Let $\Gamma$ be a smooth projective curve in $\mathbb{P}^2$ and let $U$ be an open neighborhood of $\Gamma$. Denote by $\Gamma_1,\Gamma_2,\ldots,\Gamma_n$ a finite collection of smooth curves intersecting $\Gamma$ transversally. My question is:

Does there exist a non-constant holomorphic function $\pi: U \setminus \bigcup\limits_{i=1}^n \Gamma_i \to \Gamma \setminus \bigcup\limits_{i=1}^n \Gamma_i$ ?

The situation as in the following picture. enter image description here

I can build a smooth map $\pi$ easily by partition of unity. Intuitively, $\pi$ seems like the tubular neighborhood projection of $\Gamma$ ? But unfortunately, a holomorphic tubular neighborhood does not exist in general as discussed in Is there any holomorphic version of the tubular neighborhood theorem?. And I realized that even the existence of a non-constant holomorphic map $\pi : U \to \Gamma$ (without removing any curve) is not obvious to me.

Any comment is welcome. Thanks in advance!


1 Answer 1


Let me show that such a map usually doesn't exist (even if we don't remove any additional curves). Consider the case when $\Gamma$ is a curve of degree $\ge 4$. Then one can slightly perturb $\Gamma$ in $\mathbb CP^2$ to a curve $\Gamma'$ that is not isomorphic to $\Gamma$ (so that it stays in a small neighbourhood of $\Gamma$). If the map $\pi$ would exists, we can restrict it to $\Gamma'$, and it will be extendable to the whole $\Gamma'$. So we get a holomorphic (non constant) map $\pi: \Gamma'\to \Gamma$, which is impossible, since $\Gamma$ and $\Gamma'$ are not isomorphic (and by Riemann-Hurwitz for two curves of the same genus $\ge 2$ a non-constant holomorphic map $\Gamma\to \Gamma'$ exits only if the curves are isomorphic).

This argument can be easily generalised to curves $\Gamma$ of degree $3$, i.e. cubics. The case of conics and lines should be analysed separately (and for lines $\Gamma$ such a map sometimes exists, for example, when all $\Gamma_i$'s are lines transverse to the line $\Gamma$ and intersecting at one point).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.