Timeline for group generated by unipotents in arithmetic subgroup is finitely generated
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2023 at 8:31 | comment | added | Geoffrey Janssens | Ok thank you. (I was not aware of the strike and have just read the full reasons for it and signed the petition. Hope the strike will work.) | |
| Jul 19, 2023 at 23:08 | comment | added | Moishe Kohan | There is no one reference covering this. I will write a proper answer once the strike is over (Ian Agol can also write one if he wishes). | |
| Jul 19, 2023 at 15:27 | comment | added | Geoffrey Janssens | Thank you this info! Do you maybe have some reference ? (I did try to search but was not fruitful). I am very curious how the authors use the fact that the generators are unipotent... | |
| Jul 18, 2023 at 22:45 | comment | added | Moishe Kohan | I am not sure about congruence-subgroups, but for arbitrary finite-index subgroups in lattices of $SO(3,1)$ the situation is the same as for Fuchsian groups: Each nonuniform lattice contains a finite-index subgroup such that unipotents generate an normal nontrivial subgroup such that the quotient group is not virtually cyclic. (This is a hard theorem, mostly due to Agol and Wise.) Another nontrivial (but not as hard) theorem ensures that such a subgroup cannot be finitely generated (mostly due to Stallings). | |
| Jul 18, 2023 at 9:41 | comment | added | Geoffrey Janssens | If I understood well PSL_2(Z) is an arithmetic Fuchsian group. Its principal congruence subgroups also and if its level is not a divisor of 4, then its unipotent elements generate an infinite index subgroup via the link you have added. Besides I know that PSL_2(I_d), with I_d the ring of integers in Q(\sqrt{-d}), is a Kleinian group and the PSL_2(tdqa/Q) can be seen as standard arithmetic subgroups of SO(5,1). If the coefficient ring is Euclidean those groups are generated by unipotents so there question 1 is true. But I don't know about their congruence subgroups. | |
| Jul 17, 2023 at 23:56 | comment | added | Moishe Kohan | Do you understand that in this case the group is not Fuchsian? It is the fundamental group of a 3-dimensional hyperbolic orbifold. The Fuchsian result likely extends to this case as well. | |
| Jul 17, 2023 at 21:57 | comment | added | Geoffrey Janssens | Once more thanks for your useful comment!! I however still struggle to understand what this implies for the R-rank 1 SL_2(D)'s (i.e. D= Q, imaginary quadratic extension of Q or totally definite quaternion algebra over Q). For SL_2(Z) it follows from what you linked. But is there a general result for these groups saying that infinite index normal infinite subgroups are not f.g.? | |
| Jul 11, 2023 at 15:10 | comment | added | Moishe Kohan | math.stackexchange.com/questions/168913/… | |
| Jul 11, 2023 at 14:38 | comment | added | Moishe Kohan | Take any arithmetic Fuchsian group with noncompact quotient of positive genus. Then your subgroup will be normal nontrivial of infinite index. Such a subgroup cannot be finitely generated. | |
| Jul 11, 2023 at 14:28 | comment | added | Geoffrey Janssens | Thanks for your comment, however I could not come up with an example. Would it maybe be possible to give one or a reference please? | |
| Jul 11, 2023 at 14:08 | comment | added | Moishe Kohan | For arithmetic Fuchsian groups this is in general false. | |
| Jul 11, 2023 at 11:08 | history | asked | Geoffrey Janssens | CC BY-SA 4.0 |