Let $C$ be a non-hyperbolic (smooth quasi-projective connected complex algebraic) curve. That is, $C$ is isomorphic to $\mathbb P^1, \mathbb A^1, \mathbb G\_m$, or an elliptic curve.


Let $f:X\to C$ be a family of Fano varieties, i.e., $f$ is a smooth projective morphism whose geometric fibres are (smooth projective connected) Fano varieties. (Fano means anti-canonical bundle ample.)


Is $f$ isotrivial? In other words, are all the fibres of $f$ isomorphic?


**Note.** A family $f:X\to C$ of varieties with semi-ample canonical bundle is isotrivial. This follows from the work of Campana, Kebekus, Kovacs, Lieblich, Viehweg, Zuo, et al.


**Note.** If $f$ is non-isotrivial, the relative dimension of $f$ will have to be at least three.


**Motivation.** I think it is reasonable to suspect that certain connected components of the stack of Fano varieties have only finitely many integral points over $\mathbb Z$. If this expectation holds any family of Fano varieties $f:X\to C$, where $C$ is a non-hyperbolic curve, is isotrivial over $C$.



Families of Fano varieties over non-hyperbolic curves
Let $X = SO(10)/P\_5 \subset P^{15}$ be the spinor variety. It is projectively self-dual and has codimension 5, so its generic linear section of codimension 5 is smooth, and, moreover, generic pencil of codimension 4 is smooth. On the other hand, sections of codimension 4 are parameterized by $Gr(4,16)$ which has dimension 48, while the group of automorpisms of $X$ is $SO(10)$ and so has dimension 45. So, I guess that a generic pencil of linear sections of $X$ of codimension 4 is an example of a nonisotrivial family.


EDIT --- MORE DETAILS. The Fano varieties one gets in this way are 6-folds of index 4 and degree 12. Their Hodge diamond is diagonal and the diagonal Hodge numbers are $(1,1,1,2,1,1,1)$. They are Fano just by the adjunction formula.


