Dynamic of $SL_2(\mathbb{Z}$) on $\mathbb{C}^2$ I imagine the dynamic of $SL(2,\mathbb{Z}$) on $\mathbb{C}^2$ has been studied. Does one know if it is recurrent or ergodic (with respect to the Lebesgue measure) ? Is there any explicit description of the orbits ? 
Thank you for your help, 
Selim
 A: The dynamics are ergodic with respect to Lebesgue measure. See 

G. Hedlund, Fuchsian groups and mixtures, Ann. of Math. Volume 40, Number 2 (1939) 370-383, available here. 

If you prefer something with modern terminology, I'd recommend

F. Ledrappier and M. Pollicott, Ergodic properties of linear actions of (2×2)-matrices, Duke Math. J. Volume 116, Number 2 (2003), 353-388, see here.

A: Since matrices from $SL(2,\mathbb Z)$ are real, the action on $\mathbb C^2$ is the same as the diagonal action on the product of two copies of $\mathbb R^2$. Now, the action of $SL(2,\mathbb Z)$ on $\mathbb R^2$ is well-known to be conjugate to its action on the space $H$ of horocycles in the hyperbolic plane $\mathbb H^2$. The ergodic properties of the latter action are the same as for the horocycle flow on the quotient surface.
A single horocycle is determined by a point on the boundary circle of $\mathbb H^2$ and a real parameter (the "radius" of the horocycle). Two horocycles therefore determine a geodesic in $\mathbb H^2$ (which joins the centers of these horocycles) and two points on this geodesic (where it intersects these horospheres). Therefore, the action of $SL(2,\mathbb Z)$ on $H^2$ is conjugate to the product of its action on the unit tangent bundle of $\mathbb H^2$ and the trivial action on $\mathbb R$. Finally, the action of $SL(2,\mathbb Z)$ on $U\mathbb H^2$ is obviously dissipative.
Thus, the answer is that the action of $SL(2,\mathbb Z)$ on $\mathbb C^2$ is dissipative, 
and its ergodic components ($\equiv$ orbits) are parameterized by the product of the unit tangent bundle of the quotient surface and $\mathbb R$.
Actually, the above argument shows that the action is not just dissipative, but also properly discontinuous.
Edit: Sorry, in the original answer forgot about one more copy of $\mathbb R$.
