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Oct 30, 2013 at 15:24 comment added Barbara Schapira 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.
Oct 30, 2013 at 14:51 comment added R W 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.
Oct 30, 2013 at 14:32 comment added Barbara Schapira 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)
Oct 22, 2013 at 21:14 comment added R W No-no-no - what I mean is the really trivial action - the one which does nothing.
Oct 22, 2013 at 7:22 comment added Selim G The trivial action of $SL(2,\mathbb{Z})$ on $\mathbb{R}$ is the action by linear fractional transformations ?
Oct 21, 2013 at 17:09 vote accept Selim G
Oct 21, 2013 at 16:13 vote accept Selim G
Oct 21, 2013 at 16:16
Oct 21, 2013 at 15:09 history edited R W CC BY-SA 3.0
added 73 characters in body
Oct 21, 2013 at 14:55 history answered R W CC BY-SA 3.0