Timeline for Does hypercohomology of the Koszul complex compute sheaf cohomology?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2017 at 22:15 | history | edited | 54321user | CC BY-SA 3.0 |
updated hypotheses
|
| Jun 7, 2017 at 22:01 | comment | added | 54321user | Hmm, with no common factors then. | |
| Jun 7, 2017 at 21:10 | comment | added | Jason Starr | Okay, form the Koszul complex of $(f,f^2)$. Those polynomials are not projectively equivalent, but you have the same problem. | |
| Jun 7, 2017 at 21:07 | comment | added | 54321user | and non-equal (or maybe even projectively equivalent) polynomials | |
| Jun 7, 2017 at 21:04 | comment | added | Jason Starr | The Koszul complex of $(f,f)$ is equivalent to the Koszul complex of $(f,0)$, and that is going to be a tensor product with a complex like the one in my previous comment. This issue is persistent. | |
| Jun 7, 2017 at 20:59 | comment | added | 54321user | Sure. I see how that is a problem, but if I restrict to the case of having polynomials of degrees $> 0$, my question still seems interesting. | |
| Jun 7, 2017 at 20:35 | comment | added | Jason Starr | I suggest that you compute what happens if you take $X$ equal to all of $\mathbb{P}^n$ and then you take $\underline{f}$ to be $(0,\dots,0)$, where the $\text{i}^{\text{th}}$ component is the zero homomorphism $\mathcal{O}_{\mathbb{P}^n}(d_i)\to \mathcal{O}_{\mathbb{P}^n}$, for whatever degree $d_i$ that you like. | |
| Jun 7, 2017 at 20:11 | history | asked | 54321user | CC BY-SA 3.0 |