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

  • Thread starter jackmell
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  • #1
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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,
Jack
 
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Answers and Replies

  • #2
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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,
Jack
 

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