# How to see meromorphicity of a function locally?

Given a germ of an analytic function on a (compact, for simplicity) Riemann surface, how can one see (locally) whether this is a "germ of meromorphic function"? I.e. if I do analytic continuation along various paths, how can I be sure sure that I will never see an essential singularity?

Another formulation of this question is, how can one determine whether a convergent taylor series determines a meromorphic function on the universal covering space of the Riemann surface?

The fact that there will be no essential singularity certainly implies something, e.g. when our Riemann surface is CP^1, then for a taylor series to be meromorphic, it must be rational. But how do one check this locally, in a nbhd of a point?

Thanks

P.S. I don't really know how to tag this question. Suggest a tag in comment please if possible.

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IS the monodromy theorem relevant for you? – Anweshi Jan 6 '10 at 12:23

## 1 Answer

Checking whether a function is rational locally is straightforward, since the Taylor coefficients satisfy a very strong structure theorem. There must exist complex numbers $\alpha_1, ... \alpha_k$ and polynomials $P_1, P_2, ... P_k$ such that the Taylor coefficients satisfy

$\displaystyle f_n = \sum_{i=1}^{k} P_i(n) \alpha_i^n$

for all but finitely many $n$. The degree of $P_i$ is one less than the order of the pole at $\alpha_i$. Among other things it follows that asymptotically we have $f_n \sim A \alpha^n n^k$ for some $A, \alpha \in \mathbb{C}$ and some non-negative integer $k$, and this is a very strong condition. However, I don't know what the situation is like for other Riemann surfaces.

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First of all, thanks for tagging it. Can you define exponential-polynomial form? (is the asymptotic behaviour of the coefficients a corrollary of the definition or equivalent to the definition? I assume it is the former case.) Actually I got this question from the non-simply connected case --- what I'm considering is the action of the fundamental groupoid on the structure like germs of functions. Anyway, thanks for answer the rational case! – Yuhao Huang Dec 4 '09 at 3:57
I've clarified my answer. Hope it helps! – Qiaochu Yuan Dec 4 '09 at 4:06
This is a nice question. – Idoneal Jan 12 '10 at 16:56