TITLE: Orbit of an irreducible representation of a surface group under that action of the mapping class group
QUESTION [9 upvotes]: Let $F$ be a closed oriented surface of negative Euler characteristic. Let $X^i(F)$ be the subset of the $SL_2\mathbb{C}$-character variety of the fundamental group of $F$ corresponding to irreducible representations.  The mapping class group of $F$, $\mathcal{M}(F)$ acts on $X^i(F)$.  Let $[\rho]\in X^i(F)$, is the orbit of the $[\rho]$ under the action of the mapping class group,
$$\mathcal{M}(F).[\rho]$$ Zariski dense in $X(F)$?

REPLY [3 votes]: As some further evidence for a positive answer to your question,
there is a paper of Cantat and Loray that proves a relative version of this for mapping class group the 4-holed sphere (rel boundary). In this case, the variety of representations (with fixed holonomy around each boundary curve) is a cubic surface of the form $$x^2+y^2+z^2+xyz= Ax+By+Cz+D,$$
where $A,B,C,D$ depend on the holonomy of the boundary components, and $x,y,z$ are the traces of simple loops meeting pairwise in two points. In Theorem D of the paper, they prove that there is no invariant curve. Hence any orbit must be either finite or Zariski dense. 
I believe one could leverage this theorem into a proof in the closed case. The idea would be to find for an irreducible (Zariski dense) representation $\rho$ a 4-holed sphere subsurface whose orbit under the relative mapping class group of this subsurface is Zariski dense. Then the Zariski closure would be at least two dimensional. Repeating this for other subsurfaces by induction, I believe one could show that the mapping class group orbit is Zariski dense.