# Determine all the meromorphic 1-forms on the compact surfaces with poles

I am trying to solve the following problem:

Let $$X$$ be a compact Riemann surface, and $$p\in X$$ a point. Let $$\Omega(2p)$$ be the complex vector space of meromorphic $$1$$-forms on $$X$$ that have (possibly) a pole of order at most $$2$$ at $$p$$, and are holomorphic elsewhere. Show that $$\dim \Omega(2p)=g+1.$$

(Do not use the Riemann-Roch Theorem)

My attempt:

Since we need to avoid Riemann-Roch theorem, I want to write the basis of the vector space explicitly.

All l know is that due to residue theorem on a compact Riemann surface, we have the sum of residue is $$0$$. So we have two kinds of meromorphic 1-form:

1. holomorphic everywhere
2. $$p$$ is the pole of order two, and near it, the meromorphic 1-form can be written as $$\frac{c_{-2}}{z^2}dz+\sum_{n=0}^{\infty}c_n z^ndz$$

Then I don't know how to continue. Could anyone give any hint or comments?

I am also curious what role the genus plays in this problem.

• A little remark : you only need $\dim \Omega(2p) \leq g+1$. Indeed, $g$ is by definition $\dim \Omega(0)$, and you can construct a differential form with pole of order two : just take $\omega = dz/z$ on $\Bbb P^1$ (which has a pole of order two) and then pullback it by the mean of a ramified cover $X \to \Bbb P^1$. Commented Apr 2, 2018 at 7:59
• @NicolasHemelsoet You are right! The left is to prove these differential forms is a basis, that is, every meromorphic 1-form can be written as the linear combination of them. So we can have your inequality. But this is not easy for me... Commented Apr 2, 2018 at 8:12
• Yes without Riemann-Roch I'm not sure how to do but I'm sure other people will be able to do it :) (Also there is a typo, one should take $dz/z^2$ before). Commented Apr 2, 2018 at 8:15
• @NicolasHemelsoet I am not familiar with the terminology called "ramified cover". So I might be wrong. But I think as far as I know, the cover map should be $\Bbb P^1\to X$, which is the universal cover. OK. I googled it. I was wrong. But is that okay if we do pullback via this map? The property of pole can be preserved? Could you give me some references on ramified cover? Commented Apr 2, 2018 at 8:51
• The universal cover of the elliptic curve is $\Bbb C$ and for $g \geq 2$ this is the unit disk. If there is a non-constant map $f : \Bbb P^1 \to X$ then $X$ is rational (it has genus zero). I think you might enjoy Rick Miranda's book on Algebraic curves and Riemann surfaces. Commented Apr 2, 2018 at 8:58

The only thing we need to prove is that there is exactly one meromorphic 1-form on the Riemann surfaces with a pole at $p$ of order 2 and locally looks like $$\frac{1}{z^2}+\text{analytic part}.$$