# Torsion of the Picard group for surfaces isogenous to a product

We say that a complex surface $S$ is isogenous to an (unmixed) product if there exists a finite group $G$, acting faithfully on two smooth projective curves $C_1$ and $C_2$ and freely on their product (with the diagonal action), so that $S$ is isomorphic to $(C_1 \times C_2)/G$.

If $S =(C_1 \times C_2)/G$ is any surface isogenous to a product, the two natural projections of $C_1 \times C_2$ induces two fibrations $f_1 \colon S \to C_1/G$ and $f_2 \colon S \to C_2/G$, whose smooth fibres $F_1$ and $F_2$ are isomorphic to $C_2$ and $C_1$, respectively. Moreover, since the action of $G$ on the product is free, all the singular fibres of $f_1$ and $f_2$ are multiple of smooth curves.

Let us consider now the group $\textrm{Tors(Pic}^0 S)$. Inside this group naturally lives the subgroup $H$ generated by $f_1^* \textrm{Tors(Pic}^0 C_1/G)$ and $f_2^* \textrm{Tors(Pic}^0 C_2/G)$.

Question. Are there conditions on $C_1$, $C_2$, $G$ and the actions ensuring that $H=\textrm{Tors(Pic}^0 S)?$ In other words, when the torsion part of the Picard group of $S$ is determined by the torsion parts of the bases $C_1/G$ and $C_2/G$ of the two natural fibrations?

I think the answer to the question as it is stated is negative: if I do understand correctly the question I can prove that, if $G$ is abelian and both curves $C_i/G$ are rational, this is never true.

Indeed, since ${\mathbb P}^1$ has no torsion, $H$ is trivial, so it is enough if I construct a non trivial torsion bundle on $S$.

Let $p_1,\ldots,p_r$ be the critical values of $f_1$. These are the points whose fibre are not reduced, say of multiplicity $m_i>1$; as you know very well, since $C_1/G$ is rational and $G$ is abelian, $r\geq 2$ and each $m_i$ is a divisor of the l.c.m. of the other $m_j$. In particular, up to renumbering the points, I can suppose that $d:=\gcd(m_1,m_2)>1$.

Set $n_i:=\frac{m_i}{d}$, call respectively $F'$ and $F''$ the reduced fibres over $p_i$ and $p_2$, so that $m_1F'$ and $m_2F''$ are fibres of $f_1$. Then the divisor $D:=n_1F'-n_2F''$ is a nontrivial d-torsion divisor. Indeed $dD=m_1F'-m_2F''=f_1^*p_1-f_1^*p_2$ is principal, and by the standard results on the normal bundles of a multiple fibre taken with the reduced structure $$lD_{|F'}=l(n_1F'-n_2F'')_{|F'}=ln_1F'_{|F'}$$ is trivial iff $ln_1$ is a multiple of $m_1$, that is iff $d$ divides $l$.

Let me add some comments. Even if we remove my assumptions, my construction, for each pair of $m_i$ not relatively prime on the same side, produces a torsion bundle, and I do not see any reason for it to be in $H$ even when $H$ is very big. Still these torsion bundle are determined by the fibrations, so I would modify your question as follows

Question: Under which assumptions is the group generated by $H$ and the bundles constructed above the whole torsion group of $S$?

That could be true in a wider range of cases. One could try to prove it in the regular+abelian case by using a theorem of Armstrong as it was used here for computing some fundamental groups.