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Let $G$ be a group quasi-isometric to the fundamental group of a genus 2 surface group $H$. It is well known that $G$ is quasi-isometrically rigid, i.e. $G$ and $H$ are virtually isomorphic. Does the stronger property, that $G$ and $H$ are commensurable, also hold?

If so is there a reference for this?

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1 Answer 1

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Yes (I assume that by "virtually isomorphic" you mean commensurable modulo finite kernels, which is a nonstandard misleading use of "virtually"). This is because surface groups have Serre's property D$_2$ meaning that each 2-cohomology class (in a finite abelian group with trivial action) it trivial on some finite index subgroup.

(Also a side remark: the known result on QI rigidity is stronger, since it says that every group QI to this surface group is, modulo a finite kernel, isomorphic to a cocompact lattice in the isometry group of the hyperbolic plane. I.e., there's a structural statement which does not require passing to a finite index subgroup.)

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  • $\begingroup$ Serre calls this property $D_2$ in Cohomologie galoisienne (I.2.6) (or at least it seems to me to be equivalent). $\endgroup$ Commented Sep 5, 2021 at 10:26
  • $\begingroup$ @Carl-FredrikNybergBrodda thank you, I've edited accordingly! $\endgroup$
    – YCor
    Commented Sep 5, 2021 at 10:29
  • $\begingroup$ (Just to elaborate about "misleading": for instance "virtually torsion-free" means that some finite index subgroup is torsion-free. The property that some finite index subgroup modulo some finite normal subgroup is torsion-free, is strictly weaker.) $\endgroup$
    – YCor
    Commented Sep 5, 2021 at 10:31
  • $\begingroup$ Yes virtually covirtually isomorphic would be better terminology but for whatever reason virtually isomorphic seems to be the standard usage. $\endgroup$
    – Sam
    Commented Sep 5, 2021 at 10:36
  • $\begingroup$ Thanks for your answer! $\endgroup$
    – Sam
    Commented Sep 5, 2021 at 10:37

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