and finite generation would require you to get the domains right... which is not really what I care about because I only care about the action. Probably "maximum element" is about exactly this. | |
| Feb 22, 2021 at 13:34 | comment | added | Ville Salo | I understood some of this, it suggests to me that the problems are real if trying to generalize. Actually, I did not even realize that indeed knowing the partial action of generators does not determine the entire action, I had simpler problems in mind. I think I will abandon the attempt to generalize for now. | |
| Feb 22, 2021 at 13:18 | comment | added | Ville Salo | The multiplication formula in the Birget-Rhodes construction has a typo, $h$ should be $gh$. | |
| Feb 22, 2021 at 13:13 | history | answered | Benjamin Steinberg | CC BY-SA 4.0 |