# Families of Fano varieties over non-hyperbolic curves

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 $C$.

• Even though you specify "complex", I want to point out that these are fairly easy to find in positive characteristic, e.g., using the pencil of plane curves that I describe here. You can consider $[u,v,w]=[1,1,1]$ as a point $p$ on those curves, and then you can take the moduli space of rank $r$ bundles whose determinant is isomorphic to $\mathcal{O}_C(p)$. – Jason Starr Jun 2 '14 at 12:59
• hyperbolicity properties of moduli spaces and the existence of non-isotrivial families in the sense of the hyperbolicity conjecture of ${Shafarevich}$ correspond to semi-positivity of relative canonical bundle $K_{X/C}$ , in fact if the fibers are K-stable Fano varieties then it corresponds to semi-positivity of fiberwise Kahler-Einstein metric (such relative Kahler metrics are not positive or closed in general) – user21574 Nov 17 '17 at 7:00
• To understand more about my previous comment see my comments here mathoverflow.net/questions/284809/… . My conjecture is that semi-positivity of such fiberwise Kahler-Einstein metric is corresponds the relative tangent sheaf $(T_{X/C})^{**}$ be stable in he sense of Mumford. – user21574 Nov 17 '17 at 7:03
• Another way to attack this problem is to break it up to rigidity+Hyperbolicity+boundedness of degree of direct image of relative canonical divisor see crm.umontreal.ca/Holomorphes07/pdf/viehwegslides.pdf citeseerx.ist.psu.edu/viewdoc/… and sites.math.washington.edu/~kovacs/2013/papers/… and sites.math.washington.edu/~kovacs/2013/papers/… – user21574 Nov 17 '17 at 7:45
• From Ricci flow approach the hyperbolicity conjecture of Shafarevich about the existence of non-isotrivial family corresponds to convergence of the relative Kahler Ricci flow and $C^0$-estimate and solutions remain semi-positive. arxiv.org/pdf/1709.05465.pdf . In fact, this is the right flow to give an effective way for the conjecture of Griffiths (in this case the total space we need to take $\mathbb P(E)$). In fact, Ricci flow resolves the singularities by one time $t$, but in relative Kahler-Ricci flow, we have two times $s$ and $t$. where $t\to \infty$ and $s\to 0$, – user21574 Nov 17 '17 at 18:44

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.

• Are you able to provide further details for how your construction works? For example, what are the Fano varieties in your construction, the intersection of $X$ with a linear subspace of $\mathbb{P}^{15}$ of codimension $4$? Why are these Fano? – Daniel Loughran Jun 2 '14 at 7:06
• @Daniel: I added some details. If you want to know something else I will be happy to tell you. – Sasha Jun 2 '14 at 8:50
• I am a bit skeptical. By my computation, for a pencil of codimension $q$ linear sections that contains (at most) finitely many singular members, the number of singular members equals the degree in $A_0(X)$ of the cycle class $c_1(\mathcal{O}_{\mathbb{P}^{15}}(1)|_X)^{q-1}c_{11-q}(\Omega_X(1)) + c_1(\mathcal{O}_{\mathbb{P}^{15}}(1)|_X)^{q}c_{10-q}(\Omega_X(1))$. Of course, since $X$ is homogeneous, we can compute this explicitly. Although I have not done the computation, I see no a priori reason that the degree should be zero. – Jason Starr Jun 2 '14 at 12:31
• @Jason: The reason is the large codimension of the projectively dual variety. E.g. if you take $q = 1$ then your formula should give the degree of the dual variety, if it is a hypersurface, or zero, if it has larger codimension. – Sasha Jun 2 '14 at 12:41
• @Sasha: Now I see. I did not get the "projectively self-dual until now". – Jason Starr Jun 2 '14 at 12:47

Unless you ask for minimal Fano, this is false. Take a rational curve $C$ on a Fano manifold $M$ (say, del Pezzo surface), and consider a family of blow-ups of $M$ parametrized by $C$, obtained by blowing up a point in $C$. A blow-up is often Fano, but this family is not always isotrivial (say, for an appropriate choice of del Pezzo $M$).

Update: for del Pezzo this argument does not work - sorry!

• I'm not entirely convinced by this argument. There could be some points on $C$ which when blown-up do not yield a Fano variety, for example $C$ could be a curve on a del Pezzo surface which intersects some exceptional curves. In which case the family of Fano varieties would be over a punctured curve, which could quite possibly hyperbolic. Are you able to construct examples which avoid this phenomenon? – Daniel Loughran Jun 1 '14 at 12:25
• you are right, for del Pezzo this argument does not work! Any such $C$ will meet a quadric passing through the points which are blown up, hence one of its points is not in general position. My apologies. – Misha Verbitsky Jun 1 '14 at 13:01