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Possibly this has already been asked, but it came up again in this question of Daniel Litt. Does every smooth, projective morphism $f:Y\to \mathbb{C}P^1$ admit a section, i.e., a morphism $s:\mathbb{C}P^1\to Y$ such that $f\circ s$ equals $\text{Id}_{\mathbb{C}P^1}$?

Edit. As Ariyan points out, this article proves that there are at least $3$ singular fibers of $f$ if either $Y$ has nonnegative Kodaira dimension, Theorem 0.1 of Viehweg-Zuo, or if the fibration is non-isotrivial with general fiber either general type or with $\omega_f$ semi-positive, Theorem 0.2 by Möller-Viehweg-Zuo. This suggests an approach to proving the conjecture, at least assuming the uniruledness conjecture (negative Kodaira dimension implies uniruled): take the MRC quotient and then apply the Minimal Model Program to try to reduce to these theorems. Unfortunately, both formation of the MRC quotient and the Minimal Model Program are likely to introduce singularities . . .

Second Edit. As Ben Wieland points out, this is false in the category of compact, complex manifolds. The examples are interesting (to me) because they also come up in showing "rationally connected" fibrations over a Riemann surface in the analytic category need not admit sections. Begin with the $\mathbb{C}^\times$ -torsor $T$ over $\mathbb{P}^1$ associated to any nontrivial invertible sheaf, e.g., $\mathcal{O}_{\mathbb{P}^1}(1)$. Now let $q:T\to Y$ be the fiberwise quotient by multiplication by some element $\lambda\in \mathbb{C}^\times$ of modulus $\neq 1$. Projection of $T$ to $\mathbb{P}^1$ factors through this quotient, $f:Y\to \mathbb{P}^1$. Although every fiber of $f$ is the same Hopf surface elliptic curve, there is no section: if there were, its inverse image in $T$ would be a disjoint union of sections of $T$ (since $\mathbb{P}^1$ is simply connected), and $T$ has no sections.

A Positive Answer by Paul Seidel. 22 September 2017. I received a communication from Paul Seidel that he knows how to prove this using methods of symplectic topology.

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    $\begingroup$ Those are never smooth, I think? $\endgroup$
    – R.P.
    Oct 23, 2014 at 11:30
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    $\begingroup$ @AlexDegtyarev Such fibrations are always singular. Let $f:X\to \mathbb P^1$ be a smooth proper morphism with connected fibres curves of genus one. Then the Jacobian fibration $p:J\to \mathbb P^1$ is a smooth family of elliptic curves. But the moduli of elliptic curves is affine. $\endgroup$
    – Ariyan Javanpeykar
    Oct 23, 2014 at 11:31
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    $\begingroup$ I meant to write "non-smooth". (The flat morphism $f$ has singular fibres.) $\endgroup$
    – Ariyan Javanpeykar
    Oct 23, 2014 at 11:39
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    $\begingroup$ @JasonStarr As you probably already know, if the Kodaira dimension of $Y$ is non-negative, then this follows from Theorem 0.1 in a paper of Moller-Viehweg-Zuo arxiv.org/abs/math/0512154 . Probably, one can even show that the Kodaira dimension of the fibres of $f$ must be negative. (Use a polarization and the fact that Isom-schemes are finite etale for non-birationally ruled fibres to deal with the isotrivial case. The non-isotrivial case should follow from Theorem 0.1 in loc. cit. I think.) $\endgroup$
    – Ariyan Javanpeykar
    Oct 23, 2014 at 11:59
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    $\begingroup$ @DanielLitt: I doubt that would work. First of all, if the torsor is Zariski locally trivial, then it is globally trivial by the valuative criterion. Second, the nontrivial families $A$ that I know of (e.g., Moret-Bailly's pencils) are (inseparably) dominated by constant families, $A_0\times \mathbb{P}^1$. So the torsor is covered by a torsor for a constant family. $\endgroup$
    – Jason Starr
    Oct 23, 2014 at 15:20

1 Answer 1

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Yes, using some symplectic geometry. Let's say we had $X \subset {\mathbb C}P^n \times {\mathbb C}P^1$, with projection to $\mathbb{C} P^1$ a smooth morphism (meaning, for topologists, a proper holomorphic submersion; all fibres are smooth). With the restriction of the standard Kaehler form, this becomes a symplectic fibre bundle over ${\mathbb C}P^1 = S^2$. It is known that the Gromov-Witten invariant counting sections of any such fibration (with suitable incidence conditions) is always nonzero. This is proved by reversing the orientation of $S^2$ to get another such fibration, and then considering the fibre union of those two as a degeneration of the trivial fibration. References:

PS: Slight edits, I hope this clarifies (and that I have not misunderstood the question, of course). I added another reference (the third one): see Theorem 1.5 there, and note that the map takes values in the invertible elements of quantum cohomology (hence is nonzero).

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    $\begingroup$ This is very exciting! However, I find it hard to locate the precise statements needed from the papers you linked. Could you please provide more detail and give more precise references? $\endgroup$
    – R. van Dobben de Bruyn
    Sep 23, 2017 at 3:39
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    $\begingroup$ It might help to note a few things: every symplectic fibration over the disc is trivial, therefore, symplectic fibrations over $S^2$ are determined up to symplectomorphism by the homotopy class of a clutching function (i.e. a map from $S^1$ to the symplectomorphism group of the fibre). One small technical point is that the projectivity assumption implies a reduction of this clutching function from the symplectomorphism group to its Hamiltonian subgroup. The linked papers about the Seidel homomorphism are phrased in terms of this subgroup. $\endgroup$
    – Mohammed Abouzaid
    Sep 23, 2017 at 13:30
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    $\begingroup$ Ignoring technical details, the strategy of the proof is simply to start with $M \times S^2$ with a product symplectic form and complex structure, which clearly has sections, and degenerate it to a symplectic sum of a fibration and its inverse (I think this is clearest in terms of clutching functions). In the process, you will have to deform the almost complex structure. In order to keep control of the sections you start with, you should perform the deformation within the class of fibrations such that the fibres are almost complex submanifolds. $\endgroup$
    – Mohammed Abouzaid
    Sep 23, 2017 at 13:34
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    $\begingroup$ By "keep control" I mean that you should worry about a section breaking into a multi-section and a "negative section." The class of almost complex structures above should prevent this from happening. $\endgroup$
    – Mohammed Abouzaid
    Sep 23, 2017 at 13:36
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    $\begingroup$ You still have to do some non-trivial study of fundamental classes of moduli spaces of holomorphic spheres to make sure that the "count of sections through a point" is invariant. The problem is not about the degeneration itself (that only requires Gromov compactness), but about general almost complex structures on the product $M \times S^2$. I would not be surprised if a clever argument allows one to use a minimal amount of theory to establish the existence of curves without needing an "enumerative" interpretation. $\endgroup$
    – Mohammed Abouzaid
    Sep 23, 2017 at 13:44

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