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There is some question from geometric group theory: One wonders if the following conditions are equivalent for finitely generated torsion-free nilpotent groups $\Gamma$ and $\Lambda$:

  1. $\Gamma$ and $\Lambda$ are quasi-isometric.
  2. The Mal'cev completion of $\Gamma$ is isomorphic to the Mal'cev completion of $\Lambda$.

I am wondering in what direction this question is most useful. Is it easier to find the Mal'cev completion of these groups rather than proving these are quasi-isometric? Is it the other way around? Or is there no clear answer (ie. both techniques could be useful depending on the situation)?

Thank you for your effort/answer.

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  • $\begingroup$ You mean real Malcev completion (there are also rational, complex Malcev completions: for instance two f.g. nilpotent groups are commensurable iff their rational Malcev completion are isomorphic). Anyway, this is a well-known conjecture; there was a preprint by Kyed and Petersen 1 year ago but its current status is unclear. And that 1 implies 2 is well-known and very standard (being cocompact lattices in the same locally compact group a big source of quasi-isometric groups). (...) $\endgroup$
    – YCor
    Mar 7, 2016 at 13:00
  • $\begingroup$ (...) That 2 would imply 1 is very interesting because it boils down a metric problem to an algebraic classification. Note that a stronger and simpler form of the conjecture is the equivalence, for simply connected nilpotent Lie groups G,H between: 1) $G$ and $H$ are quasi-isometric 2) $G$ and $H$ are isomorphic $\endgroup$
    – YCor
    Mar 7, 2016 at 13:02
  • $\begingroup$ Yes that was indeed what I meant. Your answer is quite pleasing, but I came up with another question, that I might not have asked explicitly above: Is it hard to construct the real Mal'cev completion of an arbitrary group as described above? $\endgroup$ Mar 7, 2016 at 15:13
  • $\begingroup$ I don't know what you call "hard". It's not immediate. The real Malcev completion is constructed in Raghunathan's book "Discrete subgroups of Lie groups", the proof is p40-42 and relies on some further basic lemmas. $\endgroup$
    – YCor
    Mar 7, 2016 at 16:30

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