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Let $X\rightarrow S$ be a (projective, flat,... or any other assumption which makes you happy) fibration of a smooth threefold over a smooth surface with connected one-dimensional fibers.

As an example you may consider a generic hypersurface of bi-degree (3,3) in $\mathbb{P}^2\times \mathbb{P}^2$ with the fibration induced from either of fibrations $\mathbb{P}^2\times \mathbb{P}^2 \rightarrow \mathbb{P}^2 $.

My question is, how we can find whether this fibrations has a section or not? Are there any known methods which might help us to answer this question for a given fibration? Or any obstruction?

Remark: It is a known theorem that any elliptic fibration with section is a resolution of a Weierstrass model, but for a given elliptic fibration as above, I don't know how to find whether it has a section or not.

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Finding sections of fibrations is in general just as difficult as finding rational points (which is notoriously hard !). In this case, if $X\to S$ is a (flat) fibration in curves of genus $>1$ and of non-zero Kodaira-Spencer class, finding a rational section is a special case of the Mordell conjecture over function fields. Szpiro gave an effective proof of this conjecture in "Propriétés numériques...", Astérisque 86 (1981) but it is in practice hard to use. If you find a section, you still have to determine if it extends. – Damian Rössler Aug 22 '11 at 23:14
For a smooth hypersurface $X$ of bi-degree $(3,3)$ in $\mathbb{P}^2 \times \mathbb{P}^2$, you can use the Lefschetz hyperplane theorem to see that the restriction map $\text{Pic}(\mathbb{P}^2 \times \mathbb{P}^2) \to \text{Pic}(X)$ is an isomorphism. So every invertible sheaf is the restriction of $\mathcal{O}(a,b)$ for integers $a$, $b$. The intersection number with a fiber over either of the two projections is then $3a$, respectively $3b$. continued – Jason Starr Aug 23 '11 at 1:44
Continued from previous comment. If you have a rational section of the projection, the closure of the image gives a Cartier divisor in $X$ whose intersection number with a fiber equals $1$. So there is no rational section. This is just one of many techniques for proving $\textbf{nonexistence}$ of rational sections / rational points. There are also a number of techniques for proving $\textbf{existence}$, the simplest of which is just guessing a solution. But as Damian R$\"{o}$ssler points out, provably (by work of Kim and Roush) there is no general algorithm deciding existence of a section. – Jason Starr Aug 23 '11 at 1:50
@Jason: The non-existence of a general algorithm only applies for fibrations of arbitrary relative dimension. In the OP's case, he wants relative dimension one. – Felipe Voloch Aug 23 '11 at 14:13

Cohomology is a useful tool for finding obstructions to the existence of sections. Note that if $p\colon\thinspace X\to S$ has a section $s\colon\thinspace S\to X$, then $p\circ s$ is the identity map on $S$, and by functoriality $s^\ast\circ p^\ast$ is the identity homomorphism on $H^\ast(S)$, where $H^\ast$ is your favourite cohomology theory. Hence if $\alpha\in H^\ast(S)$ is a nonzero class such that $p^\ast(\alpha)\in H^\ast(X)$ is zero, then $p$ does not admit a section.

Assuming the base space has the homotopy type of a CW-complex (as in the case at hand) then obstruction theory may be useful in proving that a section exists. With connected fibres, one can always construct a section over the $1$-skeleton of the base. Then the only obstruction to extending to the whole base is a cohomology class in $H^2(S;\pi_1(F))$ (cohomology of the base with coefficients in the fundamental group of the fibre).

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Do you know any reasonable example, for which the pull-back of some cohomology class is zero? – Mohammad F. Tehrani Aug 23 '11 at 18:11
A simple example (which may not fit into your framework) is the Hopf fibration $S^3\to S^2$. The fundamental cohomology class of the base pulls back to zero for dimensional reasons. I don't know enough about cohomology of projective hypersurfaces, but I'm sure there must be examples there too. – Mark Grant Aug 23 '11 at 21:28

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