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Nov 16, 2022 at 20:50 comment added Ariyan Javanpeykar 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.
Nov 16, 2022 at 8:03 vote accept CommunityBot
Nov 15, 2022 at 22:44 comment added Jason Starr Welcome new contributor. By Gritsenko-Hulek-Sankaran, the moduli spaces of genus $g$ K3 surfaces are of general type for $g$ sufficiently positive. So you cannot make the base be rational or elliptic.
Nov 15, 2022 at 20:10 history edited user494851 CC BY-SA 4.0
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Nov 15, 2022 at 19:45 answer added Sasha timeline score: 3
Nov 15, 2022 at 19:20 comment added user494851 @Sasha You are corrected, here I mean the fibre over a general point.
Nov 15, 2022 at 19:19 history edited user494851 CC BY-SA 4.0
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Nov 15, 2022 at 19:16 history edited LSpice CC BY-SA 4.0
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Nov 15, 2022 at 18:58 comment added Sasha 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?
Nov 15, 2022 at 18:45 history edited user494851 CC BY-SA 4.0
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Nov 15, 2022 at 18:45 comment added user494851 @Sasha They are the same algebraic complex K3 surface.
Nov 15, 2022 at 18:31 comment added Sasha What is the relation between $X$ in your first and second sentences?
Nov 15, 2022 at 18:17 history edited user494851 CC BY-SA 4.0
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S Nov 15, 2022 at 18:17 review First questions
Nov 15, 2022 at 22:46
S Nov 15, 2022 at 18:17 history asked user494851 CC BY-SA 4.0