Timeline for redundant relations
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2013 at 20:42 | comment | added | HJRW | @YvesCornulier - no, not for 2-generator subgroups: the number of generators of the kernel in Haglund--Wise's construction depends on the input. You need to use Rips' original construction, and hence also invoke Agol's result. | |
| Sep 14, 2013 at 19:46 | comment | added | YCor | @HJRW: Haglund-Wise is enough: they constructed a Rips machine with subgroups of right-angled Artin groups. Actually it now follows from Agol's later paper that Rips' original construction already provides subgroups of right-angled Artin groups. | |
| Sep 13, 2013 at 21:30 | comment | added | HJRW | (On the other hand, it's conceivable that it's decidable in $SL_n(\mathbb{Z})$ for any fixed $n$, though I can't say I think this is very likely.) | |
| Sep 13, 2013 at 21:20 | comment | added | HJRW | @YvesCornulier, actually, I now realise that finite presentability is now known to be undecidable for 2-generator matrix groups. Rips' original construction gives 2-generator subgroups of C'(1/6) groups, and by Agol, Haglund and Wise's theorems, they are in fact $\mathbb{Z}$-linear (and one can compute the corresponding matrices). | |
| Sep 13, 2013 at 19:59 | comment | added | YCor | Yes: in $SL_4(\mathbf{Z})$ you have an explicit direct product of 2 free groups. If $F$ is freely generated by $x,y$, and $p$ is the projection from $F$ to the cyclic group killing $y$, then let $P$ be the fibre product consisting of pairs $(u,v)$ such that $p(x)=p(y)$. Then $P$ is generated by $(x,x)$, $(y,y)$ and $(1,y)$. It's well-known to be not finitely presented, see the references in Bridson's papers. | |
| Sep 13, 2013 at 19:16 | comment | added | Igor Rivin | @YvesCornulier are there explicit non f.p. examples? | |
| Sep 13, 2013 at 8:04 | answer | added | Derek Holt | timeline score: 7 | |
| Sep 13, 2013 at 6:41 | comment | added | YCor | Do you assume beforehand that your group is finitely presentable? Lots of f.g. matrix groups over $\mathbf{Z}$ are not finitely presentable. I don't know if it's decidable whether 2 matrices in a given $SL_n(\mathbf{Z})$ generate a f.p. group. | |
| Sep 13, 2013 at 5:48 | answer | added | HJRW | timeline score: 11 | |
| Sep 13, 2013 at 1:18 | comment | added | Benjamin Steinberg | I would guess the best heuristics involve look at images under well chosen congruence subgroups. | |
| Sep 13, 2013 at 1:14 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Sep 12, 2013 at 23:32 | answer | added | Joel David Hamkins | timeline score: 4 | |
| Sep 12, 2013 at 22:50 | comment | added | Igor Rivin | @StephanKohl The matrices are over $\mathbb{Z}$... | |
| Sep 12, 2013 at 22:43 | comment | added | Stefan Kohl♦ | Over which domain(s) are your matrices, i.e. in which ring are the entries? | |
| Sep 12, 2013 at 22:35 | comment | added | Igor Rivin | @StefanKohl I agree that the general recognition of group question is very hard (as I say, most likely undecidable in general), but the question I ask is much easier (at least so it would seem). | |
| Sep 12, 2013 at 22:20 | comment | added | Stefan Kohl♦ | As far as I can tell, this is a hard problem. For example, even if your generating matrices are only $2 \times 2$ and with entries in $\mathbb{Z}$, there is no bound $B$ such that if you don't find a relation of length $\leq B$, your group is free -- cf. Mark Sapir's answer to the following question of mine: mathoverflow.net/questions/119879/free-subgroups-of-gl2-z. | |
| Sep 12, 2013 at 21:28 | history | edited | Igor Rivin | CC BY-SA 3.0 |
fixed punctuation
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| Sep 12, 2013 at 21:10 | history | asked | Igor Rivin | CC BY-SA 3.0 |