In the literature, people seem to predominantly look at lattices in nilpotent or reductive groups. Is there a result that gives a general description of a lattice in an arbitrary Lie group? Something like how a general lattice is built out of components of the reductive or nilpotent case?

  • $\begingroup$ I guess it is just that one has no general theory for solvable groups, so this gets not discussed in textbooks. $\endgroup$ – ThiKu Aug 14 at 8:48
  • $\begingroup$ A lot of information about lattices in solvable Lie groups (and arbitrary too) is given in Raghunathan's book. Also the non-solvable non-reductive case is interesting and subtle sometimes. For instance, the question when a semidirect product $S\ltimes V$, $S$ semisimple, $V$ $S$-module, has a cocompact lattice, is a subtle one. $\endgroup$ – YCor Aug 14 at 10:15
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    $\begingroup$ O. Baues (who has several papers on the subject) writes in MR: “As is well known a nilpotent Lie group admits a lattice if and only if its Lie algebra admits rational constants of structure. For solvable Lie groups the situation is more involved and the existence of lattices depends on the arithmetic properties of the restriction of the adjoint representation on the nilradical. Criteria may be found in Auslander (1973), Saito (1961).” $\endgroup$ – Francois Ziegler Aug 14 at 10:28
  • $\begingroup$ My question is about the structure of the lattices, when they exist, not about whether they exist. For instance, whether the intersection with the commutator subgroup is a lattice in the commutator subgroup. The same for the radical, the unipotent radical and so on. $\endgroup$ – Zero Aug 14 at 15:03
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    $\begingroup$ In $SO(3)\times\mathbf{R}$ there's a (cyclic) lattice whose intersection with the radical is not a lattice therein (actually this is a mistake in Raghunathan's book). However the intersection with the connected amenable radical (the subgroup generated by the connected radical and by compact simple subgroups) is always a lattice. $\endgroup$ – YCor Aug 14 at 16:22

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