Can the intersection of two non-commensurable lattices be Zariski-dense?

Question

Let $\Gamma_1$ and $\Gamma_2$ be lattices of a simple non-compact Lie group $G$. Suppose that $\Gamma_1\cap\Gamma_2$ is Zariski-dense in $G$. Can we conclude that $\Gamma_1\cap\Gamma_2$ is also a lattice?

For instance, if one consider the following lattices of $\operatorname{SL}_3(\mathbb{R})$: $$\operatorname{SU}(I_3,\sigma;\mathbb{Z}[\sqrt{d}])=\{M\in\operatorname{SL}_3(\mathbb{Z}[\sqrt{d}])\mid\sigma(M)^\top M=I_3\}$$ where $d\in\mathbb{N}$ is square-free and $\sigma$ is the non-trivial involution of $\mathbb{Q}(\sqrt{d})$. Their intersection is always finite. With suitable conjugate of these subgroups, one can make the intersection being a lattice of $\operatorname{SO}(2,1)$.

Answer 1

It appears that you are asking two different questions:

1. In the OP: Suppose that $\Gamma_1, \Gamma_2< G$ are lattices and $\Gamma_1\cap \Gamma_2$ is Zariski dense in $G$. Does it follow that $\Gamma_1, \Gamma_2$ are commensurable?

Counter-examples exist among lattices (say, one is arithmetic and the other is not) in $G=O(n,1)$ for all $n\ge 2$, see

Gromov, M.; Piatetski-Shapiro, Ilya I., Non-arithmetic groups in Lobachevsky spaces, Publ. Math., Inst. Hautes Étud. Sci. 66, 93-103 (1988). ZBL0649.22007.

You can find pdf file here. Examples are much easier to construct for $n=2$.

2. In comments: Suppose that $\Gamma_1, \Gamma_2< G$ are arithmetic lattices in a reductive Lie group and $\Gamma_1\cap \Gamma_2$ is Zariski dense in $G$. Does it follow that $\Gamma_1, \Gamma_2$ are commensurable? The positive answer is given in 1.6 (Commensurability Criterion) of the same paper.