Timeline for Is every curve on a projective three-fold a homology-theoretic complete intersection of sorts?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2023 at 10:13 | comment | added | naf | If $S$ exists for such a curve $C$, then by intersecting $S$ with various hyperplanes you would get curves not equal to $C$ having the same homology class (up to a scalar). The homology class of $C$ maps to zero in the homology of $M'$, but the class of no other curve can map to zero (assuming $M'$ is projective) since the class of any curve is always nonzero in the homology of a projective variety | |
| Dec 18, 2023 at 4:32 | comment | added | Vamsi | @naf: Thanks but sorry can you elaborate? Why does your comment imply that such an S does not exist? | |
| Dec 16, 2023 at 5:53 | comment | added | naf | @Vamsi: OK, this is indeed what my comment shows is not always possible. To get an explicit example, you can search for "Atiyah flop". | |
| Dec 16, 2023 at 1:50 | comment | added | Vamsi | I see. Sorry can you elaborate? By ampleness and Mori, the line bundle of every surface is presumably ample and hence no (r,0) ones I suppose. But why are there (1,0) curves? Sorry if I am making a stupid mistake but the ample hypersurface is CP3 and for that this property appears to be true. | |
| Dec 15, 2023 at 18:43 | comment | added | Jason Starr | That is not true. Let $M$ be a general ample hypersurface in $\mathbb{CP}^2 \times \mathbb{CP}^2$ of bidegree $(d,e)$ for $1\leq d \leq 3$ and $e\gg 0$. There exist curves $C$ in $M$ of bidegree $(1,0)$, but there are no surfaces in $M$ of bidegree $(r,0)$. | |
| Dec 15, 2023 at 16:41 | comment | added | Vamsi | I guess I am asking whether given a C, there is an S, such that a multiple of C is S cupped with omega. | |
| Dec 15, 2023 at 12:44 | comment | added | naf | Is your question asking whether the class of $C$ (in cohomology) is equal to the class of $S$ cupped with the class of $\omega$? If so, I don't think this is true: the point is that given $S$, curves obtained by intersecting $S$ with various hyperplanes will fill out all of $S$, but there exist $M$ and curves $C$ on $M$ so that any (reduced) curve on $M$ whose cohomology class is a multiple of the class of $C$ is actually equal to $C$. This happens when there is a morphism $M \to M'$ which is an isomorphism onto its image outside $C$ but maps $C$ to a point. | |
| Dec 15, 2023 at 6:12 | history | asked | Vamsi | CC BY-SA 4.0 |