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Riemann surfaces over algebraic surfaces

  1. Oct 1, 2013 #1

    tom.stoer

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    Suppose we have one polynom

    ##P(r_1, r_2, \ldots, r_n) = 0##

    in n complex variables. This defines a n-1 dimensional complex algebraic surface.

    Suppose that for each variable we have

    ##r_i = e^{ip_i}##

    with complex p.

    In the case n=1 of one variable r this results in the complex logarithm

    ##p = -i\ln r##

    and we have to deal with a Riemann surface (but only for a discrete set of solutions of P)

    What happens in the case n>1?

    Is there something like a Riemann surface (in more than one variable) over a (higher-dimensional) algebraic surface?

    What can we say about the manifold defined as the solution of

    ##P\left( e^{ip_1}, e^{ip_2}, \ldots, e^{ip_n} \right) = 0##

    in p-space?
     
    Last edited: Oct 2, 2013
  2. jcsd
  3. Oct 3, 2013 #2

    mathwonk

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    this seems a rather odd question to an algebraic geometer. by definition you are looking at an infinite analytic but non algebraic cover of an algebraic hypersurface, but only over those points where no coordinate is zero.

    i.e. you are taking an infinite analytic cover of the part of an algebraic hypersurface complementary to all the coordinate axes.

    may i ask how this arises?
     
  4. Oct 3, 2013 #3

    tom.stoer

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    The starting point is rather simple. We have

    ##c = \cos p##
    ##s = \sin p##

    and a polynomial equation

    ##\tilde{P}(c,s) = 0##

    Then we eliminate c,s via the new variable r and we get a new polynomial equation

    ##P(r) = 0##

    But b/c we started with the variables c,s we are basically interested in

    ##P\left(e^{ip}\right) = 0##

    Now you may want to look for solutions of equations formulated on higher dimensional spheres (where you need more angles) and you immediately get a higher-dimensional generalization of the polynomial equation. A rather simple example is the intersection of a sphere or an ellipsoid with a plane.
     
    Last edited: Oct 4, 2013
  5. Oct 6, 2013 #4

    tom.stoer

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    Any new thoughts?
     
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