Timeline for Kleinian groups containing an isomorphic copy of itself
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2015 at 5:01 | answer | added | Misha | timeline score: 3 | |
| May 6, 2015 at 16:28 | comment | added | Ian Agol | Yes, Scott (and Shalen) actually prove that the manifold is homotopy equivalent to a compact 3-manifold. In fact, even with torsion, it is homeomorphic to the interior of a compact orbifold by tameness - of negative euler characteristic if it is non-elementary. In higher dimensions, this strategy at least should apply to geometrically finite Kleinian groups with non-zero Euler characteristic. | |
| May 6, 2015 at 15:39 | comment | added | YCor | OK thanks. Actually it's fine if one knows in Scott's theorem that the core can be chosen to be a retract. Because then if $\Gamma$ is a torsion-free Kleinian group, then $\Gamma=\pi_1(M)$ for some hyperbolic 3-fold $M$ and if $M$ retracts onto a core $K$ then the universal covering also retracts onto the universal covering of $K$, which is thus contractible (I just mean retract, no necessarily by deformation). | |
| May 6, 2015 at 14:21 | comment | added | HJRW | @YCor, you also want Selberg's lemma (to show that they're virtually torsion-free) and the Sphere theorem (to show that the quotient is aspherical). Though perhaps the Sphere theorem is overkill... | |
| May 6, 2015 at 14:11 | comment | added | YCor | @HJRW: sure, but that Kleinian groups are virtually of type F is not just the Scott core theorem. What's a reference for this? | |
| May 6, 2015 at 10:42 | comment | added | HJRW | For reference, you seem to be asking about cofinitely Hopfian Kleinian groups. As has been pointed out, this property is much weaker than co-Hopfian, and much easier to prove. (Ian's argument above does it.) See arxiv.org/abs/1012.1785v1 and the references therein. | |
| May 6, 2015 at 10:35 | comment | added | HJRW | @YCor, there is for arbitary groups virtually of type F, which includes all Kleinian groups. | |
| May 6, 2015 at 6:34 | comment | added | YCor | @IanAgol: I don't think there's a well-defined notion of Euler characteristic for arbitrary finitely presented groups. | |
| May 6, 2015 at 4:14 | comment | added | Ian Agol | In 3-dims. (discrete groups in $PSL_2(\mathbb{C})$), either a finitely generated group has finite covolume, in which case @YCor's comment applies, or it is infinite covolume, in which case if it is non-elementary, the Euler characteristic is $<0$ (by the Scott core theorem, finitely generated groups are finitely presented, so Euler characteristic makes sense), and therefore it cannot be isomorphic to a finite-index subgroup of itself. So Rivin's answer is overkill at least in 3D (co-Hopfian is much stronger than what you're asking for). | |
| May 5, 2015 at 23:19 | comment | added | YCor | It you restrict to lattices in $PSL_2(\mathbf{C})$ (finite volume fundamental domain, e.g. compact), Mostow rigidity theorem implies that the lattice is not isomorphic to a proper subgroup of finite index. Indeed such an isomorphism would extend to an automorphism of $PSL_2(\mathbf{C})$, but automorphisms of the latter preserve the volume and cannot map a lattice to a proper subgroup of finite index. | |
| May 5, 2015 at 23:10 | comment | added | Danny Nguyen | I wasn't aware of this property. Basically I am looking for a crystallographic group with compact fundamental domain satisfying the above property. | |
| May 5, 2015 at 22:49 | comment | added | YCor | And you probably also want Zariski-dense (subgroups of $PSL_2(\mathbf{C})$ that are non-Zariski-dense for the complex Zariski topology and discrete in the ordinary topology are actually virtually abelian, parabolic or not). | |
| May 5, 2015 at 19:58 | comment | added | Danny Nguyen | Yes I meant a proper subgroup of a finitely generated group. | |
| May 5, 2015 at 19:51 | comment | added | YCor | 1) Yes, every group admits itself as a subgroup of finite index. Possibly you mean a proper subgroup... 2) Yes with a proper subgroup, namely the free group on countably many generators. Possibly you restrict to finitely generated groups? | |
| May 5, 2015 at 19:50 | vote | accept | Danny Nguyen | ||
| May 5, 2015 at 20:03 | |||||
| May 5, 2015 at 19:35 | answer | added | Igor Rivin | timeline score: 2 | |
| May 5, 2015 at 18:04 | history | asked | Danny Nguyen | CC BY-SA 3.0 |