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$SL_2(\Bbb Z)$ acts on ${\Bbb R}^2$ fixing set-wise ${\Bbb Z}^2$, so $SL_2(\Bbb Z)$ acts on ${\Bbb R}^2\setminus {\Bbb Z}^2$, and then on the universal covering space of ${\Bbb R}^2\setminus {\Bbb Z}^2$, call it $U$.

Can one not then give $U$ a hyperbolic metric so that $SL_2(\Bbb Z)$ acts by isometries?

And doing so, does one recover, up to conjugacy, the usual action of $SL_2(\Bbb Z)$ on the hyperbolic disk?

Finally then, how explicit can one make the uniformization of $U$?

Edit: Mischa's spot-on comments make this a question in search of a better question. I considered deleting, but perhaps some further interesting comments will emerge.

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    $\begingroup$ $SL(2,Z)$ does not act on the universal cover; the group acting on the universal cover is a certain extension of $SL(2,Z)$ by a free group of infinite rank. Furthermore, the action of $SL(2,Z)$ on $R^2\setminus Z^2$ has no invariant Riemannian metric (or conformal structure). $\endgroup$
    – Misha
    Apr 25, 2013 at 2:20
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    $\begingroup$ Of course, you can take a finite-index free subgroup in $SL(2,Z)$ whose action will lift to $U$, but the lift will be highly noncanonical. One of these lifts will be properly discontinuous, so it will preserve a hyperbolic metric (again, highly noncanonical), but there is nothing special about this action. $\endgroup$
    – Misha
    Apr 25, 2013 at 2:47
  • $\begingroup$ >$SL(2,Z)$ does not act on the universal cover; the group acting on the universal cover is a certain extension of $SL(2,Z)$ by a free group of infinite rank. Is my mistake only this: $SL_2(Z)$ acts on the universal covering space of $R^2\setminus Z^2 \cup \{(0,0\}$? $SL_2(Z)$ fixes $(0,0)$; I can represent elements of the universal covering space by paths (up to deformation) leaving from $(0,0)$ and terminating wherever. Then $SL_2(Z)$ acts on the paths and homotopies between them. $\endgroup$ Apr 25, 2013 at 4:28
  • $\begingroup$ David: Then the group will lift but the lift will be ugly: Fixing a common point in the "new" $U$. (This common fixed point is forced by the nontrivial center in $SL(2,Z)$.) There will be no invariant Riemannian metric (or conformal structure) for such a lift. The action of $SL(2,Z)$ on $R^2$ belongs to the field of dynamical systems and, I believe, one should ask "dynamical questions" about it rather than geometric questions. $\endgroup$
    – Misha
    Apr 25, 2013 at 5:14
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    $\begingroup$ Misha, you are not quite right. The action of $SL(2,Z)$ on $R^2∖ 0$ (rather, on the quotient of this space by the central symmetry) has a clear geometrical interpretation. It is isomorphic to the action of $SL(2,Z)$ on the space of horocycles in the hyperbolic plane (i.e., to the extension of the boundary action by the Busemann cocycle). $\endgroup$
    – R W
    Apr 25, 2013 at 6:46

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Here is the issue: $SL(2,Z)$ (linear) action on your space $X = R^2\setminus Z^2 \cup (0,0)$. You described an action that does indeed lift to the very large space of paths. But what isn't clear is that this induced action on the covering space is the standard action of the modular group on the Poincaré disc model.

[edit: removed bad example.]

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