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What is ramification for z^(p/q)(1-z)^(r/s)?

  1. Nov 19, 2013 #1
    May I ask what is the ramified branching geometry of the algebraic function:

    $$w=z^{p/q}(1-z)^{r/s},\quad (p,q,r,s)\in \mathbb{Z}\backslash\{0\}$$

    and is it computable in terms of the parameters p,q,r,s? The reason I ask is that it appears to be trivially predictable and I just want to know if that is indeed the case.

    Ok thanks,
    Last edited: Nov 19, 2013
  2. jcsd
  3. Nov 23, 2013 #2
    Hi guys,

    After studying some examples and their associated Newton polygons , I've reached the tentative conclusion that the ramified geometry is indeed simply predictable:

    Given the algebraic function:

    $$ w=z^{r/s}(1-z)^{p/q},\quad (p,q,r,s)\in\mathbb{Z}\backslash\{0\}$$

    and letting ##d=LCD(s,q)##, then the ramified geometry about the origin is simply ##d/s## s-cycles, and ##d/q## q-cycles over ##z=1##.

    Take the simple example: ##w=z^{5/2}(1-z)^{4/3}##. The lowest common denominator is of course 6. And when we expand the function about its singular points we obtain three 2-cycles about the origin and two 3-cycles about the point z=1.

    I'm pretty sure that's the case for rational exponents in general. Can anyone confirm this for me?

    Ok thanks,
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