# Algebraically integral expansions of differential forms?

Let C be a smooth, genus g curve defined over a number field K'. Suppose we have a K'-valued point P on C. We can view C as a Riemann surface; then the space of holomorphic differential forms has dimension g over C.

It seems to me that we can choose a uniformizer x at P and a basis of differential forms $\{ \omega_i \}$ such that the expansion of each $\omega_i$ in terms of x has algebraically integral coefficients; that is, there is some finite extension K of K', with ring of integers A, such that $\omega_i = \sum_{n=0}^\infty a_{i,n} x^n dx$ with all $a_{i,n} \in A$.

I would really appreciate critiques, help, or references for my reasoning. We can start with dx, which is a meromorphic differential form on X. We just need to multiply dx by rational functions with zeros in the right places to get holomorphic differential forms, but we want to choose rational functions with integral expansions in x.

First, we make an initial choice of uniformizer x. If P=(C0:C1: ... :Cn) then we know each monomial CiXj-CjXi has a zero of some order at P; we can just choose some convenient quotient of powers of these monomials to get a zero of order one and call that expression x (for now).

The poles of dx will be at various points of C with coordinates in some finite extension of K' (as will the zeros of dx). It seems to me that we can build enough rational functions from monomials involving these coordinates to get what we need. (Similar to how we chose the uniformizer x.) We can then expand these rational functions in terms of x. We may get some denominators (say powers of N) in the expansions of our functions, but then we can replace x with x/N to absorb them.

What do you think? Many thanks for the assistance!

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The result you want is true.  It seems to me that you may be making too much of it: by definition, a smooth projective genus $g$ curve over a number field $K$ has $g$ $K$-linearly independent global regular differential forms.  These will look like $\sum_i f_i dg_i$.  If you have any denominators in the coefficients of the rational functions appearing in your differential, multiplying by a suitable positive integer $N$ will get rid of them.  Aren't we done? – Pete L. Clark Feb 7 '11 at 17:11
@Pete: I think it depends crucially on the chosen uniformizer at P.  We can't just multiply the differential form by N: Consider 1/(1-x/N), which has unbounded denominators when expanded in x, but we could just use x/N instead of x.  Also, the expansion may be integral for a uniformizer x, but not integral for y=ln(x+1), which is also a local uniformizer at P.  It seems like there is some "natural" family of uniformizers that makes this work (probably rational functions), and I'm trying to nail it down precisely. – Hypocycloid Feb 7 '11 at 18:20
for one thing, you seem to be identifying a differential with its power series expansion, which is valid but certainly not necessary.  (And I don't think it's helping you out here.)  Second: $y = \ln x+1$ is not a rational function on any compact Riemann surface I know... – Pete L. Clark Feb 7 '11 at 18:24
@Pete: Thanks for the comments.  You're right; it is clear to me now that rational functions are what I want to work with.  By uniformizer at P, I mean any analytic function on a neighborhood of P with a simple zero at P. – Hypocycloid Feb 7 '11 at 19:21
@Pete: Here's my motivation, which I'll post as another question:  I'm thinking of modular curves and the equivalence of weight 2 modular forms and differential forms.  We study the q-expansions of cusp forms f, where q=e^{2pi iz/m} is a uniformizer at i\infty on the upper half-plane.  Is q a rational function on the modular curve?  Or at least an integral power series in a rational function uniformizer?  If we expand f in (z-a) at the point z=a, then we are using something like the shape z=ln(q+1); which I want to avoid. – Hypocycloid Feb 7 '11 at 19:26
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