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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

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1  
-1 How does it act? –  plusepsilon.de Oct 21 '13 at 13:34
3  
@MarcPalm I assume in the "obvious way" that 2x2 matrices act on a two-dimensional vector space?! –  Igor Rivin Oct 21 '13 at 13:47
    
By linear transformation. Namely a matrix $ \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right ) $ acts by $ \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right ) \cdot \left ( \begin{array}{c} u \\ v \end{array} \right ) = \left ( \begin{array}{c} au + bv \\ cu + dv \end{array} \right ) $. I thought it was obvious :) –  Selim G Oct 21 '13 at 13:49
    
Unless the question is not edited, I can't undo my downvote:( –  plusepsilon.de Oct 22 '13 at 13:46

2 Answers 2

up vote 8 down vote accepted

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 $\mathbb R$.

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The trivial action of $SL(2,\mathbb{Z})$ on $\mathbb{R}$ is the action by linear fractional transformations ? –  Selim G Oct 22 '13 at 7:22
    
No-no-no - what I mean is the really trivial action - the one which does nothing. –  R W Oct 22 '13 at 21:14
    
It seems to me that your answer contains some confusions. The good reference is Ledrappier-Pollicott below. The linear action of SL(2,C) on $C^2$ is (up to some details) the isometric action of SL(2,C) on the space of horospheres of $H^3$. This action is not dissipative, and admits dense orbits for example.(You can replace $\C$ by $\R$ and $\H^3$ by $\H^2$ above) –  Barbara Schapira Oct 30 '13 at 14:32
    
Did you have a chance to notice that the OP asked about the action of $SL(2,\mathbb Z)$? Or do you want to say that in this situation there is no difference between $\mathbb C$ and $\mathbb Z$? If so, then you are definitely wrong. –  R W Oct 30 '13 at 14:51
    
Sorry, I wrote my comment too fast; it contains many imprecisions. In fact, my first reaction was due to te fact that I was confused by notations $H^2$ and $\mathbb{H}^2$ that I identified in a first reading. –  Barbara Schapira Oct 30 '13 at 15:24

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.

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