Let $S=E\times C$ be a product of two curves, where $E$ is an elliptic curve and $C$ is a curve of genus at least two. Consider a foliation on $S$ generated by a global holomorphic 1form $p_1^*(\omega_1)+p_2^*(\omega_2)$, where $p_i$ is a projection map and $\omega_1$ and $\omega_2$ are nonzero holomorphic 1forms on $E$ and $C$ respectively. Is it possible to say something about algebraicity of leaves of this foliation?

$\begingroup$ Ths foliation is the pullback of a Kronecker foliation on the Albanese torus of $S$. (See Brunella's book Birational Geometry of Foliations.) If the image of the Albanese map is onedimensional. Then your foliation is the Albanese fibration. But if the image of the albanese map has higher dimension I don't know exactly what may happen... $\endgroup$– Alan MunizFeb 14, 2016 at 23:58

$\begingroup$ What may help is that if one leaf is algebraic, then all will be algebraic... $\endgroup$– Alan MunizFeb 15, 2016 at 0:01

1$\begingroup$ Isn't the Albanese map of S an embedding? It takes S=E x C to E x J(C) where J(C) is the Jacobian of C. And it is a product map; E goes to E (with the identity) and C goes to J(C) by the natural embedding? $\endgroup$– PierreFeb 15, 2016 at 4:18
2 Answers
The foliation $\mathcal F$ defined by $p_1^* \omega_1 + p_2^* \omega_2$ is everywhere transverse to the fibration $p_2 : S \to C$. One can therefore lift paths from $C$ to leaves of $\mathcal F$ in order to obtain a representation $$ \rho: \pi_1(C,b) \to \mathrm{Aut}(p_2^{1}(p)) \simeq \mathrm{Aut}(E) \, . $$ Since the foliation is defined by a closed $1$form it is a simple matter to realize that the image of this representation is contained in the subgroup of translations of $\mathrm{Aut}(E)$. It is then clear that the leaves of $\mathcal F$ are algebraic if, and only if, the image of $\rho$ is contained in a torsion subgroup of $E$.
Notice that the representation in question is rather simple. If we identify $E$ with $\mathbb C / \Gamma$ where $\Gamma$ is a lattice in $\mathbb C$ then $\omega_1$ can be identified with $\lambda [dz]$. Therefore $$ \rho(\gamma) = \lambda^{1} \cdot \int_{\gamma} \omega_2 \in (\mathbb C / \Gamma,+) \subset \mathrm{Aut}(E) \, $$ for any $\gamma \in \pi_1(C,b)$.
When $\mathcal F$ has algebraic leaves then Poincaré irreducibility Theorem (see for example Debarre's book Complex Tori and Abelian Varieties) implies that the Jacobian of $C$ is isogeneous to $E \times A$ where $A$ is an abelian variety of dimension $g(C)1$. Reciprocally when the Jacobian of $C$ has this property we can arguee as Pierre did in other answer to this question to produce a foliation having all its leaves algebraic.

$\begingroup$ Nice proof, Jorge! But it took me 10 minutes to understand that $A^{g1}$ was not the $(g1)$th power of an abelian variety... :( $\endgroup$– ACLFeb 15, 2016 at 20:57

$\begingroup$ Thanks. I have edited to prevent others from losing time... :) $\endgroup$ Feb 15, 2016 at 21:19


I don't know if this is an answer, but at least one can build examples with all leaves closed.
If one picks C such that its Jacobian is a product of several complex tori, one of which is an elliptic curve. For instance one could take C of genus 2 such that its jacobian is a product $E_1 \times E_2$ with the $E_i$ elliptic curves.
Then one considers the map $F : E_1 \times C \to E_1$ obtained by composing the natural maps
$E_1 \times C \to E_1 \times E_1 \times E_2 \to E_1 \times E_1 \to E_1$ where the last arrow is the sum.
The pullback by F of a nonzero holomorphic form on $E_1$ should give an example.