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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.

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    $\begingroup$ 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 $X \to \Bbb P^1$. $\endgroup$ Commented Apr 2, 2018 at 7:59
  • $\begingroup$ @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... $\endgroup$
    – Aolong Li
    Commented Apr 2, 2018 at 8:12
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    $\begingroup$ 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). $\endgroup$ Commented Apr 2, 2018 at 8:15
  • $\begingroup$ @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? $\endgroup$
    – Aolong Li
    Commented Apr 2, 2018 at 8:51
  • $\begingroup$ 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. $\endgroup$ Commented Apr 2, 2018 at 8:58

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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 $$\frac{1}{z^2}+\text{analytic part}.$$

For the above statement, there is a proof in this notes: Chap II, Cor 3.3.

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