Timeline for From topological actions on $\mathbb{R}^3$ to isometric actions
Current License: CC BY-SA 4.0
26 events
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| May 20, 2022 at 15:26 | history | edited | Sam Nead | CC BY-SA 4.0 |
formatting of title.
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| Apr 18, 2022 at 17:44 | vote | accept | Agelos | ||
| Apr 18, 2022 at 14:17 | answer | added | John Pardon | timeline score: 2 | |
| Mar 21, 2022 at 9:42 | history | edited | Agelos | CC BY-SA 4.0 |
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| Mar 17, 2022 at 20:39 | comment | added | Moishe Kohan | In the present form, assuming tameness of the action, the answer is positive. It is an easy corollary of the Geometrization Theorem. | |
| Mar 17, 2022 at 18:22 | answer | added | HJRW | timeline score: 3 | |
| Mar 16, 2022 at 8:24 | comment | added | HJRW | The isometry part of the question is a red herring: the counterexamples mentioned don't fail because they're topologically "wild", they just fail because there are non-geometric 3-manifolds. A better question might be: "Is every properly discontinuous topological action on a 3-manifold a limit of smooth actions?" It might be worth asking this question separately. Pardon is sometimes active on MO, and probably knows the status of the question. | |
| Mar 15, 2022 at 8:41 | comment | added | Agelos | I've updated the question: what if G is Gromov-hyperbolic (and 1-ended)? | |
| Mar 15, 2022 at 8:39 | history | edited | Agelos | CC BY-SA 4.0 |
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| Mar 13, 2022 at 1:23 | comment | added | Vitali Kapovitch | @HJRW oh right, sorry, I thought I had them all covered and forgot this one. still, graph manifolds are counterexamples to that possibility also. | |
| Mar 12, 2022 at 23:31 | comment | added | HJRW | @VitaliKapovitch: you’re right that graph manifolds provide counterexamples, but it’s not true that the fundamental group of a geometric 3-manifold is either virtually solvable or Gromov hyperbolic. The fundamental groups of Seifert fibres manifolds with hyperbolic base are neither of these. | |
| Mar 12, 2022 at 0:19 | comment | added | Vitali Kapovitch | I think $\mathbb Z^3\times G$ case when the action is by diffeos can be understood. The $\mathbb Z^3$ action should be free so we can quotient by it first and get an action of $G$ on $T^3$. by averaging this can be made isometric with respect to some smooth metric on $T^3$. then Ricci flow (possibly with surgeries) should turn this into a flat $T^3$ and since the Ricci flow preserves isometries the $G$ action will remain isometric in the limit. | |
| Mar 11, 2022 at 23:32 | comment | added | Vitali Kapovitch | Since $\mathbb Z$ can be ruled out $\mathbb Z\times G$ can also because it contains $\mathbb Z$ as a finite index subgroup and it still acts nicely. But I don't know about something like $\mathbb Z^3\times G$ where $G$ is finite. | |
| Mar 11, 2022 at 23:24 | comment | added | markvs | For finite groups everything if fine, for $\Bbb Z$ it is also fine, but for the direct product - not clear (at least for me). | |
| Mar 11, 2022 at 23:22 | comment | added | markvs | @VitaliKapovitch: You know better than I. Abelian groups with torsion are virtually torsion-free (virtually direct products of cyclic groups). But I think from the question even the groups like $\Bbb Z\times G$ where $G$ is finite are not clear. | |
| Mar 11, 2022 at 23:19 | comment | added | Vitali Kapovitch | @markvs Yes, I was perhaps too optimistic. the assumption is that the group acts on $\mathbb R^3$ properly and cocompactly by homeomorphisms (or diffeomorphisms). I feel that should be restrictive even if no assumptions on geometry are made. You are right about $\mathbb Z^4$ but what about some abelian groups with torsion? is it clear that they cannot happen? or maybe they can? – | |
| Mar 11, 2022 at 22:37 | comment | added | markvs | @VitaliKapovitch: You can ask Misha. But I do not see a reason for a positive answer in either case. There are lots of hyperbolic and solvable groups inside $\mathrm{Diff}(\Bbb R^3)$. For Abelian groups, the answer is "yes" trivially? $\Bbb Z^4$ cannot act properly and cocompactly on $\Bbb R^3$? | |
| Mar 11, 2022 at 19:37 | comment | added | Vitali Kapovitch | @Angelos I suspect the answer should be yes to both but that is a guess. Geometric group theorists might know the answer. | |
| Mar 11, 2022 at 19:27 | comment | added | Agelos | @VitaliKapovitch: I accept this as a negative answer to my question then. Any references would be appreciated. Could the answer be yes if I assumed G to be virtually solvable or Gromov Hyperbolic? | |
| Mar 11, 2022 at 19:07 | comment | added | Vitali Kapovitch | @Agelos the group is the fundamental group of the resulting closed aspherical manifold acting on the universal cover by deck transformations. it can not act isometrically discretely and cocompactly on any of the Thurston geometries because it's neither virtually solvable nor is it Gromov Hyperbolic. | |
| Mar 11, 2022 at 18:26 | comment | added | Agelos | @VitaliKapovitch: I don’t see how this yields a counterexample. What is your group, and why can’t it act isometrically on one of Thurston’s geometries? | |
| Mar 11, 2022 at 18:03 | comment | added | Vitali Kapovitch | it's certainly wrong as stated even in smooth case. There are closed aspherical 3 manifolds which are glued from several geometric pieces but are not geometric themselves. E.g. graph manifolds or more generally irreducible manifolds all of whose pieces in the JSJ decomposition are aspherical. The universal covers are $\mathbb R^3$. | |
| Mar 11, 2022 at 18:01 | comment | added | markvs | The second condition means "proper action"? | |
| Mar 11, 2022 at 17:25 | comment | added | Agelos | @YCor: By cocompact I mean that there is a compact subset K of R^3 such that the image of K under the action of G covers R^3. The other condition says that the orbit of any point has no accumulation point in R^3. | |
| Mar 11, 2022 at 17:19 | comment | added | YCor | Could you define "cocompact"? By "the orbit has no accumulation point" do you mean that it has no accumulation in the orbit, or in the ambient space? | |
| Mar 11, 2022 at 17:08 | history | asked | Agelos | CC BY-SA 4.0 |