1
$\begingroup$

Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}_t)=1$ for some closed $t\in S$.

I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.

Improve this question
$\endgroup$
6
  • $\begingroup$ What is the relation between $X$ in your first and second sentences? $\endgroup$
    – Sasha
    Commented Nov 15, 2022 at 18:31
  • $\begingroup$ @Sasha They are the same algebraic complex K3 surface. $\endgroup$
    – user494851
    Commented Nov 15, 2022 at 18:45
  • 2
    $\begingroup$ The first was complex, i.e., over $\mathbb{C}$, and the second is over the field of functions on $S$ (this would be the standard meaning of $\eta$), so they can't be the same. Or did you mean something else? $\endgroup$
    – Sasha
    Commented Nov 15, 2022 at 18:58
  • 1
    $\begingroup$ @Sasha You are corrected, here I mean the fibre over a general point. $\endgroup$
    – user494851
    Commented Nov 15, 2022 at 19:20
  • 1
    $\begingroup$ Just a little addendum to Jason Starr's comment. Namely, the moduli stack $M_d$ of polarized K3 surfaces of any fixed degree $d$ is hyperbolic (i.e., every holomorphic map $\mathbb{C}\to M_d^{an}$ is isoconstant). This implies that $S$ can't be $\mathbb{A}^1$, unless $X$ is already Picard rank one (take the trivial family). The reason $M_d$ is hyperbolic is because polarized K3 surfaces satisfy the infinitesimal Torelli property, so that hyperbolicity follows from a theorem of Griffiths-Schmid in Hodge Theory. $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 16, 2022 at 20:50

1 Answer 1

Reset to default
3
$\begingroup$

Choose any ample class in $\mathrm{Pic}(X)$, assume its degree is $d$. Let $M_d$ be the moduli space of polarized K3 surfaces of degree $d$ with appropriate level structure so that it has a universal family (alternatively one can use here the $\mathrm{Quot}$-scheme that is used to construct the moduli space). Let $x \in M_d$ be the point that corresponds to $X$. Let $S_0 \subset M_d$ be a general curve in $M_d$ through $x$, and let $S \to S_0$ be a resolution of singularities of $S_0$. Then the pullback to $S$ of the universal family from $M_d$ is a family with the required properties.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What is a general curve means? $\endgroup$
    – user494851
    Commented Nov 15, 2022 at 20:11
  • $\begingroup$ $M_d$ is a quasiprojective variety of dimension $19$. So, you can choose an embedding into a projective space, and intersect with $18$ general hyperplanes passing through $x$. $\endgroup$
    – Sasha
    Commented Nov 16, 2022 at 4:39

You must log in to answer this question.