# Families of curves on compact complex surfaces and algebraicity

Let $S$ be a compact complex manifold of dimension $2$ and assume that there exists a two-dimensional family of curves on $S$. Is it true then that the algebraic dimension of $S$ is $2$, i.e. that $S$ is Moishezon (and hence projective)?

By a two-dimensional family of curves I mean a complex analytic subset $X \subset S \times T$ where $T$ is some compact complex space of dimension 2, and such that fibres $X_t$ are complex spaces of dimension 1 (possibly singular and reducible), and such that $X_t$ and $X_s$ are distinct for $t \neq s$. One can assume that the projection of $X$ on $T$ is flat or smooth, if it helps.

• What do you mean by a "two-dimensional family of curves" ?
– js21
Apr 17, 2017 at 9:53
• @js21: by a two-dimensional family of curves I mean a complex analytic subset $X \subset S \times T$ where $T$ is some compact complex space of dimension 2, and such that fibres $X_t$ are complex spaces of dimension 1 (possibly singular and reducible). Apr 17, 2017 at 10:34
• Just to be sure: is $X$ closed in $S\times T$? Also, you need to assume somehow that the family is "really'' two-dimensional. Apr 17, 2017 at 13:08
• @Laurent Moret-Bailly: yes to both concerns. Fibres are distinct for distinct points in $T$ Apr 17, 2017 at 13:25
• @DimaSustretov Are you assuming $T$ is an algebraic surface? Also, do I understand correctly that $X\to T$ is a family of curves? Is the genus of this family at least two? Apr 17, 2017 at 19:48

Write your analytic family of curves as $$\{X_t\} = Z + \{M_t\},$$ where $Z$ is the fixed part (i.e., the maximal effective divisor contained in any member of the family) and $\{M_t\}$ is a $2$-dimensional analytic family with at most isolated base points (the mobile part).
Then, calling $M$ the cohomology class of $M_t$, which is by definition independent on $t$, we claim that $M^2 >0$. In fact, given any irreducible $M_t$ and a general point $p$ on it, our assumption on the dimension implies that we can find an irreducible $M_s$, with $s \neq t$, such that $p \in M_s$, hence $p \in M_t \cap M_s$ and this proves our claim.
Remark. If the general element of the mobile part $\{M_t\}$ is not irreducible, the result is in general false. For instance, take a surface $S$ with algebraic dimension $1$, and consider the fibration $f \colon S \to \mathbb{P}^1$ given by the essentially unique meromorphic function on $S$ (if necessary, blow-up some point on $S$ in order to make it a holomorphic map). Writing $F_{u}$ for the fibre over $u \in \mathbb{P}^1$, the linear system $|F_0 + F_{\infty}|$ provides a $2$-dimensional family of curves on $S$, whose members are all disjoint unions of two fibres.