Timeline for From topological actions on $\mathbb{R}^3$ to isometric actions
Current License: CC BY-SA 4.0
24 events
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| Mar 22, 2022 at 16:47 | comment | added | HJRW | (I guess I'm assuming that all topological 3-dimensional orbifolds can be smoothed. I think this is probably true. Again, if your action is really piecewise linear this isn't going to be a problem.) | |
| Mar 22, 2022 at 16:46 | comment | added | HJRW | @Agelos: That's right. Since $G$ is one-ended with no $\mathbb{Z}^2$ subgroups, the orbifold and geometrisation theorems imply that the orbifold is hyperbolic, so $G$ is a finite extension of a Kleinian group. One consequence of Agol's Virtual Haken theorem is that Kleinian groups are also "good in the sense of Serre", from which it follows that $G$ is residually finite and hence virtually torsion-free. | |
| Mar 22, 2022 at 12:46 | comment | added | Agelos | @HJRW: indeed, my proof produces an equivariant embedding. Moving up to 3D, if G acts on R^3 so that the quotient is homeomorphic to a 3-orbifold, is virtual torsion-freeness still a "byproduct"? | |
| Mar 22, 2022 at 12:25 | comment | added | HJRW | @Agelos: If your proof produces an equivariant embedding of the Cayley graph into the plane then the action on the plane can easily be made piecewise linear, so all the difficulties arising from only assuming the action to be by homeomorphisms go away. The quotient is then visibly an orbifold, and the result follows from the fact that orbifolds admit geometric structures. Virtual torsion-freeness is then a byproduct (and one would still probably use Selberg's lemma to handle the case of triangle orbifolds). | |
| Mar 22, 2022 at 11:40 | comment | added | Agelos | ... But the fact that G is isomorphic to a Kleinian group used above is the 2D version of my original question. To obtain a proof more likely to extend to 3D, we could instead use the fact that G has an 1-ended planar (hyperbolic) Cayley graph (which the aforementioned paper proves without using that G is Kleinian), and try to deduce that G is virtually-torsion-free from that. Can you do this? | |
| Mar 22, 2022 at 11:37 | comment | added | Agelos | For n=2, it is indeed true that every discrete group acting cocompactly and properly discontinuously on R^n is virtually torsion-free. Here is another way to prove it. It is known that every such group G is isomorphic to a Kleinian function group (see e.g. ams.org/journals/tran/2020-373-07/S0002-9947-2020-08026-X/…), in particular, G is a group of matrices over C, and we can apply Selberg's Lemma. Hyperbolicity is not needed. | |
| Mar 21, 2022 at 14:03 | comment | added | HJRW | Actually, I guess I left out an important detail in dimension 2, namely that the action should be smoothable. I suspect this is true and classical (some suitable extension of the Jordan curve theorem) but I don't know a reference. | |
| Mar 21, 2022 at 14:01 | comment | added | HJRW | @YCor: For n=2 this is true, using the classification of compact, 2-dimensional orbifolds and the fact that all orbifold groups are "good in the sense of Serre". (This latter implies that finite extensions of orbifold groups are residually finite, hence also virtually torsion free.) For n=3 and a smooth action it's probably also true by a suitable orbifold version of geometrisation and the fact that 3-manifold groups are also good. The case of a non-smooth action seems interesting, and one should probably ask Pardon. For $n\geq 4$ I suspect it's false, even in the smooth case. | |
| Mar 21, 2022 at 9:48 | comment | added | YCor | @Agelos Say, this one: for $n=2$, is every discrete group acting properly cocompactly continuously on $\mathbf{R}^n$ virtually torsion-free? What if assumed in addition to be hyperbolic? | |
| Mar 21, 2022 at 9:44 | comment | added | Agelos | @Ycor: could you clarify which "same question" you want to have solved for an action on the plane instead of 3-space? | |
| Mar 21, 2022 at 9:38 | comment | added | Agelos | I don't mind assuming the action to be properly discontinuous (I'll update the question). This answer is already looking appealing, thanks HJRW! It would be nice to see an aswer dropping the virtual torsion-freeness assumption. | |
| Mar 18, 2022 at 10:37 | comment | added | HJRW | @YCor: I think that's a reasonable hope (and certainly some versions are known), but one needs a smooth action. | |
| Mar 18, 2022 at 10:33 | comment | added | YCor | My idea was whether there is an "orbifold" version of the Thurston-Perelman hyperbolization theorem. Actually one could try to (inconditionally) solve the same question beforehand for an action on the plane instead of the 3-space. | |
| Mar 18, 2022 at 10:15 | comment | added | HJRW | @YCor: Well, I think the other idea for a proof conditional on the Cannon conjecture might work in that case, although there are a few things to check, as I explained in the first version of this answer. These two approaches are not as disconnected as they might at first seem -- there are (still pretty speculative) approaches to the Cannon conjecture conditional on the hypothesis that every hyperbolic group is residually finite. | |
| Mar 18, 2022 at 8:53 | comment | added | YCor | Yes, I referred to the last paragraph while the virtual torsion-free assumption was used beforehand. It's intriguing how this assumption could be removed. | |
| Mar 18, 2022 at 6:01 | comment | added | HJRW | @YCor: Good point! That’s another nice way to conclude. Though I think you need the virtually torsion-free assumption to get to produce the copy of H^3 in the first place. | |
| Mar 17, 2022 at 22:23 | comment | added | YCor | I think Tukia's 1986 "quasiconformal" paper (DOI link) directly implies that every discrete group quasi-isometric to $X=H^n(\mathbf{R})$ (for $n\ge 3$) has a geometric action on $X$, without passing to any finite index subgroup. (This is also true for $n=2$ but was proved later by other authors.) | |
| Mar 17, 2022 at 20:48 | history | undeleted | HJRW | ||
| Mar 17, 2022 at 20:48 | history | edited | HJRW | CC BY-SA 4.0 |
Rewritten -- my first answer was confused.
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| Mar 17, 2022 at 18:58 | history | deleted | HJRW | via Vote | |
| Mar 17, 2022 at 18:51 | comment | added | HJRW | @YCor wait: no, I’m not saying that! Sorry, I think I was forgetting that Bestvina—Mess predates geometrisation. I’ll edit later this evening. | |
| Mar 17, 2022 at 18:47 | comment | added | YCor | Are you saying that the Cannon conjecture is unknown even in the special case of a 3-manifold group? | |
| Mar 17, 2022 at 18:44 | comment | added | YCor | Minor point: the OP doesn't assume properness, but a weaker discreteness assumption, satisfied, for instance, by the action of $\mathbf{Z}$ by powers of $(x,y)\mapsto (2x,y/2)$ on $\mathbf{R}^2-\{(0,0)\}$. | |
| Mar 17, 2022 at 18:22 | history | answered | HJRW | CC BY-SA 4.0 |