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Determining Riemann surface geometry of algebraic functions

  1. Dec 20, 2011 #1
    Hi, given the algebraic function:


    how can I determine the geometry of it's underlying Riemann surfaces? For example, here's a contrived example:


    That one has a single sheet manifold, a double-sheet and a triple-sheet manifold as best as I can describe it and illustrated by the real part of w(z) shown below and will of course give rise to the expansion



    which is my interest in this problem and I've not split up the factoring into separate single-valued terms, i.e., gold and purple are multi-valued. However, is there a method to determine this manifold geometry for the general case just by an analysis of the function [itex]f(z,w)[/itex]? I know the Newton Polygon procedure can determine this indirectly but I believe that has limitations and will not work for certain functions. I was wondering is there is another way. That is, given [itex]f(z,w)[/itex], is there some (exact) algebraic operation on it I can perform and obtain the sheet orders which in the contrived case above would be 1,2, and 3?


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    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 21, 2011 #2


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    well in the first place, any example like the second explicit one, is trivial, because it is a graph. i.e. there is a holomorphic map from the complex w- line isomorphically onto the zero locus,

    sending w to (w-1)(w-2)^2(w-3)^3. the map in the other direction taking (w,z) to w, is inverse to it, so your example is isomorphic to the complex w - line.

    i have written detailed notes on this topic but they are not on my website yet.

    the book i taught out of is the one by rick miranda. it is excellent.

    here are some of my notes. there is another one i want to attach but the browser will not let me since it says i have already attached it to the thread "what is the genus of a riemann surface?" maybe you can find that thread here.


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    Last edited: Dec 21, 2011
  4. Dec 22, 2011 #3
    Thanks for helping me on this Mathwonk. I see I have a long way to go and am planning to study algebraic geometry. For now though, would you have time to show me how


    could be analyzed to conclude the factorization of [itex]f(z,w)[/itex] near z=0 can be written as

    Last edited: Dec 22, 2011
  5. Dec 22, 2011 #4


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    well that looks unlikely, since the product of those factors is quadratic in w, not cubic. where are you getting these answers you ask me to verify? where are you trying to factor these equations, in the power series k[[z,w]]?

    no i see you are using fractional powers of z. maybe that changes the total degree of w, but i don't se why instantly. have you looked at walker's little book, algebraic plane curves? he treats fractional power series. im sorry im not expert enough to answer this off the top of my head.

    maybe when i have ,more time i can look at it more but im sorry im too busy for the time being to learn this theory before answering.
  6. Dec 23, 2011 #5
    That's ok Mathwonk. I figured you are busy. Thanks for the Walker reference and the other references above. They're helpful. Also, I'm afraid I used an abuse of notation above. That's really three factors:

    [tex]f(z,w)=\left(w-g(z)\right)\left(w-h(\epsilon_1z^{1/2})\right)\left(w-h(\epsilon_2 z^{1/2})\right)[/tex]

    however the two factors of [itex]h(z^{1/2})[/itex] come from the same 2-valued manifold, just different single-valued determinations of it. Just thought my condenced notation was cleaner-looking.

    And the expansion comes from Newton-Puiseux's Theorem:

    The quotient field [itex]\mathbb{C}((z^*))[/itex] of convergent fractional power series (which I assume includes all analytic functions) is algebraically closed so for [itex]f(z,w)=a_n(z)w^n+\cdots+a_0(z)[/itex], with [itex]a_i(z)\in \mathbb{C}((z^*))[/itex], we have

    [tex]f(z,w)=\prod_{i=1}^n \left(w-g_i(z^{1/k_i})\right)[/tex]

    with [itex]g_i(z^{1/k_i})\in\mathbb{C}((z^*))[/itex]. One reason I knew the expansion was in terms of [itex]g(z)[/itex] and [itex]h(z^{1/2})[/itex], is because I drew it:

    (Not entirely sure I'm quoting that theorem precisely. This is all very new to me.)

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    Last edited: Dec 23, 2011
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