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Math Help - 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 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.)
    Last edited by Bruno J.; June 25th 2010 at 12:23 PM.
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    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 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?
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