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Is there a geometric interpretation of a surface subgroup being Zariski dense? Or, conversely, given a $\Pi_1$ injective surface in a 3-manifold, is there a geometric or topological requirement on the surface so that the corresponding group is Zariski dense?

The theorems I've seen that show existance of Zariski dense surface subgroups (in SL(3,Z) and SL(4,Z) for example) do it in an algebraic/number theoretic way and I'm wondering if there is another way of visualizing this problem. Specific examples would be greatly appreciated.

Thank you.

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    $\begingroup$ Zariski-dense in $G$ makes sense when $G$ is given as a subgroup of a linear group, but in the $\pi_1$ of a 3-manifold it is a priori undefined. $\endgroup$
    – YCor
    Commented Mar 21, 2018 at 23:26

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I am interpreting the question as meaning that the OP asks for a geometric interpretation of a Zariski dense surface subgroup in $SL(n, \mathbb{Z})$ for $n=3, 4.$ The existence of such has been shown by Long, Reid, and Thistlethwaite, in a series of papers, using methods of $3$-dimensional topology (and a healthy dose of algebra.)

A non-Zariski dense subgroup of $SL(3, \mathbb{Z})$ cannot be a surface subgroup (since it would have to live in $SL(2),$ and I don't think any interesting $SL(2)$s intersect $SL(3, \mathbb{Z})$ in a nontrivial way (where $SL(2, Z)$ is not viewed as interesting, since obviously it has not closed surface subgroups).

In any cAse, it is a result of Aoun that a random subgroup of $SL(n, \mathbb{Z})$ is free, and a result of yours truly that it is Zariski dense, so it is clear that surface subgroups are rare. Perhaps if someone could figure out a concrete geometric interpretation, it would help us find them, but I am not too hopeful. In fact, it is not clear to me that there is any way of showing at present that there are fundamental groups of hyperbolic $3$-manifolds which are not subgroups of $SL(4, \mathbb{Z}).$

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    $\begingroup$ Actually, $SL(3,Z)$ does contain non-Zariski dense surface subgroups, they are certain uniform lattices in $SO(2,1)$, contained in $SO(2,1)\cap SL(3,Z)$. $\endgroup$
    – Misha
    Commented Mar 22, 2018 at 21:36

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