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Thread: Derivative of an automorphic rational function

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Derivative of an automorphic rational function

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

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

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

    Problem 1 Show that we can write


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

    Problem 2 Show that, for any $\displaystyle p \in S^2$, the order of $\displaystyle j'(z)$ at $\displaystyle p$ is equal to the number of nontrivial elements in the stabilizer of $\displaystyle p$ in $\displaystyle G$. (Remember that the order of $\displaystyle f$ at a point $\displaystyle p$ is the least integer $\displaystyle n$ such that $\displaystyle f(z)/(z-p)^n$ is holomorphic at $\displaystyle p$.)
    Last edited by Bruno J.; Jun 25th 2010 at 11:23 AM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Here's an extra problem for those of you who are familiar with algebraic functions.

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