Timeline for Is every countable discrete group a subgroup of a non discrete Lie group?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2020 at 13:09 | vote | accept | Ali Taghavi | ||
| Aug 13, 2020 at 18:09 | answer | added | Moishe Kohan | timeline score: 6 | |
| Aug 13, 2020 at 15:26 | answer | added | Sami Douba | timeline score: 6 | |
| Aug 13, 2020 at 14:11 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 6 characters in body
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| Aug 13, 2020 at 14:03 | comment | added | Ali Taghavi | @MoisheKohan Ok thanks,I try to understand the reason. | |
| Aug 13, 2020 at 14:02 | comment | added | Moishe Kohan | $BS(2,3)=<a,b| ab^2a^{-1}=b^3>$ does not embed in a connected Lie group. | |
| Aug 13, 2020 at 13:59 | comment | added | Ali Taghavi | @KevinCasto Please read my previous comment. | |
| Aug 13, 2020 at 13:59 | comment | added | Ali Taghavi | @MoisheKohan So I realize that the ocean of Lie groups is wider and deeper than I imagined. Among various example you and Kevin pointed ot, what is a precise example of a countable group not embaddable in a connected lie group of dimension at least 1. | |
| Aug 13, 2020 at 13:49 | comment | added | Moishe Kohan | If you mean nondiscrete, then the firs comment by Kevin answers the question. As for examples, take any centerless nonresidually finite group such as BS(2,3), Baumslag Solitar group. Or take any group with undecidable word problem if you want a more exotic example. | |
| Aug 13, 2020 at 13:37 | comment | added | Ali Taghavi | @MoisheKohanThank you. By locally connected I was (somehow) meant : non discrete.May be some example for your last sentemse?this is what I search for. | |
| Aug 13, 2020 at 13:31 | comment | added | Moishe Kohan | Every Lie group is locally connected, so the answer to your question is obviously yes since every countable discrete group is a Lie group. On the other hand, many countable discrete groups do not embed (even algebraically) in connected Lie groups. | |
| Aug 13, 2020 at 12:48 | comment | added | Ali Taghavi | @KevinCasto Thank you for this link. Befor I start reading this link, do you say that the link contains an answer to the question that every countable group is (isometrically) a subgroup of a connected Lie group? | |
| Aug 13, 2020 at 12:07 | comment | added | Kevin Casto | In that case there is a huge literature on this subject. See e.g. this question: mathoverflow.net/questions/110208/… I suppose technically you need to worry about nonlinear Lie groups like $\widetilde{SL_2}(\mathbb R)$ but I suspect the properties listed in the first answer to that question also apply to discrete subgroups of this. | |
| Aug 13, 2020 at 11:50 | comment | added | Ali Taghavi | @KevinCasto Infact one feel that $\mathbb{R}$ intrincically contains $\mathbb{Z}$. The same for$S^1$ containing finite cyclic group. So accoring to your interesting comment one impose the following question: Let G be a countable discrete group. Is $G$ a subgroup of a connected Lie group? | |
| Aug 13, 2020 at 11:39 | comment | added | Ali Taghavi | @KevinCasto Thank you for your attention. To be honnest I was considering $\mathbb{Z}\subset \mathbb{R}$ or $\mathbb{Z}/2\mathbb{Z} \subset S^1$. Could you please help me to impose some extra conditions to have a minimal number of connect compenent. I had intention to ask "A connected Lie group". I am not sure if there is a connected Lie group containing the example in the "Remark". | |
| Aug 13, 2020 at 11:26 | comment | added | Kevin Casto | I assume you want more conditions, or else you can just take $G \times \mathbb R$? | |
| Aug 13, 2020 at 11:08 | history | asked | Ali Taghavi | CC BY-SA 4.0 |