Is the property of left invariant orderability for finitely generated groups preserved by quasi-isometrics? More precisely, if $G$ is a left orderable (finitely generated) group and $H$ is a torsion-free group quasi-isometric (in the sense of Gromov) to $G$, can we conclude $H$ is left orderable? If the answer is no in the general case, what about 3-manifold groups?
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asked Sep 7, 2014 at 3:54
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3$\begingroup$ $\mathbb Z$ and $\mathbb Z\times (\mathbb Z/2)$ are quasi-isometric groups, both are fundamental groups of 3-manifolds, $\mathbb Z$ is left orderable, but $\mathbb Z\times (\mathbb Z/2)$ is not. $\endgroup$– Anton PetruninCommented Sep 7, 2014 at 4:11
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$\begingroup$ Thanks! I edited the question to exclude trivial counterexamples. $\endgroup$– Mahdi Teymuri GarakaniCommented Sep 7, 2014 at 4:32
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10$\begingroup$ All closed hyperbolic 3-manifold groups are quasi-isometric; however there are both orderable and non-orderable such groups. $\endgroup$– Ian AgolCommented Sep 7, 2014 at 4:39
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4$\begingroup$ Note that all the previous examples are virtually left-orderable (VLO). There are candidates to show that for torsion-free groups, VLO is not QI-invariant. For instance, Burger-Mozes groups and irreducible cocompact lattices in $SL_2(\mathbf{Q}_p)^2$ are QI to products of two free groups; still I don't know if they are all known to be not left-orderable. Same question for irreducible cocompact lattices in $SL_2(\mathbf{R})^2$, which are QI to products of 2 surface groups. $\endgroup$– YCorCommented Sep 7, 2014 at 10:54
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3$\begingroup$ finally I found examples not VLO: let $H,H'$ be two torsion-free, bilipschitz groups, with $H$ LO and $H'$ not LO and perfect (e.g. virtually abelian, or cocompact Kleinian). Then the wreath products $G=H\wr\mathbf{Z}$ and $G'=H'\wr\mathbf{Z}$ are QI, but $G$ is LO while $G'$ is not VLO. $\endgroup$– YCorCommented Sep 7, 2014 at 21:03
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2 Answers
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This answer adds some details, and some references, following Ian's comment.
First the references:
https://arxiv.org/abs/math/0211110, https://arxiv.org/abs/math/0203192v2
Now, for the answer: In the Boyer, Rolfsen, Wiest paper (first reference) we find:
Theorem 1.8 For each of the eight 3-dimensional geometries, there exist closed, connected, orientable 3-manifolds with the given geometric structure whose fundamental groups are left-orderable. There are also closed, connected, orientable 3-manifolds with the given geometric structure whose groups are not left-orderable.
As Ian mentions, for any pair of closed three-manifolds $M$ and $N$ with the same Thurston geometry, the fundamental groups $\pi_1(M)$ and $\pi_1(N)$ are quasi-isometric.
In the second reference, by Calegari and Dunfield, there is a table of closed hyperbolic rational homology three-spheres, of low volume, with the orderability of the fundamental group given as orderable, non-orderable, or unknown. It looks like this is a difficult property to determine. Also, in this situation, orderability appears to be rare (but possible!).
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There are also lots of amenable examples: by a theorem of P. Linnell and D. Witte-Morris (in this paper), an amenable group is left-orderable if and only if it is locally indicable (i.e. any non-trivial subgroup surjects onto $\mathbb Z$). It is not hard to see that this implies that any crystallographic group with non-solvable holonomy is not left-orderable; on the other hand any such group contains a free abelian group of finite index, and the latter is certainly orderable. One can also get examples more complicated olycyclic examples.
A simpler crystallographic example is given by the subgroup of ${\rm Isom}(\mathbb{R}^3)$ generated by three half-turns with disjoint axes of orthogonal directions (its abelianization is finite; in fact it has a presentation given by $$ \langle a,b | ab^2a^{-1}=b^{-2},\, ba^2b^{-1} = a^{-2}\rangle $$ (see for example the last section in this paper of Bowditch).
