Timeline for $K3$ surfaces in $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 1, 2022 at 16:57 | comment | added | Basics | In the question, it is not assumed that $\rho$ necessarily comes from an automorphism of $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$ . | |
| May 1, 2022 at 16:26 | comment | added | Jason Starr | Yes, I was assuming that the automorphism comes from $\mathbb{P}^1\times \mathbb{P}^1\times \mathbb{P}^1$. If the automorphism does not arise in this way, then there is no reason that the normal bundle should be $\rho$-invariant. | |
| May 1, 2022 at 15:36 | comment | added | Basics | Why is the normal bundle $\rho$-invariant? Are you assuming that $\rho$ comes from an automorphism of $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$? | |
| May 1, 2022 at 15:01 | comment | added | Jason Starr | You added a hypothesis about a specified primitive ample divisor class. For most choices of primitive ample divisor class, the answer is "no". The normal bundle of the surface in $\mathbb{P}^1\times \mathbb{P}^1\times \mathbb{P}^1$ is already one $\rho$-invariant divisor class. If you specify another $\rho$-invariant divisor class that is not a (rational) multiple of the class of the normal bundle, then necessarily the "invariant Picard rank" is at least $2$. | |
| May 1, 2022 at 11:48 | answer | added | Joe Silverman | timeline score: 2 | |
| May 1, 2022 at 10:26 | comment | added | Basics | @Jason star, no other requirements are put on, hence you answered the question! Being excited by such a quick answer, I changed the question a bit, which I would like to know the answer originally. | |
| May 1, 2022 at 10:26 | history | edited | Basics | CC BY-SA 4.0 |
added 12 characters in body
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| May 1, 2022 at 10:07 | comment | added | Jason Starr | What kind of automorphism? There is an obvious cyclic automorphism of order $3$ of $\mathbb{P}^1\times\mathbb{P}^1\times \mathbb{P}^1$ with rank $1$ invariant subgroup of the Picard group., It looks like the usual Noether-Lefschetz argument will give that a sufficiently general $(2,2,2)$-hypersurface that is stabilized by the automorphism will have the same Picard group (and will have $6$ fixed points). However, if your automorphism is free on the K3 surface, obviously this forces the K3 surface to have "invariant Picard rank" equal to $10$. | |
| May 1, 2022 at 8:02 | history | edited | Basics | CC BY-SA 4.0 |
edited title
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| May 1, 2022 at 7:52 | history | asked | Basics | CC BY-SA 4.0 |