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By a nilmanifold I mean a quotient $M =\Gamma \backslash G$ of a connected, simply-connected nilpotent real Lie group $G$ by the left action of a maximal lattice 􀀀, i.e. a discrete cocompact subgroup.

Suppose we have two maximal lattices $\Gamma_1$ and $\Gamma_2$ of $G$. What can be said about the nilmanifolds $\Gamma_1 \backslash G$ and $\Gamma_2 \backslash G$? Are they homeomorphic? Or perphaps one is a covering space of the other?

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  • $\begingroup$ In general, they are not homeomorphic, and one is not a covering space of the other. I guess it should be a nilmanifold which admits two finite covers homeomorphic to the given two. $\endgroup$
    – Anton Petrunin
    Apr 3, 2016 at 10:58
  • $\begingroup$ maximal lattice" is a strange assumption: no nontrivial simply connected nilpotent Lie group $G$ admits a maximal lattice. This is quite obvious in the abelian case (e.g., $G=\mathbf{R}$). In the general case, if $\Gamma$ is a lattice in $G$ and $Z$ is the center of $G$ (which is nontrivial), then $\Gamma\cap Z$ is a lattice in $Z$ and $M=\{x\in Z:2x\in\Gamma$ is such that $M\Gamma$ is a lattice strictly containing $\Gamma$. $\endgroup$
    – YCor
    Apr 3, 2016 at 20:59

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The manifold $\Gamma_1/G$ is homeomorphic to $\Gamma_2/G$ implies that $\Gamma_1$ is isomorphic to $\Gamma_2$ since $G$ is $1$-connected and $\Gamma_1,\Gamma_2$ are discrete, thus $\pi_1(\Gamma_1/G)=\Gamma_1$ and two homeomorphic manifolds have isomorphic fundamental groups. This implies that $\Gamma_1$ is isomorphic to $\Gamma_2$. In fact the converse is also true: $\Gamma_1$ is isomorphic to $\Gamma_2$, implies that $\Gamma_1/G$ is homeomorphic to $\Gamma_2/G$. This is a consequence of:

Raghunathan. Discrete subgroups of Lie groups Corollary 2. p.34

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