As is well known, the universal covering space of the punctured complex plane is the complex plane itself, and the cover is given by the exponential map.
In a sense, this shows that the logarithm has the worst monodromy possible, given that it has only one singularity in the complex plane. Hence we can easily visualise the covering map as given by the Riemann surface corresponding to log (given by analytic continuation, say).
Seeing how fundamental the exponential and logarithm are, I was wondering how come I don't know of anything about the case when two points are removed from the complex plane.
My main question is as follows: how can I find a function whose monodromy corresponds to the universal cover of the twice punctured complex plane (say ℂ∖{0,1}), in the same way as the monodromy of log corresponds to the universal cover of the punctured plane.
For example, one might want to try f(*z*) = log(*z*) + log(*z*-1) but the corresponding Riemann surface is easily seen to have an abelian group of deck transformations, when it should be F2.
The most help so far has been looking about the Riemann-Hilbert problem; it is possible to write down a linear ordinary differential equation of order 2 that has the required monodromy group.
Only trouble is that this does not show how to explicitly do it: I started with a faithful representation of the fundamental group (of the twice punctured complex plane) in GL(2,ℂ) (in fact corresponding matrices in SL(2,ℤ) are easy to produce), but the calculations quickly got out of hand.
My number one hope would be something involving the hypergeometric function 2F1 seeing as this solves in general second order linear differential equations with 3 regular singular points (for 2F1 the singular points are 0, 1 and ∞, but we can move this with Möbius transformations), but I was really hoping for something much more explicit, especially seeing as a lot of parameters seem to not produce the correct monodromy. Especially knowing that even though the differential equation has the correct monodromy, the solutions might not.
I'd be happy to hear about any information anyone has relating to analytical descriptions of this universal cover, I was quite surprised to see how little there is written about it.
Bonus points for anything that also works for more points removed, but seeing how complicated this seems to be for only two removed points, I'm not hoping much (knowing that starting with 3 singular points (+∞), many complicated phenomena appear).
Best Answer
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.