Let $G$ be a semisimple algebraic group, and let $\Gamma$ be a finitely generated subgroup. Given the generators of $\Gamma$, is there a good way to determine if $\Gamma$ is Zariski-dense in $G$?


For example, let $\Gamma \subseteq \mathrm{PSL}\_2$ where $ \displaystyle
\Gamma = \left\langle \gamma\_1 , \gamma\_2 \right\rangle$ and $\gamma\_1 = \begin{pmatrix} 1 & i \newline 0 & 1 \end{pmatrix}$ and $\gamma\_2 = \begin{pmatrix} 1 & 0 \newline
 1 & 1 
 \end{pmatrix}$.



Is there a good way to show that a subgroup is Zariski-dense?
1. What do you mean by "good"? In practice, a theorem of either Lubotzky or Weigel (depending on who you ask) states that if a congruence quotient is surjective for SOME prime (bigger than 3, say), then the subgroup is Zariski-dense, so generally checking for a couple of primes gives a certificate.
2. In $PSL\_2,$ Zariski-dense is equivalent to non-elementary, so your group is Zariski-dense.


**EDIT** To answer the OP's comment: in this paper: Zariski density and genericity
I Rivin - International Mathematics Research Notices, 2010, I show that two "random" matrices generate a Zariski dense subgroup. R. Aoun in this paper: Transience of algebraic varieties in linear groups and application to generic Zariski density
R Aoun - Arxiv preprint arXiv:1103.0944, 2011 - arxiv.org


Generalized it (at the cost of losing the effectiveness). Aoun also showed (in a different paper) that a random subgroup is free (which, in higher rank) means that it is not a lattice. In rank one a corresponding result (not yet written up fully) was shown by Elena Fuchs, myself, and Peter Sarnak: there you show that generically the Hausdorff dimension of the limit set is small, so again you don't get a lattice generically (of course lattices are Zariski dense by the Borel Density Theorem).


You might also look at [these slides](http://www.slideshare.net/rivinmath/hot-topics-11519548): 


