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It's a consequence of the uniformization theorem for simply connected Riemann surfaces that the universal cover of $\mathbb{C}\setminus(\mathbb{Z}\oplus i\mathbb{Z})$ ($\mathbb{C}$ punctured at all the integral lattice points) is the upper half plane $\mathcal{H}$.

How should I think about this map? How does the map behave near the missing lattice points?

A related question is this: $\mathcal{H}$ is also the universal cover for a punctured torus, whose fundamental group is $F_2$, the free group on two generators. By comparing the punctured torus to the wedge of two circles, I feel like the universal cover for the punctured torus, ie $\mathcal{H}$, ought to be deformation-retractable to an infinite 4-regular tree. Ie, the infinite 4-regular tree ought to be able to be embedded in $\mathcal{H}$ such that the vertices of the tree all lie on the boundary of $\mathcal{H}$. What does this tree look like in $\mathcal{H}$?

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  • $\begingroup$ I actually mean $\mathbb{C}$ minus all the lattice points. $\endgroup$
    – Will Chen
    Sep 5, 2013 at 20:24
  • $\begingroup$ Ah, sorry, I should have read more carefully. $\endgroup$ Sep 5, 2013 at 20:58
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    $\begingroup$ I did not understand the second question. The universal cover of the punctured torus is the open unit disc. $\endgroup$ Sep 5, 2013 at 22:24
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    $\begingroup$ Alexandre Eremenko: The Riemann mapping theorem shows that the open unit disc is conformally equivalent to the upper half plane. $\endgroup$
    – Will Chen
    Sep 5, 2013 at 22:59
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    $\begingroup$ The infinite 4-regular tree embeds nicely in the unit disk as the dual graph to the tesselation by regular ideal quarilaterals described by Prof. Eremenko below. This is because the puntured torus is covered by your space $\mathbb{C} \setminus (\mathbb{Z} + i \mathbb{Z})$. Of course, you can lift any deformation-retraction of the punctured-torus onto the wedge of two circles to a deformation-retraction of the unit disk onto the infinite tree. $\endgroup$ Sep 7, 2013 at 0:17

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On the first question (the universal cover of the complement of a lattice). The missing points are in the image, so it is not the map that "behaves" but the inverse map. The inverse map behaves in a very simple way: it has infinitely many "logarithmic singularities" over each missing point. "How to think about the map" is not a well defined question. But the way I think about it is this. Consider the circular quadrilateral inscribed in the unit disc, say with vertices 1,-1,i,-i; the sides are arcs of circles orthogonal to the unit circle. All angles of this quadrilateral are 0. There is a conformal homeomorphism of this circular quadrilateral onto a (rectilinear) square with vertices 0,1,1,1+i, sending vertices to vertices. By Schwarz's symmetry principle, applied very many times, the map extends to a map from the unit disc to the plane minus the lattice. This is our universal covering map. You can make a picture. You can express it in terms of special functions (it is a ratio of two solutions of a very special Heun equation, linear differential equation of second order with regular singular points at 1,-1,i,-1.

EDIT: I am not sure what exactly you want to know, in your question you mention visualization, rather than computation, but once I computed this map, arXiv:1110.2696. It can be expressed in terms of hypergeometric functions. And I also asked a MO question related to it: Maximum of a function of one variable.

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  • $\begingroup$ This makes things very clear. $\endgroup$
    – Lubin
    Sep 6, 2013 at 11:01

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