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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?

Yiorgos S. Smyrlis
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2 Answers2

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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, $\text{Aut}(\mathfrak h) \cong \text{PSL}_2(\mathbb R)$.

Bruno Joyal
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  • Can you explain something to me Bruno? It seems as though what you're using is the Uniformization Theorem (the classification of simply connected Riemann surfaces). How does that follow from the big/small Picard theorems? The proof I know uses Green's functions/differential geometry. – Alex Youcis Feb 15 '15 at 08:43
  • He assumes the uniformization theorem and then uses the fact that an analytic function from $\tilde{X}$ will omit too many values unless it's constant. But yes, the majority of the strength is from uniformization.

    I'm not sure about an alternate proof of the uniformization theorem, but classifying admissible analytic maps between spherical/flat/hyperbolic complex surfaces can be done using some normal families and dynamics (esp. hyperbolic compactness theorem). So just exhibit the hyperbolicity of a 2+ punctured sphere directly.

    Lee's Topological Manifolds takes a stab at this.

     – John Samples Feb 11 '16 at 11:30
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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.

Curious student
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