Can the fundamental group of a closed hyperbolic 3-manifold act freely on the plane by homeomorphisms? Freely and cocompactly? Freely, cocompactly, and preserving orientation?
$\begingroup$
$\endgroup$
6
-
3$\begingroup$ @IanAgol: no, left-orderable countable groups do not necessarily act freely continuously on the line. A group acting freely continuously on the line is abelian. $\endgroup$– YCorFeb 2, 2016 at 14:50
-
$\begingroup$ Steven, please specify if you require an action by homeomorphisms or preserving some further structure. Also, please define cocompact, whose meaning is clear if the action is proper but has several non-equivalent meanings otherwise (namely one can require, or not, that the quotient be Hausdorff). $\endgroup$– YCorFeb 2, 2016 at 14:53
-
1$\begingroup$ @YCor: By homeomorphisms. For cocompactness, let's just say every point can be taken into some fixed compact set. $\endgroup$– Steven FrankelFeb 2, 2016 at 14:53
-
1$\begingroup$ Are free planar actions of surface groups understood? $\endgroup$– Eric S.Feb 2, 2016 at 16:58
-
1$\begingroup$ @EricS. See my answer. This is all that is known. $\endgroup$– Igor RivinFeb 2, 2016 at 19:39
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
Not an answer, exactly - I am guessing this is open, but there is some very nice work by Frederic le Roux on related matters:
MR2851071 (2012j:37068) Reviewed
Le Roux, Frédéric(F-PARIS11-M)
Free planar actions of the Klein bottle group. (English summary)
Geom. Topol. 15 (2011), no. 3, 1545–1567.
37E30 (57M60)
He gives some torsion free groups that do NOT admit a free action, so maybe the methods adapt to the Kleinian case.
answered Feb 2, 2016 at 15:16