Given a unimodular Lie group $G$ and a discrete subgroup $\Gamma\subseteq G$, under what conditions does there exists a discrete subgroup $H$ s.t. $\Gamma\subseteq H$ and $G/H$ has finite volume? Also, can someone give an example of a discrete subgroup $\Gamma$ of a unimodular Lie Group which cannot be extended to a lattice?

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    $\begingroup$ If you take a discrete cyclic subgroup in $SL_3(\mathbf{R})$ generated by an element whose trace is transcendental, then it cannot be extended to a lattice since any lattice is arithmetic (Margulis) and hence trace of its elements are algebraic. $\endgroup$ – YCor Jan 31 '13 at 15:11

The same argument as in Yves' comment applies whenever you have a discrete subgroup of a semisimple Lie group without factors locally isomorphic to $SL(2,{\mathbb R})$, since Mostow Rigidity (or even local rigidity) suffices here (this observation is due to Selberg). The best result in the positive direction is due to Robert Brooks: If $\Gamma$ is a geometrically finite subgroup of $PO(3,1)$ then it admits arbitrarily small deformations $\Gamma_i$ (isomorphic to $\Gamma_i$) all of which are subgroups in lattices. In view of the later density theorem of Brock and Bromberg, one can replace geometric finiteness assumption here with merely "finitely generated".

In the case when $G=PO(2,1)$ then one can do the same without deformation, you just need to assume that $\Gamma$ is finitely generated (unlike Brooks' result, this observation is elementary: If $S$ is a finite area hyperbolic surface/orbifold with geodesic boundary, then the double of $S$ across the boundary is a finite area hyperbolic surface without boundary). However, if you allow infinitely generated discrete subgroups of $PO(2,1)$, then again there are counter-examples: It suffices to take infinite area hyperbolic surfaces with non-discrete length spectrum: They cannot cover finite area hyperbolic surfaces.

Also, if you are willing to re-embed $\Gamma$ in another Lie group, then it suffices to assume that $\Gamma$ is a subgroup of $G(O)$, where $G$ is an algebraic group (say, over the integers) and $O$ is the ring of integers of a number field $F$ (no need to assume discreteness here). Then $G(O)$ is a lattice in a (finite) direct product of Lie groups $$ \prod_{\nu} G(F_\nu), $$ where each $F_\nu$ is either ${\mathbb R}$ or ${\mathbb C}$ (this construction is called restriction of scalars). In the case of groups of rank $\ge 2$ this is essentially optimal by Margulis' arithmeticity theorem.

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  • $\begingroup$ A quick follow up. When you re-embed $\Gamma$ into another Lie group, does the original group $G$ also embed into that group? $\endgroup$ – Davis Jan 31 '13 at 17:20
  • $\begingroup$ @Davis: The group of, say, real points $G({\mathbb R})$ does embed in the new Lie group $L$, but this embedding (in general) does not agree with the embedding of $\Gamma$ as a subgroup of a lattice in $L$. $\endgroup$ – Misha Jan 31 '13 at 17:41

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