Let $ \Gamma $ be a finitely generated torsion free group. Does there always exist a Lie group $ G $ containing $ \Gamma $ (or really a subgroup isomorphic to $ \Gamma $) such that $ G/\Gamma $ is compact?
In other words, can every finitely generated torsion free group be realized as a cocompact lattice in some Lie group?
Best Answer
Yes and no.
Why yes: Every finitely-generated group $\Gamma$ is countable and every countable group with discrete topology is a Lie group. Clearly, $\Gamma$ is a (cocompact) lattice in $\Gamma$.
However, I do not think you are interested in Lie groups with infinitely many components. Hence, suppose that $G$ is a Lie group with finitely many components and $G_0< G$ is the identity component. For every lattice $\Gamma< G$, $\Gamma_0:= \Gamma\cap G_0$ is a lattice in $G_0$; the index $|\Gamma: \Gamma_0|$ is, of course, finite.
Connected Lie groups need not be linear (over real numbers), but one can use the adjoint representation $Ad_{G_0}$ to obtain a faithful matrix representation of $G_0/Z(G_0)$, where $Z(G_0)$ is the center of $G_0$.
In particular, the quotient of $\Gamma_0$ by a subgroup of its center (namely, $\Gamma_0\cap Z(G_0)$) is a matrix group. There are many algebraic restrictions on finitely generated matrix groups, for instance, they are residually finite, hence, always admit proper finite index subgroups.
Thus, to get an example of a finitely-generated group which is not a lattice in any Lie group with finitely many components, you can take any infinite finitely generated group without proper finite index subgroups, e.g. the Higman group given by the presentation:
$$ \langle a, b, c, d| aba^{-1}=b^2, bcb^{-1}=c^2, cdc^{-1}=d^2, dad^{-1}=a^2\rangle. $$
There are many other examples, for instance, due to the fact that lattices have finite virtual cohomological dimension, etc.