1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '16 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 '16 at 20:59
7
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.