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 $D={\mathrm{zeroes}}(g)$?

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    $\begingroup$ 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. $\endgroup$ Oct 15, 2018 at 16:40
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    $\begingroup$ My smooth divisor is given by a meromorphic function already. Will your be given by a meromorphic function? $\endgroup$
    – Bugs Bunny
    Oct 15, 2018 at 16:47
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    $\begingroup$ 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. $\endgroup$ Oct 15, 2018 at 18:18
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    $\begingroup$ @DonuArapura Isn't $f$ only defined on a blow up of $X$ a priori? $\endgroup$ Oct 15, 2018 at 18:57
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    $\begingroup$ 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.) $\endgroup$ Oct 15, 2018 at 21:00

1 Answer 1


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 $X\setminus \{x\}$.

  • $\begingroup$ 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... $\endgroup$
    – Bugs Bunny
    Oct 17, 2018 at 16:30
  • $\begingroup$ 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. $\endgroup$
    – Bugs Bunny
    Oct 17, 2018 at 16:35
  • $\begingroup$ 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. $\endgroup$ Oct 19, 2018 at 20:55
  • $\begingroup$ 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... $\endgroup$
    – Bugs Bunny
    Oct 22, 2018 at 13:14
  • $\begingroup$ I think I misled you with the first comment of $N_f=\emptyset$. I am sorry for that. $\endgroup$
    – Bugs Bunny
    Oct 22, 2018 at 13:18

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