Let $X$ be a complex manifold with $f$ be a meromorphic function on it, locally it's represented as quotient of two holomorphic function, i.e. $f_x = \frac{a_x}{b_x}$ with $a_x,b_x \in \mathcal{O}_{X,x}$ on the stalk. We can define a principle divisor associated to this meromorphic function as $$\text{div}(f) = \sum \text{ord}_Y(f)[Y]$$ where we take the sum over all irreducible hypersurfaces $Y\subseteq X$.
Prove that if $\text{div}(f) \ge 0$ then it's in fact holomorphic function
I know for the Riemann surface the proof is not hard since the hypersurface are set of points, however, I have no idea how to prove for the complex manifold case?
Best Answer
As you have observed, if a meromorphic function $f$ has $\mathrm{div}(f) \geq 0$ then it cannot have poles along any hypersurfaces. But by Hartogs' Theorem, any function holomorphic outside a compact submanifold of codimension $\geq 2$ has a unique holomorphic extension across the submanifold. So $f$ must be holomorphic on all of $X$.
Edit: Actually one don't need to use Hartogs' theorem here. By definition a meromorphic function is locally a quotient of holomorphic functions, hence it may only have poles along hypersurfaces. More precisely, given a point $x \in X$, one can find a hypersurface $Y \subseteq X$ containing $x$ together with a irreducible holomorphic germ $g$ at $x$ defining $Y$ around $x$. Recall $\mathrm{ord}_Y(f)$ is the largest integer $k$ for which $f \cdot g^{-k}$ is a holomorphic germ at $x$. Since $\mathrm{div}(f) \geq 0$, we conclude that $f$ is a holomorphic germ at $x$. This shows $f$ is holomorphic on all of $X$.
Edit 2: As pointed out in the comment, the statement $\mathrm{ord}_Y(f)$ being the largest integer $k$ for which $f \cdot g^{-k}$ is holomorphic is true only in dimension $1$ case. Here is an argument to address this issue.
Given $x \in X$, suppose the meromorphic function $f$ is defined locally at $x$ by quotient $g/h$. It is already holomorphic at $x$ if $h(x) \neq 0$. Otherwise we can factor $h$ into irreducibles $h = h_1 \cdots h_n$ where $h_1(x) = \cdots = h_k(x) = 0$ and $h_i(x) \neq 0$ for $i > k$. Then we can look at irreducible hypersurfaces defined locally by $h_i$'s. By assumption $\mathrm{ord}_{V(h_i), x}(g) \geq \mathrm{ord}_{V(h_i), x}(h) \geq 1$ for $1 \leq i \leq k$. Therefore $f \cdot h_i^{-1}$ is holomorphic at $x$ for all $i$, so is $f/h$.