MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth projective variety. Is there a criterion (apart from the definition) for the existence of a projective curve $C$ and a proper surjective morphism $\pi:X \to C$?

share|cite|improve this question
Note that you can always blow up $X$ so that such a map $\pi$ exists, even with $C=\mathbb{P}^1$ and very mildly singular fibers (a "Lefschetz pencil"). This is enough for many applications, since for example cohomology of $X$ injects into the cohomology of the blow-up. – Piotr Achinger Jan 17 '13 at 19:40
Just for the record, the fact that you can choose $C=\mathbb P^1$ is not too deep since you can just map $C$ to $\mathbb P^1$ and replace $\pi$ with the composition. – Sándor Kovács Jan 17 '13 at 22:05… – M P Jan 18 '13 at 11:44

Another interesting theorem in this direction is Castelnuovo-de Franchis theorem. It says that if you have two linearly independent holomorphic 1-forms $\omega_1,\omega_2$ with $\omega_1\wedge\omega_2=0$ on $X$, then there exists a morphism $f:X\to C$ with $C$ a smooth curve of genus at least 2 and forms $\omega_i'$ on $C$ such that $\omega_i=f^*\omega_i'$.

share|cite|improve this answer
This is essentially saying that the condition below holds for the canonical divisor $D=K_X$ and then the induced morphism is a pluricanonical morphism, so $C$ is the canonical model and hence necessarily of general type. – Sándor Kovács May 11 at 15:37

Yes. This is equivalent to the existence of a non-trivial divisor $D$ on $X$ such that there exists two distinct members of the linear system $|D|\ni D_1,D_2$ that are disjoint: $D_1\cap D_2=\emptyset$.

share|cite|improve this answer

By a theorem of Gromov and Schoen if the fundamental group of X is a proper amalgamated product or HNN extension then X maps surjectively to a curve .

share|cite|improve this answer
Furthermore, $X$ maps surjectively to a hyperbolic orbicurve. – Misha Jan 17 '13 at 23:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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