# Derivative of an automorphic rational function

• Jun 18th 2010, 06:20 PM
Bruno J.
Derivative of an automorphic rational function
Suppose you have a finite subgroup $G$ of $PSL(2, \mathbb{C})$, viewed as the group of Mobius transformations acting on the Riemann sphere $S^2$. An example of such a subgroup is given by the group of "harmonic" transformations, which take $z$ to $z, \frac 1 z, 1-z, \frac{1}{1-z}, \frac{z}{1-z} \mbox{ or to } \frac{1-z}{z}$. Take two points $a, b \in \mathbb{C}$ which have a trivial stabilizer in $G$, and whose orbits are disjoint and do not contain the point $\infty$. Form the rational function

$j(z)=\prod_{g \in G}\frac{g(z)-a}{g(z)-b}$.

This function has the property of being invariant under $G$, i.e. $j(z)=j(g(z)) \: \forall z \in S^2 \: \: \forall g \in G$.

Problem 1 Show that we can write

$j(z)=C\prod_{g \in G}\frac{z-g(a)}{z-g(b)}$, for some constant $C$.

Problem 2 Show that, for any $p \in S^2$, the order of $j'(z)$ at $p$ is equal to the number of nontrivial elements in the stabilizer of $p$ in $G$. (Remember that the order of $f$ at a point $p$ is the least integer $n$ such that $f(z)/(z-p)^n$ is holomorphic at $p$.)
• Jun 19th 2010, 07:34 AM
Bruno J.
Here's an extra problem for those of you who are familiar with algebraic functions.

Show that the Riemann surface of the inverse function $z=z(j)$ is a ramified cover of $S^2$, whose automorphism group is isomorphic to $G$. What do the ramification points and indices correspond to?