Timeline for 2 K3s and cubic fourfolds containing a plane
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2020 at 7:52 | comment | added | IMeasy | @Sasha: sure it was a typo! $(d+2)/2$, thanks. But still the question stays open. | |
| Sep 23, 2020 at 4:25 | comment | added | Sasha | @IMeasy: Do you mean degree $d$ and genus $d/2 + 1$? | |
| Sep 22, 2020 at 19:44 | comment | added | IMeasy | Still that degree 8 (and sectional genus 5) K3 must have something to do with the cubic fourfold. In the $\mathcal{C}_d$ cases where there is an associated K3, the surface has degree $d$ and genus $2d-2$. This is why I suspect that when such a cubic 4fold has an associated K3 (for example if it is in $\mathcal{C}_8\cap \mathcal{C}_{14}$ or 26), then this K3 may be related to the octic. | |
| Sep 21, 2020 at 17:19 | comment | added | Sasha | FM partners of a given K3 form a discrete set in the moduli space, so a general member of a non-trivial family of K3 (like intersections of three quadrics from your question) cannot be a partner of the K3 associated with the quadric fibration. | |
| Sep 21, 2020 at 16:25 | comment | added | abx | Indeed what is well-defined is a net of 2-dimensional quadrics in $X$ (Weil divisors). This does not define a K3 in $\Bbb{P}^5$. | |
| Sep 21, 2020 at 15:33 | comment | added | IMeasy | @RP: good point, I nedd to think about it but probably you are right. But still the question makes (some) sense - I guess. | |
| Sep 21, 2020 at 14:04 | comment | added | R.P. | Is the octic K3 uniquely determined up to isomorphism? I mean we can choose the L_i differently, which changes the Q_i, which conceivably -- to me -- changes the K3. | |
| Sep 21, 2020 at 13:51 | history | asked | IMeasy | CC BY-SA 4.0 |