Applications of the Little and Great Theorems of Picard

Question

I have completed the two famous theorems of Picard, presenting their proofs in a graduate course of Complex Analysis, but I have not managed to discover a good number interesting applications.

List of applications (rather straight-forward though):

1. If a meromorphic function on $\mathbb C$ misses three values, then it is constant.

2. The equation $f^3+g^3=1$ has non-trivial meromorphic in $\mathbb C$ solutions only if $n\le 3$.

3. If $f$ is entire and one-to-one, then it is linear.

4. If $f,g$ are entire and $g'=f(g)$, then $f$ is linear or $g$ is constant.

Could you provide any interesting applications of these theorem?

Answer 1

An interesting consequence is that if $X = \mathbf{P}^1(\mathbb C)-\{p_1, \dots, p_n\}$ is the Riemann sphere with $n$ punctures, then $\widetilde X$, the universal covering space of $X$, is the upper-half plane for $n\geq 3$. Indeed, $\widetilde X$ must be simply-connected, and by Riemann's theorem it is either $\mathbf{P^1}(\mathbb C)$, $\mathbb C$, or the upper-half plane $\mathfrak h$. It cannot be $\mathbf{P^1}(\mathbb C)$ because $X$ is not compact; it cannot be $\mathbb C$ by Picard's theorem; therefore it is $\mathfrak h$.

Corollary: For each $n \geq 3$, $\text{PSL}_2(\mathbb R)$ contains a copy of $F_n$, the free group on $n$ generators (and therefore also for $n=2$).

Indeed this follows from the theory of covering spaces; the group of deck transformation of $\widetilde X \to X$ is $\pi_1(X)\cong F_n$. On the other hand, $\text{Aut}(\mathfrak h) \cong \text{PSL}_2(\mathbb R)$.

Answer 2

In general, a Holomorphic map from $\mathbb C\to \mathbb C $ does not have a fixed point. But using the Little Picard theorem, we can prove that even though $f$ does not have a fixed point, $f\circ f$ does have a fixed point until unless $f$ is translation function.

More precisely, the statement is as follows:

Theorem: Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be holomorphic. Then $f \circ f: \mathbb{C} \rightarrow \mathbb{C}$ always has a fixed point unless $f$ is a translation $z \mapsto z+b, b \neq 0$.

Proof. Suppose $f \circ f$ has no fixed points. Then $f$ also has no fixed points (if it has then that also fixed point of $f \circ f$), and it follows that $$g(z):=\left.\frac{f(f(z))-z} {f(z)-z}\right. $$ is entire function. This function omits the values $0$ and $1$ ( If it does then that implies flex point) ; hence, by Picard, there exists a $c \in \mathbb{C} \backslash\{0,1\}$ with $$ f(f(z))-z=c(f(z)-z), \quad z \in \mathbb{C} $$ By differentiation and rearranging terms gives $$f^{\prime}(z)\left[f^{\prime}(f(z))-c\right]=1-c .$$ Since $c \neq 1$ (if equals to 1 then it implies fixed point), $f^{\prime}$ has no zeros and $f^{\prime}(f(z))$ is never equal to $c$. Thus $f^{\prime} \circ f$ omits the values 0 and $c \neq 0$ by Picard, $f^{\prime} \circ f$ is therefore constant. It follows that $f^{\prime}=$ constant, hence that $f(z)=a z+b .$ Since $f$ has no fixed points, $a=1$ and $b \neq 0$

So we have proved that when $f\circ f$ has no fixed point, then $f$ is translation.