# algebraic leaves of foliation on a product of two curves

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 1-form $p_1^*(\omega_1)+p_2^*(\omega_2)$, where $p_i$ is a projection map and $\omega_1$ and $\omega_2$ are nonzero holomorphic 1-forms on $E$ and $C$ respectively. Is it possible to say something about algebraicity of leaves of this foliation?

• Ths foliation is the pull-back 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 one-dimensional. 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... Feb 14, 2016 at 23:58
• What may help is that if one leaf is algebraic, then all will be algebraic... Feb 15, 2016 at 0:01
• 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? Feb 15, 2016 at 4:18

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.

• Nice proof, Jorge! But it took me 10 minutes to understand that $A^{g-1}$ was not the $(g-1)$th power of an abelian variety... :-(
– ACL
Feb 15, 2016 at 20:57
• Thanks. I have edited to prevent others from losing time... :) Feb 15, 2016 at 21:19
• Muito bonito Jorge. Feb 18, 2016 at 15:00
• Merci @Pierre . Feb 19, 2016 at 13:28

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 pull-back by F of a nonzero holomorphic form on $E_1$ should give an example.