Timeline for Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 12, 2021 at 18:43 | answer | added | Dinisaur | timeline score: 0 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Aug 4, 2010 at 7:44 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added Edit 4.
|
| Aug 3, 2010 at 9:12 | comment | added | Pierre-Yves Gaillard | Thanks for such a clear and precise answer. Your interventions in this thread have been quite useful to me. I hope the general problem will be solved soon. | |
| Aug 3, 2010 at 8:56 | comment | added | Keivan Karai | This is what he is alluding to in a remark 6.18. He says clearly that the two results by Wang and Kazhdan are needed in proving that every lattice is finitely generated. | |
| Aug 3, 2010 at 8:53 | comment | added | Keivan Karai | Several results in Chapter XIII of Raghunathan show that lattices with some extra properties (e.g. when the ambient group has a rank one factor) are finitely generated (or presented). In Remark 13.21, says that "Corollary 12.20 together with theorem 6.15 and the theorems of Kazhdan and Wang show that any lattice in a semi-simple Lie group is finitely generated." Corollary 13.20 is the fact I mentioned in the parentheses above, and the paper of Kazhdan is the one in which he introduces property T and shows that any lattice in a semi-simple Lie group of higher rank has property T, hence f.g. | |
| Aug 3, 2010 at 6:04 | answer | added | Pierre-Yves Gaillard | timeline score: 6 | |
| Aug 2, 2010 at 19:31 | comment | added | Pierre-Yves Gaillard | I agree that it was silly of me to invoke lattices in this connection, and that elementary arguments were sufficient. [I mentioned this in EDIT 3.] I don't have Raghunathan's book at hand, but I'll look at Thm. 6.11. About finite generation of lattices, you say that Raghunathan only "handles the case of rank one groups". But he writes on p. 100: "any lattice in a connected Lie group is finitely generated". He seems to consider any lattice of any connected Lie group. | |
| Aug 2, 2010 at 18:36 | comment | added | Keivan Karai | On a different note, the fact that a (non-uniform) lattice in a connected semi-simple Lie group is finitely generated is a deeper fact and considerably more difficult to prove. Raghunathan handles the case of rank one groups. In hight rank case (say for $SL_n(\mathbb R)$ for $n \ge 3$) the first proof is given by Kazhdan and uses property T. | |
| Aug 2, 2010 at 18:36 | comment | added | Keivan Karai | you do not need a full lattice to show that there is a copy of $\mathbb Z$. However, assuming the existence of a lattice, the problem becomes easy. Suppose $\Gamma$ is a lattice in $G$: Apply Theorem 6.11 in Raghunathan to a find a torsion-free subgroup of finite index in $\Gamma$. This group is not trivial, since otherwise $\Gamma$ would be finite; Now, choose $\gamma \in \Gamma -\{ 1 \}$ to get a discrete copy $\langle g \rangle $. | |
| Aug 2, 2010 at 17:52 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added EDIT 3
|
| Aug 2, 2010 at 12:52 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added parenthesis about "more elementary arguments".
|
| Aug 2, 2010 at 12:34 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added EDIT 2.
|
| Aug 2, 2010 at 10:49 | answer | added | Keivan Karai | timeline score: 1 | |
| Aug 2, 2010 at 7:51 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added EDIT 1. ; Post Made Community Wiki
|
| Aug 2, 2010 at 6:49 | answer | added | C.S. | timeline score: 2 | |
| Aug 2, 2010 at 6:03 | history | asked | Pierre-Yves Gaillard | CC BY-SA 2.5 |