# A Proof of Residue Theorem on a Compact Riemann Surface

Usually a proof of the Residue Theorem on a Compact Riemann Surface uses the crucial fact that Holomorphic forms are closed. I tried to write a proof and somehow I didn't use that fact anywhere. Can someone please check if my proof is ok? Also, if it is correct I don't exactly understand how I was able to avoid that.

The Residue Theorem: $$X$$ Compact Riemann Surface. $$a_1,\dots, a_n$$ are distnct points in $$X$$. Let $$X'=X-\{a_1,\dots, a_n\}$$. Then for every holomorphic $$1$$-form $$\omega\in\Omega(X')$$, one has $$\sum_{k=1}^{n}Res_{a_k}(\omega)=0$$

Proof. Choose coordinate neighborhoods $$(U_k,z_k)$$ of the $$a_k$$ such that $$U_j\cap U_k\neq\emptyset$$ if $$j\neq k$$. Also we may assume that $$z_k(a_k) = 0$$ and each $$z_k(U_k)\subset\mathbb{C}$$ is an open disk of radius $$1$$. For every $$k = 1, ... , n$$ choose a function $$f_k$$ with compact support Supp$$(f_k)\subset U_k$$ such that there exists an open neighborhood $$U'_k\subset U_k$$ of $$a_k$$ with $$f_k\big| U'_k = 1$$ and $$U'_k$$s are balls of the form $$U'_k=\{x\in U_k:|z_k(x)|<\epsilon\}$$ (for small enough $$\epsilon>0).$$

Set $$g := 1 - (f_1 + ... + f_n).$$ Then $$g\big|U'_k= 0.$$ Thus $$g\omega$$ may be continued to the point $$a_k$$ by assigning it the value zero, and may thus be considered as an element of $$\mathcal{E}^1 (X)$$. So by Stokes' Theorem we have $$\int\int_X d(g\omega)=0$$. $$\exists R\in (\epsilon,1)$$ such that Supp$$(f_k)\subset \{|z_k| and $$f_k\big|\{|z_k|<\epsilon\}=1$$. Define $$\Delta_k:=\{x\in U'_k:|z_k(x)|< R\}$$ $$\therefore \int\int_X d(g\omega)=\int\int_{X-\cup_{i=1}^{n}\Delta_i}d(g\omega)=0$$

$$\Rightarrow -\int_{\cup_{i=1}^{n}\partial\Delta_i}g\omega=0$$ (Using Stokes')

$$\Rightarrow \sum_{i=1}^{n}\int_{\partial\Delta_i}g\omega=0$$

$$\Rightarrow \sum_{i=1}^{n}\int_{\partial\Delta_i}\omega=\sum_{i=1}^{n}\int_{\partial\Delta_i}(f_1+\dots+f_n)\omega=0$$ (as $$f_k$$ vanish outside $$\Delta_k)$$

$$\Rightarrow \sum_{k=1}^{n}$$ Res$$_{a_k}(\omega)=0$$

• how do you obtain $\int\int_X d(g\omega)=0$ Sep 15, 2019 at 4:21
• $\int\int_X d(g\omega)= \int_{\partial X} g\omega =0,$(using Stokes); because $X$ has no boundary. Sep 15, 2019 at 4:23
• So that the "boundary of $X$" (that I made precise as $\partial U$) is homologous to 0 ie. my answer Sep 15, 2019 at 4:24
• Yeah, I was wondering that I didn't have to use that holomorphic forms are closed. Sep 15, 2019 at 4:26
• As I tried to make clear in my answer that $\omega$ is closed is needed only to relate the residues to a path/surface integral (the residue at $a$ is defined as integral other $\epsilon$-circle around $a$ and it is when $\omega$ is closed that it is the same as $\int$ over a large loop) Sep 15, 2019 at 4:28

Let $$X$$ be a compact Riemann surface and $$\omega$$ a smooth one-form on $$X-p_1,\ldots,p_N$$, let $$U$$ be an open $$\subset X$$ such that : $$U$$ contains all the $$p_n$$, it is homeomorphic to a simply connected domain in $$\Bbb{C}$$ and $$U\cup\partial U=X$$.

From here see $$\partial U$$ not as a subset of $$X$$ but as a closed-loop in $$X$$.

Topologically $$X$$ is obtaining by gluing different pieces of $$\partial U$$ together, that is $$\partial U = \bigcup_{j=1}^J \gamma_j$$ and each $$\gamma_j$$ is glued with some $$\gamma_{\sigma(j)}$$.

The complex topology induces an orientation, and as curves in $$X$$,

$$\gamma_{\sigma(j)}$$ must be $$\gamma_j$$ traversed in opposite direction because otherwise the points "on the left side of $$\gamma_j$$" would appear twice in $$U$$.

Thus $$\partial U$$ is homologous to $$0$$

$$2\int_{\partial U} \omega=\sum_j \int_{\gamma_j}+\int_{\gamma_{\sigma(j)}} \omega=\sum_j 0=0$$

And hence if $$\omega$$ is a meromorphic one-form ($$\omega=gdf$$ for two meromorphic functions) then

$$2i\pi\sum Res_{p_n}(\omega) = \int_{\partial U} \omega=0$$

• Thanks for your answer. But can you please point out if my proof is correct or wrong? Sep 15, 2019 at 4:20