Residue theorem on a compact Riemann surface

(The pole set is closed and discrete, hence finite.) Suppose $\omega \mathrm{\ne }0$. The pole set $S$ is discrete: each $p\in S$ has a chart neighbourhood off which the local expression of $\omega$ is holomorphic except at $p$, by [F1] and [F2]. It is also closed: if $x\in X\setminus S$ and $h$ is a chart expression of $\omega$ near $\phi (x)$, then $h$ is holomorphic at $\phi (x)$ because $x$ is not a pole, and by [F2] the poles of $h$ are isolated, so a neighbourhood of $\phi (x)$ contains none of them, whence a neighbourhood of $x$ is disjoint from $S$. As a closed subset of the compact surface, $S$ is compact by [F8]; a discrete compact space is finite, because the family of singletons is an open cover admitting a finite subcover only when the space is finite. Hence $S=\{p_1,\dots ,p_N\}$ is finite.

(The integral of $\omega$ along a path is chart-independent.) For a piecewise $C^1$ path $\lambda$ in $X$ whose trace lies in a chart $(U,\phi )$ define ${\int }_{\lambda }\omega :={\int }_{\phi \circ \lambda }hdz$ with $h$ the chart expression. If another chart $(U^{\mathrm{\prime }},{\phi }^{\mathrm{\prime }})$ contains the trace, put $\tau =\phi \circ ({\phi }^{\mathrm{\prime }}{)}^{-1}$ on ${\phi }^{\mathrm{\prime }}(U\cap U^{\mathrm{\prime }})$ and $w={\phi }^{\mathrm{\prime }}\circ \lambda$; by [F1] the transition law gives $h_{\phi }(\tau (w)){\tau }^{\mathrm{\prime }}(w)=h_{{\phi }^{\mathrm{\prime }}}(w)$, and by [F7] and the chain rule the parametrized integrals of $h_{\phi }$ over $\phi \circ \lambda =\tau \circ w$ and of $h_{{\phi }^{\mathrm{\prime }}}$ over $w$ agree; so the definition is independent of the chart used. The integral is additive over concatenations and changes sign under reversal. The face edges used below admit piecewise $C^1$ parametrizations: each Puiseux endpoint germ becomes $C^1$ after the parameter substitution $t=s^m$ supplied in the planar lemma, and holomorphic chart changes preserve piecewise $C^1$ regularity.

(Chart cellulation subordinate to the poles.) Apply [F3] with $F=S$: the cells ${\mathrm{\Delta }}_1,\dots ,{\mathrm{\Delta }}_m$ cover $X$, each lies in a chart $({\phi }_i,U_i)$, the interiors are pairwise disjoint, each pole lies in the interior of exactly one cell, and the cell boundaries have the finite common subedge system of [F3].

(Each cell is graph-bounded in its chart.) Fix $i$. By [F4] there is an orientation-preserving similarity ${\sigma }_i$ of the chart plane and a graph-bounded region $T_i$ with ${\phi }_i({\mathrm{\Delta }}_i)={\sigma }_i(T_i)$; write ${\gamma }_i$ for the positively oriented boundary contour of ${\phi }_i({\mathrm{\Delta }}_i)$, i.e. the image under ${\sigma }_i$ of the boundary contour of $T_i$. The induced orientation of ${\mathrm{\Delta }}_i$ is the complex orientation of $X$, and ${\phi }_i$ is holomorphic hence orientation-preserving. The finite common boundary subdivision of [F3] splits ${\gamma }_i$ into contour contributions from subedges, each of which occurs on precisely two cells with opposite orientations.

(A chart domain in which the only poles are the interior ones.) Fix $i$, put $R_i:={\phi }_i({\mathrm{\Delta }}_i)={\sigma }_i(T_i)$, and let $h_i$ be the chart expression of $\omega$ on the chart domain of ${\phi }_i$, with pole set $P_i$; by [F2] and [F1], $P_i$ is closed and discrete in the chart domain. The compact set ${\gamma }_i^{\ast }=\mathrm{\partial }{R}_i$ is disjoint from $P_i$, because the poles of $\omega$ lie in face interiors and ${\phi }_i$ maps the boundary of ${\mathrm{\Delta }}_i$ onto ${\gamma }_i^{\ast }$; hence the distance from ${\gamma }_i^{\ast }$ to $P_i$ is positive (read as $+\mathrm{\infty }$ if $P_i$ is empty). Since the compact set $R_i$ lies in the open chart image $D_i:={\phi }_i(U_i)$, its distance from $\mathbb{C}\setminus D_i$ is also positive. Choose ${\delta }_i>0$ smaller than half of both distances and put ${\mathrm{\Omega }}_i:=\{q:\mathrm{dist}(q,R_i)<{\delta }_i\}$; it is open, contains $R_i$ and lies in $D_i$, so $h_i$ is defined throughout ${\mathrm{\Omega }}_i$. Every pole in ${\mathrm{\Omega }}_i$ lies in $R_i^{\circ }$: if a pole $q$ were outside the closed set $R_i$, then $\mathrm{dist}(q,{\gamma }_i^{\ast })=\mathrm{dist}(q,R_i)<{\delta }_i$, contradicting $\mathrm{dist}(q,{\gamma }_i^{\ast })\ge \mathrm{dist}({\gamma }_i^{\ast },P_i)>2{\delta }_i$, and $q\in \mathrm{\partial }{R}_i$ is impossible because ${\gamma }_i^{\ast }\cap P_i=\mathrm{\varnothing }$.

(Summing over the cells.) By step 3.1 each of the finitely many subedges occurs in the boundary of exactly two cells, traversed with opposite orientations; split each ${\gamma }_i$ at the subedge vertices and use additivity, chart independence and reversal from step 1.2. Every subedge contribution occurs twice with opposite signs, so ${\sum }_{i=1}^m{\int }_{{\gamma }_i}{h}_{i}dz=0.$

(Plane residue theorem on each face.) Fix $i$. The contour ${\gamma }_i$ has trace in ${\mathrm{\Omega }}_i\setminus P_i$, and it is null-homologous in ${\mathrm{\Omega }}_i$: for $q\mathrm{\notin }{\mathrm{\Omega }}_i$ we have $q\mathrm{\notin }{\sigma }_i(T_i)$, so [F5] gives $n({\gamma }_i,q)=0$. Applying [F6] to $h_i$ on ${\mathrm{\Omega }}_i$ and $\mathrm{\Gamma }={\gamma }_i$, whose only poles in ${\mathrm{\Omega }}_i$ are the images of the poles of $\omega$ lying in ${\mathrm{\Delta }}_i^{\circ }$, gives ${\int }_{{\gamma }_i}{h}_{i}dz=2\pi i\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\sum }_{p\in S\cap {\mathrm{\Delta }}_i^{\circ }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\mathrm{Res}}_p(\omega ),$ because the index of ${\gamma }_i$ at each of those poles is $1$ by [F5].

(Conclusion.) Summing the identities of step 5.1 over $i$ and using step 4.2 gives $0={\sum }_{i=1}^m{\int }_{{\gamma }_i}{h}_{i}dz=2\pi i{\sum }_{p\in S}{\mathrm{Res}}_p(\omega ),$ since each pole lies in exactly one cell interior by step 2.1; dividing by $2\pi i\mathrm{\ne }0$ gives ${\sum }_{p\in X}{\mathrm{Res}}_p(\omega )=0$, and the residues vanish off the finite set $S$. For $\omega =0$ the assertion is the convention recorded in the statement. All choices made were finite, so no choice principle was used.