Timeline for Picard Group of Projective Bundle over an Integral scheme
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 22, 2019 at 4:48 | answer | added | Y_q | timeline score: 2 | |
| Apr 11, 2011 at 6:57 | vote | accept | HNuer | ||
| Apr 9, 2011 at 17:38 | comment | added | HNuer | Thanks for the suggestion about the Euler characteristic. I'm only aware of, but comfortable yet with, the generalized Grothendieck-Riemann-Roch theorem so I came up with a more elementary argument just using semicontinuity and you comment about the Euler char. | |
| Apr 9, 2011 at 17:34 | answer | added | HNuer | timeline score: 2 | |
| Apr 8, 2011 at 19:09 | answer | added | Georges Elencwajg | timeline score: 8 | |
| Apr 8, 2011 at 14:11 | comment | added | Piotr Achinger | If you don't want to use intersection theory, note that the function $y\mapsto \chi(X_y, \mathcal{M}_y)$ is locally constant (Hartshorne III 12). Then use Riemann-Roch. | |
| Apr 8, 2011 at 14:00 | comment | added | Francesco Polizzi | Sorry, I vas thinking of $\textrm{rank}(\mathcal{E})=3$. If $\textrm{rank}(\mathcal{E})=n+1$, the degree of the restriction to a fiber is the $n$-th root of $\mathcal{L}^n \cdot F$. A good reference is Hartshorne, but the most complete treatment of intersection theory is given in Fulton's book. | |
| Apr 8, 2011 at 13:14 | comment | added | HNuer | Fair enough. I just haven't seen the degree defined that way before so I'm not yet comfortable with it. Care to provide a good reference? Couldn't one argue something similar, namely that since the fibres are isomorphic the induced invertible sheaves must be isomorphic? I'm not being precise here, I admit, since that doesn't mean that the induced invertible sheaf by some random isomorphism will be the invertible sheaf induced by restriction to the other fiber, but I believe this idea should lead to a precise answer, no? | |
| Apr 8, 2011 at 13:04 | comment | added | Francesco Polizzi | Yes it follows from the degree being 0. It seems to me that the degree is the natural way to treat this kind of problems | |
| Apr 8, 2011 at 12:56 | comment | added | HNuer | Doesn't effectiveness then follow from the degree being 0? Is there a way of seeing this wihtout the notion of degree? Thanks for your help. | |
| Apr 8, 2011 at 12:53 | comment | added | Francesco Polizzi | The degree of $\mathcal{L}:=\mathcal{M} \otimes \mathcal{O}_X(-m)$ restricted to a fibre $F$ is given by the square root of the intersection number $\mathcal{L} \cdot \mathcal{L} \cdot F$. This is clearly independent on the fibre, since all the fibres are algebraically equivalent. It follows that $\mathcal{L}$ has degree $0$ when restricted on each fibre. Since the fibre is a projective space, the restriction must be trivial. | |
| Apr 8, 2011 at 12:43 | comment | added | HNuer | How do you see that it's effective and of degree 0 on every fibre? | |
| Apr 8, 2011 at 12:40 | comment | added | Francesco Polizzi | It is effective and of degree $0$ on every fibre, so the restriction is trivial for every $y$. | |
| Apr 8, 2011 at 12:10 | history | asked | HNuer | CC BY-SA 3.0 |