About the order of a meromorphic function

Question

In Gunning's book, lectures on Riemann surfaces, the order of a meromorphic function is defined as the order of the first non-zero coefficient in Laurent series of the function. About this assertion, "the order of a meromorphic function f is non-zero only at a discrete set of points ", I don't understand the meaning, as we all know, locally a meromorphic function f is a quotient of two holomorphic functions, say f=g/h, so ord(f)=Z(h) where Z(h) is the zero set of h, Z(h) is non-zero only at a discrete set of points? There must be some wrong of my understanding. So where is the wrong ? Any help will be greatly appreciated. Thanks

Answer 1

By definition of a Riemann surface $X$ and a meromorphic function on $X$,

if $p \in X$ then there is a neighborhood $U $ of $p$ such that there is a local biholomorphism $\phi : U \to V \subset \mathbb{C}$, and $f$ is meromorphic at $p$ iff $f \circ \phi^{-1}$ is meromorphic at $\phi(p)$.

The order of the zero/pole of $f$ at $p$ is defined to be the order of $f \circ \phi^{-1}$ at $\phi(p)$ (this is consistent since the order of a zero or pole doesn't change under biholomorphic maps).

Yes, the zeros and poles of a meromorphic function on $X$ are isolated (be careful with the definition of isolated : $0$ and $\infty \not \in\mathbb{C}^* $, that's why the zeros of $\sin(1/z)\sin(z)$ which is meromorphic on $\mathbb{C}^*$ can accumulate at $0$ and $\infty$).

That's why compact Riemann surfaces are much more restrictive : on those Riemann surfaces, there are no such "points" where the zero/poles can accumulate.

On Riemann surfaces, you can't say anymore that a meromorphic function is globally the quotient of two holomorphic functions (on a compact Riemann surface, there are no non-constant holomorphic functions).