Let $G_1$ and $G_2$ be two finitely generated groups which are quasi-isometric in the sense of geometric group theory.
Are their rational cohomology rings $H^{\ast}(G_i; \mathbb Q)$ necessarily isomorphic?
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asked Sep 23, 2015 at 0:50
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5$\begingroup$ For finitely generated nilpotent groups the answer is positive by work of Shalom and Sauer. $\endgroup$– YCorCommented Sep 23, 2015 at 9:04
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2$\begingroup$ The infinite dihedral group $\mathbb{Z}/2\mathbb{Z}\ast \mathbb{Z}/2\mathbb{Z}$ and the integers $\mathbb{Z}$ are quasi-isometric, so Yves' observation does not hold for virtually nilpotent groups. $\endgroup$– Ian AgolCommented Sep 23, 2015 at 13:46
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$\begingroup$ It's not an "observation" but a deep theorem. $\endgroup$– YCorCommented Nov 5, 2015 at 0:57
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2 Answers
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No. For example, $F_2$ is quasi-isometric to $F_3$ because the latter is finite index in the former, but their rational $H^1$s differ.
answered Sep 23, 2015 at 2:30
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$\begingroup$ And here is a machine for producing lots of counterexamples. Let $F$ be a finitely generated group on which a finite group $G$ acts. Then $H^{\bullet}(F \rtimes G, \mathbb{Q}) \cong H^{\bullet}(F, \mathbb{Q})^G$, but $F$ is finite index in $F \rtimes G$ so again they are quasi-isometric. $\endgroup$ Commented Sep 23, 2015 at 3:19
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$\begingroup$ Ian Agol's example in the comments can be obtained by taking $F = \mathbb{Z}$ and $G = \mathbb{Z}_2$ acting by $-1$. $\endgroup$ Commented Sep 23, 2015 at 19:59
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No. Any two groups acting properly and cocompactly by isometries on the same locally compact space $X$ are quasi-isometric to $X$ and hence to each other (Milnor-Svarc Lemma).
For example all fundamental groups of closed hyperbolic $3$-manifolds are quasi-isometric to hyperbolic 3-space. But not all closed hyperbolic 3-manifolds have the same rational cohomology.
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$\begingroup$ For a survey on quasi-isometric rigidity questions (and in particular positive answers in some special cases) see math.utah.edu/pcmi12/lecture_notes/kapovich.pdf $\endgroup$– ThiKuCommented Sep 23, 2015 at 2:35