Suppose $f$ is a meromorphic function on a compact Riemann surface. How do I go about showing that the sum of the orders of the poles and zeros of $f$ is 0? This really surprises me because I cannot see a reason as to why there should be any relation between the poles and the zeros of a meromorphic function. Perhaps a compact Riemann surface gives some sort of symmetry structure? I am a bit lost – any help is appreciated.
Is the sum of the orders of the zeros and poles of a meromorphic function 0
complex-analysisriemann-surfaces
Best Answer
Let $X$ be a compact Riemann surface and $f$ be meromorphic on $X$. Its divisor of zero/poles $$div(f) = \sum_{l=1}^L n_l P_l$$ let $U \subset X$ be a fundamental domain : an open $U \subset X$ which is homeomorphic to a simply connected domain and such that $X = U \cup \partial U$. Choose it such that $f$ has no zero/pole on $\partial U$ and think to $\partial U$ not as a subset of $X$ but as the curve enclosing $U$. With the residue theorem / argument principle $$ \deg(div(f))=\sum_{l=1}^L n_l = \int_{\partial U} \frac{df}{2i \pi f}$$ Topologically $X$ is obtained by gluing some edges of $\partial U$ together, that is $\partial U = \bigcup_{j=1}^J\gamma_j$ and each piece of curve $\gamma_i$ is glued to some other $\gamma_j$. The complex topology induces an orientation, thus when we glue $\gamma_i,\gamma_j$ together, they must appear in reverse direction in the boundary of $U$ (since otherwise $U$ would contain twice the points "on the left side" of $\gamma_i$). Whence $\int_{\gamma_i}+\int_{\gamma_j} =0$ and $$\int_{\partial U} \frac{df}{2i \pi f} = 0$$