Timeline for Are genus zero Gromov Witten Invariants on Del-Pezzo surfaces enumerative?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| May 29, 2015 at 14:22 | vote | accept | Ritwik | ||
| May 28, 2015 at 22:33 | answer | added | Jim Bryan | timeline score: 6 | |
| May 28, 2015 at 14:19 | comment | added | Ritwik | @Jason: could you explain one thing about your remark; you said you computed $<[point]>_{\beta}$ for $\beta = L-E_1-0E_2$. How did you compute this? Using the recursion formula given by Kontsevich on page $29$ or using something else? | |
| May 28, 2015 at 14:13 | comment | added | Ritwik | @ulrich: I said that I obtained $N_{L-E_1-0E_2}$ as $-1$. I used what you are saying (i.e. $N_{L-E_1-E_2} =1$) as one of the inputs to the recursion formula (the other two being $N_{E_1} =1$ and $N_{E_2} =1$.) To apply Kontsevich's formula I decomposed the class precisely how Jason said. Let me check my calculation once more, since Jason obtained $+1$. I must have made some mistake. | |
| May 28, 2015 at 12:22 | comment | added | user47305 | It looks like the reason for the guess is this: on page 28 they assert "the cone of effective classes $B$ is generated by [...] all exceptional classes for $r \geq 2$". This is false (it is only generated over $\mathbb Q$), but as checked in the Batyrev--Popov paper Artie mentioned in the other thread, the anticanonical on a degree $1$ del Pezzo is the only counterexample. | |
| May 28, 2015 at 11:42 | comment | added | Jason Starr | @Ritwik: I agree with ulrich. First of all, $L-E_1$ is not indecomposable on the blowing up of $\mathbb{P}^2$ at two points. Indeed, it is $(L-E_1-E_2) + E_2$. Second, by my computation, for $\beta = L-E_1$, the Gromov-Witten invariant $\langle [\text{point}] \rangle_\beta$ equals $+1$, not $-1$. This is one of the enumerative Gromov-Witten invariants that arise in the Koll\'ar - Ruan theorem. | |
| May 28, 2015 at 11:14 | comment | added | naf | $L - E_1 - E_2$ is an indecomposable class, so $N_{L - E_1 -E_2}$ must be an input to the formula. How do you conclude that it is $-1$? | |
| May 28, 2015 at 10:43 | comment | added | Jason Starr | I do not know what led to that expectation. They may have believed that for every del Pezzo surface S and for every indecomposable class $\beta$ that supports genus $0$ curves, there exists a birational, projective morphism $S\to \mathbb{P}^2$ such that $\beta$ is the pullback of the line class. I believe this is true for $\beta$ the class of a minimal free rational curve, except when $S$ is a degree $1$ del Pezzo surface. Testa discusses this in his thesis. | |
| May 28, 2015 at 9:55 | comment | added | Ritwik | @Jason: Since $N_{3L-E_1-\ldots -E_8} =12$, the general expectation that $N_{\beta} =1$ for all indecomposable $\beta$ is not always correct, as the authors expect in their paper (page 29)? Perhaps they tacitly excluded this case in their remark? | |
| May 28, 2015 at 9:44 | comment | added | Ritwik | @Jason: I am sorry, I just realized your remark was for the surface blown up at $8$ points, not at one point (which I verified using Kontsevich's formula). In either case, my question still stands; how does one explain those negative numbers, i.e. for example how does $N_{L-E_1-0E_2} =-1$ make sense? | |
| May 28, 2015 at 9:26 | comment | added | Ritwik | @Jason: $N_{3L-E}$ is indeed $12$; I checked that this is consistent with Kontsevich's formula. However, when you consider the surface blown up at two points, I get for example $N_{L-E1-0E2} =-1$. Does this value make any sense? Assuming I applied his formula correctly, does this mean the numbers $N_{\beta}$ are not necessarily enumerative for surfaces with more than one blow up? | |
| May 28, 2015 at 8:58 | comment | added | Jason Starr | For the del Pezzo surface of degree $1$, $N_{3L - E}$ equals $12$. Each of the $12$ genus $0$ curves is unobstructed. More generally, for every minimal free genus $0$ class $\beta$ on a uniruled variety, there is an associated enumerative Gromov-Witten invariant with one point insertion. This is the basis for the Koll'ar - Ruan theorem on symplectic invariance of uniruledness. Of course for rational connectedness this is much harder, with the best results due to Zhiyu Tian. You might also consult the thesis of Damiano Testa. | |
| May 28, 2015 at 6:16 | history | asked | Ritwik | CC BY-SA 3.0 |