Lifting a representation from a discrete subgroup Let $\pi_0$ be an irreducible representation of a discrete subgroup $\Gamma$ of the reductive group $G$. My question in general is

What can be said of representations $\pi$ of $G$ extending $\pi_0$, in the sense that $\pi$ acts on $\Gamma$ as $\pi_0$?

I would like to know technical possibilities in this direction, even assuming freely further properties on $\pi_0$. What if in particular if $\pi_0$ is finite dimensional?
The base case $\Gamma = SL(2, \mathbb{Z})$ and $G=SL(2, \mathbb{Q})$ is already of interest for me.
 A: If $\Gamma$ is a lattice in a connected semisimple $\mathbb{R}$-algebraic group $G$ without compact factors, then the Borel density theorem says that $\Gamma$ is Zariski dense in $G$.  This implies that there can be at most one extension of a representation of $\Gamma$ to a rational representation of $G$.
Of course, extensions need not exist.  For finite-dimensional representations, however, you can often get information using the Margulis superrigidity theorem.  This says that if $\Gamma$ is a lattice in a semisimple Lie group $G$ satisfying some technical assumptions and if $V$ is a finite-dimensional representation of $\Gamma$, then there exists a finite-index subgroup $\Gamma'$ of $\Gamma$ such that the action of $\Gamma'$ on $V$ extends to a rational representation of $G$.
Passing to a finite-index subgroup is necessary here.  For instance, if the action of $\Gamma$ on $V$ factors through a finite group, then you first have to pass to a subgroup $\Gamma'$ that acts trivially on $V$ (and the extension to $G$ is then the trivial action).
The most important technical assumption in the superrigidity theorem is that $G$ is of higher rank.  It applies to $\text{SL}(n,\mathbb{Z})$ in $\text{SL}(n,\mathbb{R})$ for $n \geq 3$, but not for $n=2$.
