Timeline for How to compute the genus of the (singular) intersection of three quadratics in $\mathbb{C}P^4$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2021 at 15:29 | comment | added | Will Sawin | I believe Hartshorne contains (1) how to compute the arithmetic genus of a complete intersection, (2) resolving a node singularity reduces the arithmetic genus by 1, (3) for a smooth irreducible curve the arithmetic genus matches the geometric genus. | |
| Jan 12, 2021 at 15:24 | comment | added | Zhaoting Wei | @WillSawin "This is a valid deduction if the curve is irreducible and the singularities are all nodes." Do you know where I can find this result? | |
| Jan 10, 2021 at 15:39 | comment | added | Will Sawin | This is a valid deduction if the curve is irreducible and the singularities are all nodes. Can you check this? | |
| Jan 8, 2021 at 20:44 | history | edited | Zhaoting Wei | CC BY-SA 4.0 |
added 22 characters in body
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| Jan 8, 2021 at 20:44 | comment | added | Sasha | Not two, but four! | |
| Jan 8, 2021 at 20:43 | comment | added | Zhaoting Wei | @Sasha Yes the only possibilities are $x_1=\pm x_3$ and $x_2=\mp x_4$ so there are two components. I made some changes on the problem. | |
| Jan 8, 2021 at 20:22 | comment | added | Sasha | The curve is not indecomposable, so what do you mean by its genus? | |
| Jan 8, 2021 at 20:03 | history | asked | Zhaoting Wei | CC BY-SA 4.0 |