Timeline for What are the indecomposable classes on a del-Pezzo surface?
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| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 28, 2015 at 6:20 | history | edited | Ritwik | CC BY-SA 3.0 |
explained that I have posted another question on mathoverflow
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| May 27, 2015 at 20:02 | answer | added | user47305 | timeline score: 2 | |
| May 27, 2015 at 14:25 | comment | added | user47305 | Sure, maybe it's a minor oversight. But certainly Manin and Kontsevich both know more about curves on del Pezzos than I ever will, so more likely I misunderstand what $N_\beta$ is. | |
| May 27, 2015 at 14:23 | comment | added | Ritwik | Sure. Let me think about your observations in the meantime...Kontsveich and Mannin do say that the numbers are ``expected'' to be one...but I imagine they would have thought about this counter example to their expectation. | |
| May 27, 2015 at 14:17 | comment | added | user47305 | I can't get chat log-in to work. I will try again later today. | |
| May 27, 2015 at 14:02 | comment | added | Ritwik | Let us continue this discussion in chat. | |
| May 27, 2015 at 14:01 | comment | added | user47305 | @Ritwik An algebraic curve on $X_8$ representing $3L - \sum E_i$ is the strict transform of a cubic in $\mathbb P^2$ that goes through the $8$ points. There is a 1-dimensional family of such cubics, but most of them are not parametrized by $\mathbb P^1$, and so don't contribute to $N_\beta$, as I understand it. The ones that count for the $N_\beta$ are the cubics that are rational, of which there are 12 in a generic pencil of cubics (these are the 12 cubics with a node). | |
| May 27, 2015 at 13:58 | comment | added | user47305 | @Ritwik An effective divisor on a del Pezzo is just an algebraic curve in $X_k$ (but maybe of higher genus). What Batyrev-Popov call "$M_{eff}(X_r)$" is the same thing as "$B$" in the Kontsevich-Manin paper, as far as I can tell. They show that all these classes are sums of $(-1)$-curve classes, hence decomposable, with the possible exception of the anticanonical on the $8$ point blow-up. But I have never thought about these sorts of questions, and it is entirely possible that I'm mistranslating something as I think about it! | |
| May 27, 2015 at 13:57 | comment | added | Ritwik | @Mark and Artie: I just want to be sure I understand the paradox correctly. Let $\beta := 3L -E1-\ldots -E_8$. Assuming this is indecomposable, by Kontsevich and Mannin, $N_{\beta}$ is expected to be one. You think $N_{\beta}$ ought to be $12$. Why is that? | |
| May 27, 2015 at 13:40 | comment | added | Ritwik | @Mark: I looked at the paper you sent (Cor 3.3 in particular). How is it answering my question? I am asking for a procedure to decide when is a class indecomposable. Cor 3.3 is talking about effective divisors. | |
| May 27, 2015 at 13:39 | comment | added | user47305 | @Ritwik There is no curve representing the class $L-E_1 -\cdots- E_8$ unless the 8 points are collinear. That is the class of a line going through all eight points. | |
| May 27, 2015 at 13:38 | comment | added | Ritwik | @ Mark: As per the definition of indecomposable; I said a class $\beta$ is indecomposable if it can be written as $\beta_1 + \beta_2 + \ldots $ where there are holomorphic representatives for each $\beta_i$. I did not say that they need be able to pass through a prescribed number of generic points. | |
| May 27, 2015 at 13:37 | comment | added | user5117 | @Ritwik: Mark's previous comment (which was in response to what appears to be the identical comment from you, now deleted) remains true. | |
| May 27, 2015 at 13:36 | comment | added | Ritwik | @Artie and Mark: I think the anticanonical class for k= is decomposable. The class is 3L−E1−…−E8. This can be written as 3(L−E1−…−E8)+2E1+…+2E8. This is a decomposition (there are holomorphic curves of class (L−E1−…−E8) I think). | |
| May 27, 2015 at 13:32 | comment | added | user5117 | @Mark: I see. I don't know how to reconcile that "expectation" with your observation. | |
| May 27, 2015 at 13:30 | comment | added | user47305 | @Ritwik There are no such curves if the points are general. The only way that class can be effective is if all $8$ points are collinear. | |
| May 27, 2015 at 13:28 | comment | added | user47305 | I'm looking after claim 5.2.3 on page 29 -- "It is expected that all these values are 1" | |
| May 27, 2015 at 13:27 | comment | added | user5117 | To the OP: I skimmed the Kontsevich-Manin notes, but I couldn't find the claim that $N_\beta=1$ for an indecomposable class --- where is it written? As Mark observes, there are 12 nodal cubics through 8 general points, which seems in conflict with this claim. | |
| May 27, 2015 at 13:08 | comment | added | user47305 | @Artie PS: Sorry for deleting my earlier comment. I think you're right that Batyrev-Popov seems to do what the question asks. I'm a little confused about the anticanonical class for $k=8$, which seems to me to be indecomposable with $N_\beta = 12$, although Kontsevich-Manin suggest that $N_\beta = 1$ on a del Pezzo. Hopefully the poster can sort us out. | |
| May 27, 2015 at 12:48 | comment | added | user5117 | Dear Mark, of course, I was forgetting that we are talking about maps from $\mathbf P^1$, not just looking at numerical classes. My mistake! | |
| May 27, 2015 at 12:43 | comment | added | user47305 | @Artie Yes, I mean a cubic through all of them with a node somewhere -- in that case the class should be represented by a map from $\mathbb P^1 \to X_k$ as needed (I think?). It's true that the class $3L - \sum E_i$ is represented by smooth cubics, but $N_\beta$ is only counting rational curves in the class, which should come from the singular cubics in the pencil. Anyway, I suspect I am misunderstanding something and embarrassing myself. | |
| May 27, 2015 at 12:36 | comment | added | user5117 | Dear Mark, I also know nothing about Gromov-Witten theory, so I won't say anything about that part. I am a little confused about your earlier mention of nodal cubics, though; given 8 general points there won't be a cubic through all of them and with a node at one. Did you mean just that the cubic has a node somewhere? Then I agree such a curve exists, but it has the same class as a smooth cubic through the 8 points. On the other hand, in your last comment you just wrote the class of a smooth cubic, so maybe I am misunderstanding you. In any case, Batyrev--Popov seems to answer OP's question. | |
| May 27, 2015 at 5:12 | comment | added | user5117 | @Mark: see arxiv.org/abs/math/0309111, especially Corollary 3.3. | |
| May 27, 2015 at 2:17 | comment | added | user47305 | I guess when $k = 8$ you have the class of a nodal cubic through the points. So there are more than just $(-1)$-curves. (I think for that class $N_\beta = 12$, though I could be wrong.) | |
| May 27, 2015 at 2:11 | comment | added | user47305 | If $k \geq 2$ the extremal rays on the cone of curves are all the classes of $(-1)$-curves (which are in particular rational), and I guess these classes should be indecomposable. Any other effective class can be written as a positive $\mathbb Q$-linear combination of these, but maybe not as an integer combination, I suppose. Do you know any examples of indecomposable classes that aren't just $(-1)$-curves? | |
| May 26, 2015 at 17:30 | history | asked | Ritwik | CC BY-SA 3.0 |