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).


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.


The function you want is, as David mentioned, a modular function. It appears in almost any proof of the Little Picard Theorem (I believe that Ahlfors' text on complex analysis has some discussion) and can be constructed using abstract nonsense by finding a conformal mapping of the domain

1 > Re(z) > 0, |z-1/2| > 1/2

onto the upper half-plane and then bouncing it around using Schwarz reflection. More concretely the group of Gamma(2) of 2x2 integer matrices that reduce to zero mod 2 acts freely (except for minus the identity matrix) on the upper half-plane by linear fractional transformations, and this gives an explicit action of the free group on two generators by deck transformations.

This leads into modular functions and forms which are, to put it mildly, well-studied.

Most introductory texts on modular forms usually have a thorough discussion and proof of this covering - for example, Goro Shimura's "Introduction to the Arithmetic Theory of Automorphic Functions".

EDIT: whoops, if I'm going to say "Schwarz reflection" I want half of the fundamental domain.

  • $\begingroup$ Aha! That's what screwed me up. I had the mental image of half the fundamental domain in my head, and that lead me to the wrong group. By the way, if someone would vote Tyler's answer up, I'd appreciate it. He definitely deserves to be above me, since he got the details right and I didn't. $\endgroup$ – David E Speyer Oct 17 '09 at 5:18
  • $\begingroup$ Very nice. I believe the correct modular function is called the Lambda modular function; it is indeed described in Ahlfors' book on complex analysis. Does this idea work for free groups on more generators? I don't know much about the algebraic properties of congruence subgroups, but it definitely seems plausible as the upper half plane is going to be the universal cover again. Although the fundamental domains are going to look pretty complicated. Is there a computer package or something that deals with this (algebra of congruence subgroups and plots of corresponding fundamental domains)? $\endgroup$ – Sam Derbyshire Oct 17 '09 at 15:39
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    $\begingroup$ Incidentally if you use the z -> 2z map to replace \Gamma(2) with the isomorphic group \Gamma_0(4), the map is given by the trace of an element of order 4 (conjugacy class 4A) acting on the monster vertex algebra. It can be expressed (up to a constant shift of 24) as \Delta(2z)^2/(\Delta(z)\Delta(4z)). If you want covering maps for free groups with more generators, you will have to account for moduli, i.e., the maps aren't necessarily taken to each other by Mobius transformations. $\endgroup$ – S. Carnahan Oct 17 '09 at 23:30
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    $\begingroup$ I think MAGMA and SAGE can work with congruence subgroups. H. Verrill has written a java program that draws pictures of fundamental domains. $\endgroup$ – S. Carnahan Oct 17 '09 at 23:35

I am sorry to revive such an old topic. But it really needs to be mentioned that this universal cover of the twice punctured plane was constructed by PICARD, and as one of the comments says, this was used in the proof of little Picard (and also big Picard). This is one of Picard's most famous constructions, so you will excuse my emphasizing it. This is explicitly constructed in Ahlfors' book on complex analysis.


The function you want is a modular function. The universal cover of C \ {0,1} is the upper half plane; a fundamental domain is { z : 0 < Re(z) < 1, |z-1/2| > 1/2 }. This is also a fundamental domain for the action of \Gamma_0(2) on the upper halfplane, where \Gamma_0(2) is the group of integer matrices whose lower left hand entry is even. There is a standard construction of a modular function which has this symmetry; but I'm forgetting the terminology. Scott will probably come along soon and fill in the details I'm missing.

A good reference for this sort of thing is Conformal Mapping, by Zeev Nehari.

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    $\begingroup$ Isn't the fundamental domain -1 < Re(z) < 1, |z - 1/2| > 1/2, |z + 1/2| > 1/2, using Gamma(2) level structures (integer matrices which are the identity mod 2)? $\endgroup$ – Tyler Lawson Oct 17 '09 at 4:03
  • $\begingroup$ You are right, I am wrong. It is $\Gamma(2)$ which is isomorphic to the free group on 2 generators, not $\Gamma_0(2)$. If you want to write an answer explaining in more detail, I'll gladly vote it up. $\endgroup$ – David E Speyer Oct 17 '09 at 4:12
  • $\begingroup$ Ok good, I was getting worried, but didn't dare to say anything. I started thinking of modular forms as soon as I realised the covering space was the upper half plane, but I didn't really see much point in it. I tried coming up with holomorphic modular functions never attaining 0 or 1, but that obviously failed. I'm very interested in hearing about how it goes in this case though; I love modular forms! $\endgroup$ – Sam Derbyshire Oct 17 '09 at 4:43

I think you're looking for polylogarithms. It's not just one function you're after; the fundamental group of the once-punctured plane is cyclic and doesn't have a ton of interesting representations, while the fundamental group of the twice-punctured plane is free of rank two and thus has a lot more beef.

Note, though, that the polylogarithms all have unipotent monodromy; you should think of them as seeing, not quite the full fundamental group of C - 0,1, but the pro-unipotent algebraic envelope of this. But as we learned from Deligne's monograph on the fundamental group of P^1 minus three points, there's a huge amount of content even here.

Not sure what reading to recommend for an introduction to this story, except for Deligne's paper itself, but if you search for things containing some combination of polylogarithm, multizeta, and iterated integral you'll find tons; among these, find something that suits your taste....


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