Reverse residue theorem without using Serre's duality

Question

In V. Talovikova's text "Riemann-Roch Theorem", a key part of the proof of Riemann-Roch theorem is the following proposition (4.6 in the text):

Let $\{a_1, \dots,a_n\}$ be a set of points in a compact Riemann surface $X$ and let $\{f_1,\dots,f_n\}$ be a set of principles parts (i.e., Laurent series with no positive exponents). Then the following are equivalent:

1. There exists a meromorphic function $f$ on $X$ such that the principle part of $f$ around $a_i$ is $f_i$, and $f$ has no additional poles.

2. $\sum_{i=1}^n \text{Res}(f_i\omega,a_i)=0$ for every holomorphic $1-$form $\omega$ on $X$

In the text, the proof of $2 \to 1$ is not mentioned, but it is stated that it "follows from the Serre Duality". Yet the Serre Duality follows from Riemann-Roch theorem, which relies on this proposition!

Is there a proof of this proposition which does not rely on Serre Duality or on Riemann-Roch Theorem?

This is V. Talovikova's text: https://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Talovikova.pdf

Answer 1

If you allow to use some sheaf cohomology, all you need is a non-degenerate pairing between the $(0,1)$-Dolbeault cohomology $H^{0,1}_{\bar{\partial}}(X)$ and the space of holomorphic forms $\Omega^1(X)$:

$$ H^{0,1}_{\bar{\partial}}(X)\times\Omega^1(X)\to \mathbb C.\tag{1}\label{1}$$

The non-degeneracy comes from the fact that any class $\alpha\in H^{0,1}_{\bar{\partial}}(X)$ is represented by an anti-holomorphic 1-form $\bar{\omega}$, and $\int_X\bar{\omega}\wedge\omega=0$ iff $\omega=0$.

Now, take an open cover $\{U_{\alpha}\}$ of $X$ such that each $U_{\alpha}$ contains at most one $a_i$. Let $g_{\alpha}$ be a meromorphic function on $U_{\alpha}$ whose principal part is the corresponding $f_i$. Take a bump function $\eta_{\alpha}$ that is constant $1$ in a neighborhood of $a_i\in U_{\alpha}$ and has compact supported in $U_{\alpha}$.

Let $g=\sum_{\alpha}g_{\alpha}\eta_{\alpha}$, then $\phi=\bar{\partial}g$ is a $C^{\infty}$ $(0,1)$-form on $X$ that vanishes around each $a_i$. If $\phi=\bar{\partial}h$ for some global $C^{\infty}$ function $h$ on $X$, then $f=g-h$ is a meromorphic function satisfying the condition 1.

Therefore, to prove $2\to 1$, it suffices to show the Dolbeault cohomology class $[\phi]\in H^{0,1}_{\bar{\partial}}(X)$ is zero. By nondegeneracy of the pairing \eqref{1}, it suffices to show that for all $\omega\in \Omega^1(X)$, $$I(\omega)=\int_X\phi\wedge\omega=0.$$

This is exactly the condition 2. To see this, use the fact that $\phi$ vanish in a neighborhood $V_i$ of $a_i$ for each $i$, and that $\bar{\partial}g\wedge \omega=dg\wedge \omega=d(g\omega)$, we find the following by Stokes' theorem $$ I(\omega)=\int_{X\setminus \cup_iV_i}d(g\omega)=\sum_i\int_{\partial_{V_i}}f_i\omega=\sum_i\text{Res}_{a_i}(f_i\omega),$$ which is zero by condition 2.