How to design algebraic function with particular ramified covering?

[jackmell]

Hi,

I wish to study (non-trivial) algebraic functions which have varied cycles at a few places other than the origin. It's not hard to search for a particular covering at the origin. For example, the function

$$f(z,w)=(8 z^2)\text{}+(-10 z-4 z^2-6 z^3)w+(-10 z)w^2+(-3 z^2-6 z^3)w^3+(-9 z^2+4 z^4)w^4+(6-5 z+z^3)w^5$$

has a 1-cycle and 4-cycle at the origin. I just randomly searched for one. However, it seems it's not so easy to find one with higher cycles than a 2-cycle outside the origin or maybe I'm missing something. I'd like to find some with more than one such point. I can certainly make a translation say:

$$g(z,w)=f(z+1,w)$$

and now the function $g(z,w)$ will have a 1-cycle and a 4-cycle at the point z=1. But that's as far as I can go.

How would I design one with say two such cycle types say at the points -1 and 1?

I don't think just making the substitution g(z-1,w) would do it.

So is there a way to design an algebraic function $f(z,w)$ so that it will have a particular ramified covering at several points outside the origin or is there an easy way to search for ones other than just randomly?

Edit: Just thought of a more cioncise way of asking my question:

How can I find a non-trivial algebraic function of degree at least 5 in $p_n(z)$ and w so that it has a non-trivial ramified covering above several of it's non-zero singular points and by non-trivial, I mean coverings which are more ramified than a set of single-cover branches with one 2-cycle branch?

Thanks,
Jack

 

[jackmell]

Hey guys. Turns out to be a trival matter. Let's take a simple example:

$$f(z,w)=p_0+p_1 w+p_2 w^2+p_3 w^3$$

and suppose we wish to have it fully-ramify at the points $z=1,-1$. A sufficient condition for full ramification of the translated function $f(z\pm 1,w)$, is for the lower Newton-leg of it's first polygon to have a slope of $1/3$. Thus, one possibility is:

$$f(z\pm1,w)=(z+q_1)+(z+q_2)w+(z+q_3)+(a+q_4)w^3$$

where each $q_i$ is a suitable polynomial in $z$. Take $p_0(z)=a+bz+cz^2$. Then one solution is: $a+b(z\pm 1)+c(z\pm 1)^2=\alpha z+\beta z^2$ or $a\pm b+c=0$. Let's take b=0, a=1 and c=-1 so that for the first term, $p_0(z)=1-z^2$. We can do the same for the others and arrive at a suitable function:

$$f(z,w)=(1-z^2)+(1-z^2)w+(1-z^2)w^2+z w^3$$

with the desired ramification geometry at the points z=-1 and 1.