Let $X$ ba a smooth hyperelliptic curve of genus $g$, and let $f:X\rightarrow X$ be the hyperelliptic involution. Consider a $K3$ surface $S$ with an involution $g$ without fixed points. The quotient $F = S/g$ is an Eniques surface. 


Now $f\times g:X\times S\rightarrow X\times S$ does not have fixed points, and the quotient $(X\times S)/(f\times g)\rightarrow F$ is a family of hyperelliptic curves parametrized by $E$ whose fibers are all isomorphic to $X$.


My question is: why isn't the family $(X\times S)/(f\times g)\rightarrow F$ trivial?



Non trivial family of hyperelliptic curves
If it was, $Y:=(X\times S)/(f\times g)$ would be isomorphic to $X\times F$. One way to see this is not the case is to look at 3-forms: $X\times F$ has no nonzero holomorphic 3-forms (because $H^0(F,\Omega ^2\_F)=0$). On the other hand, let $\omega $ be the generator of $H^0(S,\Omega ^2\_S)$; for any $\alpha \in H^0(X,\Omega ^1\_X)$, the form $pr\_1^\*\alpha \wedge pr\_2^\*\omega $ on $X\times S$ is invariant under $f\times g$, hence comes from a 3-form on $Y$.


This is a typical example of an *isotrivial* family : it is not trivial, but becomes trivial after the étale base change $S\rightarrow F$. 


