Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Little Picard's theorem is the following: Suppose $f:\mathbb{C}\rightarrow \mathbb{C}$ is entire. Then either

1) $f$ is constant

2) $f$ is surjective or

3) $f$ is onto $\mathbb{C}-\{p\}$ for some point $p\in \mathbb{C}$.

Said another way, an entire function which misses 2 points in the codomain is actually constant.

The exponential function $f(z) = e^z$, which is never $0$, shows that in general all 3 cases can arise.

As someone who knows a bit of differential geometry and algebraic topology, the proof I like (indeed, the only one I remember) goes as follows: Consider $X= \mathbb{C}-\{p,q\}$ where $p$ and $q$ are distinct complex numbers. Then an entire map $f:\mathbb{C}\rightarrow X$ lifts to the universal cover $\mathbb{D}(0,1)$ (the unit disc around $0$ in $\mathbb{C}$) of $X$. The lift of $f$ is an entire bounded function, and hence, by Liouville's theorem, is constant. This easily implies $f$ itself is constant. $\square$

The part that's hazy, to me, is the justification that the universal cover of $X$ is (biholomorphic to) the unit disc. Of course, by uniformization, the universal cover is biholomorphic to either $\mathbb{C}$, $\mathbb{D}(0,1)$, or $S^2$. Since $S^2$ is compact and $X$ is not, it cannot be $S^2$. Interestingly, if we remove only a single point from $\mathbb{C}$ (which we assume wlog is $0$), then $e^z$ is a covering map from $\mathbb{C}$ to $\mathbb{C}-\{0\}$. In other words, the universal cover a once punctured plane is $\mathbb{C}$ but the universal cover of a twice or more punctured plane is the unit disc.

Is there good intuition as to why 2 is the crucial number of punctures for which the universal cover is no longer biholomorphic to $\mathbb{C}$?

Relatedly (and nicer since it would allow me to avoid using uniformization, which I feel is way overkill for this),

Is the universal covering map $\pi:\mathbb{D}(0,1)\rightarrow X$ easy, or even possible, to explicitly write down?

By translating, rotating, and scaling, one can obviously assume that $\{p,q\} = \{0,1\}$

A quick google search shows the answer to the second question should be "yes", and that $\pi$ can be expressed in terms of modular functions. Sadly, I don't know anything at all about modular functions, but I couldn't find anyting that tells how to express $\pi$ with modular functions.

share|improve this question
    
That's a nice argument, but uniformization is a couple of orders of magnitude harder to prove that Picard's theorem, no? –  Mariano Suárez-Alvarez Jul 10 '11 at 17:58
    
@Mariano: You're right - that's a major reason for my second question. –  Jason DeVito Jul 10 '11 at 19:04
    
One can construct that covering map using riemann's mapping theorem, some hyperbolic geometry (to construct a tiling), Schartz's reflection principle to do analytic continuation and lots of determination, I think; maybe Ahlfors does something similar in his book, iirc. –  Mariano Suárez-Alvarez Jul 10 '11 at 20:21
1  
Alternatively, another way to do this if we didn't know the answer is: find two elements of PGL(2) which generate a free group, and construct a differential equation on the complement of 0 and 1 which has them as monodromy—that would prove the function exists implicitly. –  Mariano Suárez-Alvarez Jul 10 '11 at 20:47
    
Not really an answer, but: there are easy one-line proofs of the Little Picard theorem that do not invoke uniformisation, in the case where $f$ can be extended to a holomorphic map on the Riemann sphere. This leaves only the case where $f$ has an essential singularity at $\infty$, where we see that the Picard theorems are strengthenings of the Casorati–Weierstrass theorem. –  Zhen Lin Jul 11 '11 at 1:06

1 Answer 1

up vote 8 down vote accepted

$\mathbb{C} - \{ 0, 1 \}$ can be thought of as the modular curve $Y(2) \cong \mathbb{H}/\Gamma(2)$ parameterizing elliptic curves together with a basis for their $2$-torsion. This parameterization takes a point $\lambda \in \mathbb{C} - \{ 0, 1 \}$ to the elliptic curve $y^2 = x(x - 1)(x - \lambda)$ together with the ordered basis $(0, 0), (1, 0)$ (say) for the $2$-torsion. Explicit formulas for the covering map can be found, for example, in Dolgachev's Lectures on Modular Forms, Lecture 9.

As for intuition, I suppose one could say the following: generically we should expect a hyperbolic structure. $\mathbb{C}$ is not generic as it has a group structure, and $\mathbb{C} - \{ 0 \}$ is also not generic as it also has a group structure, but $\mathbb{C} - \{ 0, 1 \}$ has no obvious group structure so we can expect generic behavior.

share|improve this answer
    
I really like the intuition! It will take me quite some time to understand the material leading up to and including lecture 9. –  Jason DeVito Jul 10 '11 at 19:18
    
@Jason: well, all you need to know (as far as getting an explicit formula) is Dolgachev's notation for theta functions (which I admit I haven't gone through myself). He expresses $\lambda$ as a quotient of the fourth powers of two such functions in Lecture 9; that's fairly explicit. –  Qiaochu Yuan Jul 10 '11 at 22:10
    
This is the same approach taken in Ullrich's Complex Made Simple, with a lot more details :) He shows that the upper half plane covers $\mathbb{C}-\lbrace0,1\rbrace$, and then later gives the same proof the OP did for Little Picard. –  user641 May 11 '12 at 23:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.