Let $V$ be a connected smooth complex projective curve of negative Euler characteristic. Can there exist a connected smooth complex algebraic curve $U$ such that there is a non-constant holomorphic map $U\to V$ but no non-constant holomorphic map from the compactification of $U$ to $V$? Note that we are not merely asking that the map $U\to V$ does not extend.

EDIT: the question considers the smooth compactification of $U$.

  • $\begingroup$ Your body seems quite a bit stronger than the kind of question that might be implied by your title. $\endgroup$ – LSpice Jul 4 at 3:23
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    $\begingroup$ @LSpice I agree but I don't think there is a title that would fit in one line and be accurate to the body. $\endgroup$ – vrz Jul 4 at 8:36
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    $\begingroup$ Sometimes you impose technical hypotheses like smoothness to rule out trivial examples. There is no need to explain them. But when you impose substantive hypotheses, you should say why. In particular, why did you impose the negative Euler characteristic? Does this rule out some trivial example? But it does rule out the transcendental example of an elliptic curve covered by $G_m$, when of course $P^1$ does not map to an elliptic curve. Surely the point is not to denigrate this example, but to celebrate it and ask about higher genus versions. Motivation by example makes questions clearer. $\endgroup$ – Ben Wieland Jul 4 at 13:28
  • $\begingroup$ The curve V is hyperbolic and thus satisfies the property that every holomorphic map from a reduced variety to it is in fact algebraic; see arxiv.org/abs/1806.09338 the answer to your question is this “no” $\endgroup$ – Ariyan Javanpeykar Jul 5 at 20:08
  • $\begingroup$ The algebraicity property mentioned in my comment above holds for all hyperbolic compact varieties by a theorem of Kwack; references in the arXiv link above. $\endgroup$ – Ariyan Javanpeykar Jul 5 at 20:09

Just turning my comments into an answer:

Following the OP, let $V$ be a smooth projective connected curve with negative Euler characteristic (i.e., genus at least two) over $\mathbb{C}$. Then $V$ is hyperbolic in the sense that Kobayashi's pseudometric is a metric. In particular, by a theorem of Kwack, it is "Borel hyperbolic", i.e., for every reduced finite type scheme $S$ over $\mathbb{C}$, every holomorphic map $S^{an}\to V^{an}$ is algebraic.

This implies that the answer to the OP's question is no. Indeed, let $\varphi:U^{an}\to V^{an}$ be a non-constant holomorphic map with $U$ a smooth curve. Then there is a morphism of varieties $f:U\to V$ such that $f^{an} = \varphi$. Such a morphism extends to a morphism $\overline{U}\to V$ by the valuative criterion for properness.

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The title and the body of the question seem to ask two different things. Let me give an answer to the second one. Since you are not asking that the compactification of $U$ is smooth, we can build up an example as follows.

Let $\bar{U}$ be a curve with one cusp $p$ such that its normalization is a projective genus $2$ curve $V$, and let $\nu \colon V \to \bar{U}$ be the normalization map. If $U=\bar{U} - \{p\}$ is the smooth locus of $\bar{U}$ and $q=\nu^{-1}(p)$, the restriction $\nu \colon V-\{q\} \to U$ is an isomorphism, and so is $\nu^{-1} \colon U \to V-\{q\}$.

Composing with the inclusion $V-\{q\} \to V$, we get a smooth algebraic map $U \to V$, whose image is $V - \{q\}$. This cannot be extended to a holomorphic map $\bar{U} \to V$, since $\bar{U}$ and $V$ are not biholomorphic.

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  • $\begingroup$ I mean for the compactification to be smooth, apologies. $\endgroup$ – vrz Jul 4 at 8:37
  • $\begingroup$ thanks for the comment. If you edit the question, please indicate the change you have made, so that this answer keep making sense. $\endgroup$ – Francesco Polizzi Jul 4 at 8:39

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