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Given a family of complex compact surfaces $X \to B$ over a smooth curve $B$ and a one-dimensional family of effective Cartier divisors $\{D_t\}$ in the central fiber $X_0$. For simplicity, let's just assume that $X$ and $X_0$ are smooth. I would like to know what are sufficient conditions that guarantee the existence of a one-dimensional family of (effective Cartier) divisors $\{\Delta_t\}$ in $X$ such that:

  1. $\{\Delta_t\} \cap X_s$ is again a family of divisor in $X_s$ for all $s \in B$;
  2. $\{\Delta_t\} \cap X_0$ coincide with $\{D_t\}$.

If such an extension doesn't exist, can we just do some base change of $B$ to ensure the existence of $\{\Delta_t\}$?

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  • $\begingroup$ Can $X_0$ be singular ? Then which definition of divisors on $X_0$ do you use ? $\endgroup$
    – Cantlog
    Commented Aug 31, 2013 at 21:18
  • $\begingroup$ Thanks for asking, I just edited my question, $\endgroup$
    – JacobI
    Commented Aug 31, 2013 at 22:22
  • $\begingroup$ Consider the constant family $D_t=$ one given point of $X_0$. To deform it to $X$, $X\to B$ must have a section. So the answer is no in general. But if you allow an extension of the family of parameters $\{ t\}$, then the answer will be yes. $\endgroup$
    – Cantlog
    Commented Sep 1, 2013 at 11:06

1 Answer 1

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Before trying to deform the family of divisors, it makes sense to first consider deforming even a single divisor $D_0$ from $X_0$ to the family $X$. This is not possible in general: the cycle class $c_0$ of $D_0$ is a $(1,1)$-form in $H^2(X,\mathbb C)$, but when you deform the cycle class to nearby fibres $X_s$ (via the Gauss--Manin connection) the class $c_s \in H^2(X_s,\mathbb C)$ may no longer be a $(1,1)$-class. Indeed, this is the unique obstruction to deforming $D_0$ in the family. (This is called the variational Lefschetz $(1,1)$-theorem; it is variation on the usual Lefschetz $(1,1)$-theorem, and can be proved in a similar way.)

A basic example would be to let $E_s$ be a non-trivial family of elliptic curves for which $E_0$ is CM, by $\mathbb Z[\sqrt{-D}]$ say, and then to let $X_s := E_s \times E_s$. In $X_0$ we have a divisor $D_0$ given by the graph of $\sqrt{-D}$, which does not deform to the nearby $X_s$, since $E_s$ will not be CM for nearby curves $E_s$ (since the family is non-trivial).

Note that this cohomological obstruction can't be removed just by making a base-change in the parameter space $B$.


Just to record it, another example can be obtained by considering families of $K3$ surface. The generic Neron--Severi rank of a K3 is $1$, but it can get all the way up to $20$. So if we start with a K3 of NS rank $> 1$ and deform it, the NS rank will drop back to $1$ at nearby K3's, and so there will be divisors on the original surface which don't deform.

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