Is there any elementary (i.e. without using analytical methods like the theory of Riemann surfaces or more elaborate results from differential geometry) way to show that the universal covering of the compact oriented surface of genus $g>0$ is homeomorphic to $\mathbb R^2$?
[Math] Universal covering of compact surfaces
at.algebraic-topologygt.geometric-topology
Related Solutions
Others have already given a satisfactory qualitative description as a modular function under a suitable congruence group. Since the quotient in question is necessarily genus zero, there are explicit formulas for such functions.
I was mistaken in my comment to Tyler's answer. The function I provided there is invariant under a larger group than the one we want, and yields the universal cover for the complex plane with one and a half punctures.
The Dedekind eta product eta(z/2)^8/eta(2z)^8 is not only invariant under Gamma(2) (= F2), but it maps H/Gamma(2) bijectively to the twice punctured plane. You will have to post-compose with a suitable affine transformation to move the punctures to zero and one. An alternative description is: eta(z)^24/(eta(z/2)^8 eta(2z)^16).
This function arises in monstrous moonshine: eta(z)^8/eta(4z)^8 + 8 is the graded character of an element of order four, in conjugacy class 4C in the monster, acting on the monster vertex algebra (a graded vector space with some extra structure). It is invariant under Gamma0(4), which is what you get by conjugating Gamma(2) under the z -> 2z map.
Other elements of the monster yield functions that act as universal covers for planes with specific puncture placement and orbifold behavior. For the general case of more than 2 punctures, you have to use more geometric methods, due to nontrivial moduli. I think you idea of using hypergeometric functions is on the right track. I think Yoshida's book, Hypergeometric Functions, My Love has a few more cases worked out.
One of the main "gains" of the Teichmuller theory approach is that you're dealing with a ball. So you're in a situation where you can readily make analytic arguments using fixed-point theory.
Thurston's homotopy-classification of elements in the mapping class group "reducible, (pseudo) anosov, or finite-order" is one example. His argument proceeds roughly along these lines (no real details included): the mapping class group acts on Teichmuller space tautologically. Thurston defined a compactification of Teichmuller space (the "projective measured lamination space") such that the action of the mapping class group extends naturally. In particular, the compactification is a compact ball/disc. So given any element of the mapping class group, you can ask what kind of fixed points it has in this ball. Thurston's theorem is that the fixed point is in the interior if and only if the mapping is finite-order (in the mapping class group). You can think of this part as an elaboration of the theorem that isometry groups of hyperbolic manifolds are finite. There are exactly two fixed points on the boundary (and the automorphism acts as a translation along a line connecting the two points) if and only if the mapping is (isotopic to) a pseudo-anosov. A necessary and sufficient condition to be reducible is that your automorphism of the projective measured lamination space is not of the other two types, i.e. it could have one fixed point on the boundary or any number, so long as it is not precisely two acting as a translation from one to the other.
The proof of geometrization for manifolds that fibre over the circle is of course closely related.
These techniques were used to show mapping class groups satisfy the Tits alternative (which linear groups satisfy) so it was one of the big chunks of "evidence" leading people to ask the question of whether or not mapping class groups are linear.
Another application would be the resolution of the Nielsen Realization problem: http://en.wikipedia.org/wiki/Nielsen_realization_problem
The list goes on. But these are really applications of Teichmuller space to other things -- specifically not Moduli space.
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Best Answer
You can build a certain covering space of the surface $S$ rather explicitly as a nested union of closed discs $D_1 \subset D_2 \subset D_3 \subset \cdots$, each contained in the interior of the next, from which it follows that the union is $\mathbb{R}^2$ and, being simply connected, is therefore the universal covering space. To construct these discs, start by representing your surface in the usual fashion as the quotient of a $4g$-gon, with oriented 1-cells assigned labels, and with side pairings respecting orientations and labellings. The disc $D_1$ is a single copy of the $4g$-gon $Q$ with appropriate side labels, and the map $D_1 \to S$ is the one given by the quotient map. Then $D_2$ is constructed by attaching additional copies of $Q$ to the periphery of $D_2$, exactly as they ought to be: one copy of $Q$ attached to each edge of $D_1$, and additional copies of $Q$ attached to each vertex so as to fill out $4g$ copies of $Q$ attached around each vertex of $D_1$. Et cetera.