Timeline for Subgroups of RAAGs vs. subgroups of RACGs
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 26, 2021 at 8:35 | vote | accept | AGenevois | ||
| Jul 26, 2021 at 8:35 | answer | added | AGenevois | timeline score: 1 | |
| Jul 14, 2021 at 21:20 | comment | added | HJRW | @YCor, Anthony’s answer also answers this subquestion, since all its noncyclic quotients are dihedral. I suspect the fundamental group of the surface of Euler characteristic -1 also fails to be residually torsion-free nilpotent, but don’t know a proof. | |
| Jul 14, 2021 at 19:39 | answer | added | AGenevois | timeline score: 7 | |
| Jul 13, 2021 at 6:59 | comment | added | YCor | @MarkSapir isn't the (sub)question rather whether torsion-free f.g. subgroups of RACG residually (torsion-free nilpotent)? | |
| Jul 13, 2021 at 6:14 | answer | added | HJRW | timeline score: 11 | |
| Jul 12, 2021 at 18:58 | comment | added | AGenevois | @MattZaremsky: Thank you. In fact, RAAGs satisfy a strong Tits alternative: every finitely generated subgroup either is abelian or it surjects onto $\mathbb{F}_2$. For RACGs, you have something similar, every finitely generated subgroup either is virtually abelian or it surjects onto a non-elementary free product of $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$. So you get a similar obstruction: for finitely generated subgroups, the abelianisation cannot be cyclic. | |
| Jul 12, 2021 at 18:11 | comment | added | Matt Zaremsky | @MarkSapir: That sounds right, so I think my one trick for seeing something doesn't embed in a RAAG also makes it not embed in a RACG, and hence is not much help for this question. | |
| Jul 12, 2021 at 16:48 | comment | added | markvs | @MattZaremsky: Aren't RACG residually nilpotent? Also see link.springer.com/content/pdf/10.1007/s00013-019-01332-7.pdf, Theorem 1 (ii). | |
| Jul 12, 2021 at 16:12 | comment | added | Matt Zaremsky | @AGenevois: The proof I know that $B_4$ (even $B_3$) doesn't embed into any RAAG (as told to me by Thomas Koberda) is that thanks to residual torsion-free nilpotence, any non-abelian subgroup of a RAAG must surject onto $\mathbb{Z}^2$, and of course braid groups can't do this. | |
| Jul 12, 2021 at 16:03 | comment | added | YCor | Sorry for adding mess, indeed my comment was irrelevant since $B_4$ is not a RAAG. | |
| Jul 12, 2021 at 15:49 | comment | added | AGenevois | @MattZaremsky: I see. Is it easy to prove that $B_4$ does not embed in a RAAG? | |
| Jul 12, 2021 at 15:33 | comment | added | Matt Zaremsky | I think @YCor's point was that $B_4$ can't embed in a RAAG, so one could wonder whether it embeds in a RACG and hence serves as a counterexample, but "probably" $B_4$ also can't embed in a RACG. (Actually all of this could be said already for $B_3$.) (For the record I can't think of any other candidate counterexamples.) | |
| Jul 12, 2021 at 15:05 | comment | added | AGenevois | @YCor: I don't see the connection between your comment and my question. Am I missing something? | |
| Jul 12, 2021 at 15:03 | comment | added | AGenevois | @MarkSapir: Sure, but that only proves that torsion-free subgroups of RACGs are virtually subgroups of RAAGs. My question asks precisely whether or not the virtually can be removed. | |
| Jul 12, 2021 at 14:13 | comment | added | YCor | I guess the braid group $B_4$ is not isomorphic to any subgroup of any Coxeter group. (If it were, we would deduce its linearity, which was hard to establish. Anyway that's not a proof). | |
| Jul 12, 2021 at 14:12 | comment | added | YCor | f.g. RAACG are virtually torsion-free. But not orbitrary RAACG, since these include the group of finitely supported permutations of the integers. | |
| Jul 12, 2021 at 13:37 | comment | added | markvs | RAACGs are not virtually torsion-free? Did you see this: core.ac.uk/download/pdf/82768543.pdf | |
| Jul 12, 2021 at 13:10 | history | asked | AGenevois | CC BY-SA 4.0 |