# Are there enough meromorphic functions on a compact analytic manifold?

Let $$X$$ be a compact complex analytic manifold, $$D\subset X$$ an irreducible smooth divisor, given as zeroes of a global meromorphic function $$f\in {\mathfrak M} (X)$$. Are there enough other meromorphic functions defining $$D$$?

Here is a precise question: can one find, for each $$x\in X$$, a meromorphic function $$g\in {\mathfrak M} (X)$$ such that $$g$$ is defined (i.e., not $$\infty$$) at $$x$$ and $$D={\mathrm{zeroes}}(g)$$?

• One can construct a compact complex manifold $X'$ with no non constant meromorphic functions. Now blow up a point to obtain $X$ with a smooth divisor and no nonconstant meromorphic functions. So no. Oct 15, 2018 at 16:40
• My smooth divisor is given by a meromorphic function already. Will your be given by a meromorphic function? Oct 15, 2018 at 16:47
• Now that I understand what you're asking. View $f$ as holomorphic map $X\to \mathbb{P}^1$. Then compose $f$ with an automorphism of $\mathbb{P}^1$ fixing $0$ and sending $f(x)$ to a finite value. Oct 15, 2018 at 18:18
• @DonuArapura Isn't $f$ only defined on a blow up of $X$ a priori? Oct 15, 2018 at 18:57
• Yes, right. This works for $x$ off a set of codim $\ge 2$. (I admit I'm not really thinking about this for more than a few seconds at a time.) Oct 15, 2018 at 21:00

First, let's clarify some possible confusion. A compact complex manifold does not admit non-constant holomorphic functions, so, assuming that $$X$$ admits non-constant meromorphic functions, you actually want the divisor of such a function to include poles, not just zeros. Infinity is thus a legitimate value. The points at which the function is (truly) undefined are called indeterminacy points. Following pretty much the notation and approach of Encyclopedia of Mathematics, \url{https://www.encyclopediaofmath.org/index.php/Meromorphic_function} this is the distinction:

Let $$\Omega$$ be a complex manifold (at this stage, we do not require compactness, algebraicity or anything more, and I will stick to the notation $$\Omega$$ to emphasize that the description which follows is local). Let $$\mathcal{O}$$ be the sheaf of germs of holomorphic functions on $$\Omega$$, and for each point $$x \in \Omega$$ let $$\mathcal{M}_x$$ denote the field of fractions of the ring $$\mathcal{O}_x$$ (the stalk of the sheaf $$\mathcal{O}$$ over $$x$$). Then $$\mathcal{M}=\bigcup \mathcal{M}_x$$is naturally endowed with the structure of a sheaf of fields, called the sheaf of germs of meromorphic functions in $$\Omega$$. A meromorphic function in $$\Omega$$ is defined as a global section of $$\mathcal{M}$$, i.e., a continuous mapping $$f: x \to f_x$$ such that for all $$x \in \Omega$$, $$f_x \in \mathcal{M}_x$$. The polar set $$P_f$$ (of codimension $$1$$) and the set of indeterminacy $$N_f \subset P_f$$ (of codimension at least $$2$$) are defined as follows: Let $$f_x=\varphi_x/\psi_x, \quad \varphi_x, \psi_x \in \mathcal{O}_x$$, with $$\psi_x$$ not identically $$0$$. Then $$x \in P_f$$ if $$\psi_x(x)=0$$ and $$x \in N_f$$ if $$\varphi_x(x)=\psi_x(x)=0$$. So at each point $$x \in P_f\setminus N_f$$ (a pole) one can define the value of $$f$$ to be $$\lim_{y \to x}f(y)=\infty \in \mathbb{P}^1$$. I cannot think of any general condition that would imply $$N_f = \emptyset$$. Even on an algebraic manifold this is not guaranteed. See the comment below.

Now to answer the question: in a compact complex manifold a meromorphic function (if admissible) is uniquely determined by its divisor, up to multiplication. If you are asking for another" meromorphic function with the same divisor, all you get will be a scalar multiple of the original one.

To broaden the context a bit, given a divisor $$D$$ in $$\Omega$$, finding a meromorphic function $$f$$ in $$\Omega$$ with prescribed divisor $$D$$ is one of Cousin problems (the multiplicative one). Its solvability depends on some cohomological conditions on the manifold. Finally, a question how many meromorphic functions are there" (not what you are asking) can be also approached from the point of view of Siegel's theorem: if $$X$$ is a compact, connected, complex manifold of dimension $$n$$ and $$\mathcal{M}(X)$$ denotes the field of (globally defined) meromorphic functions on it, then the transcendence degree of $$\mathcal{M}(X)$$ over $$\mathbb{C}$$ does not exceed $$n$$.

Edit: If you are going to change $$f$$ into $$g$$ while keeping the zero part $$Z_f$$ of the divisor of $$f$$, remember that on a manifold $$Z_f=P_{1/f}$$. So you want to solve the (additive) Cousin problem of finding a meromorphic function $$h$$ on $$X$$ with prescribed polar part (that of $$1/f$$). This will be solvable in general if $$H^1(X,\mathcal{O})=\emptyset$$. If $$H^1(X,\mathcal{O})\neq \emptyset$$, a solution might be still possible in specific cases. Then $$Z_{1/h}=Z_f$$. If you want the polar set $$P_{1/h}=Z_h$$ to avoid a specific point $$x \in X\setminus Z_f$$, solve your problem on $$X\setminus \{x\}$$.

• Thank you, Margaret, for a very detailed answer. Are there any known sufficient conditions that could make $N_f=\emptyset$? I suppose it is true on an algebraic variety but I want more... Oct 17, 2018 at 16:30
• In a sense, I am trying to change $f$ into $g$ changing only part of the divisor: divisor of zeroes stays the same, divisor of poles moves away from $x$, giving me a value $g(x)$. If I understood your answer fully, there are situations when it is impossible. Oct 17, 2018 at 16:35
• Algebraicity does not help in avoiding indeterminacies. Consider $X=\mathbb{P}^2$ and $f(z_0:z_1:z_2)=p(z_0,z_1,z_2)/q(z_0,z_1,z_2)$, where $p,q$ are two homogeneous polynomials of the same degree $d \geq 1$. If the hypersurfaces $p=0, q=0$ do not have a common component, they will intersect at $d^2$ points by Bezout theorem, so $f$ will always have indeterminacy points. Oct 19, 2018 at 20:55
• Off course, it does. I believe I can multiply! E.g., $\frac{z_1-z_2}{z_1-z_3}$ and $\frac{z_1-z_2}{z_1+z_3}$ have the same divisor of zeroes, and at least one of these two functions is defined at every point. This is roughly the behaviour I hoped to get in the analytic setting... Oct 22, 2018 at 13:14
• I think I misled you with the first comment of $N_f=\emptyset$. I am sorry for that. Oct 22, 2018 at 13:18