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Let $X_k$ be $\mathbb{P}^2$ blown up at $k$ points (where $k$ is $0$ to $8$). Let $\beta \in H_2(X_k, \mathbb{Z})$ be the homology class given by $$ \beta := n L + m_1 E_1 + \ldots + m_k E_k $$ where $L$ is the homology class of a line and $E_i$ are the exceptional divisors. Also, define $$ \delta_{\beta} := < c_1(TX_k), ~\beta>-1= 3n + m_1 + \ldots m_k-1. $$

Let $N_{\beta}$ be the number of genus zero curves in the class $\beta$ passing through $\delta_{\beta}$ generic points.

$\textbf{Questions:} $ I have two questions. In their paper

http://www.ihes.fr/~maxim/TEXTS/WithManinCohFT.pdf

Kontsevich and Mannin give a recursive formula to compute $N_{\beta}$ (page $29$).

1) Is it known that the numbers $N_{\beta}$ that one gets from their formula are actually the enumerative numbers (i.e. they are actually the honest count of curves through the right number of generic points)? A priori, Gromov Witten Invariants need not be enumerative and I suspect the formula given by Kontsevic and Mannin are for the genus zero GW invariants. In particular, on page $26$ of their paper (second last paragraph), they make the remark

"We expect that $N_{\beta}$ counts the number of rational curves in the homology class $\beta$ passing through $\delta_{\beta}$ points, at least in unobstructed problems. "

This remark seems to suggest that at the time of writing the paper they did not know if the numbers are actually enumerative. Is this presently known (i.e are genus zero GW Invariants on Del-Pezzo surfaces enumerative)? The answer is yes for $\mathbb{P}^2$.

2) My second question is how does one actually compute $N_{\beta}$ using their recursive formula? One needs enough initial conditions for the recursion. On page $29$ (just after they state the formula) they say that $N_{\beta}$ is ``expected'' to be one for all indecomposable $\beta$. This seems to imply that $$N_{3L-E_1-E_2-\ldots- E_8} = 1.$$

But as observed by Mark in this post

What are the indecomposable classes on a del-Pezzo surface?

it seems that this number ought to be the same as the number of rational planar cubics through $8$ generic points, i.e. $12$. So what have I misunderstood here?

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    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    May 28, 2015 at 8:58
  • $\begingroup$ @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? $\endgroup$
    – Ritwik
    May 28, 2015 at 9:26
  • $\begingroup$ @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? $\endgroup$
    – Ritwik
    May 28, 2015 at 9:44
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    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    May 28, 2015 at 10:43
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    $\begingroup$ 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. $\endgroup$
    – user47305
    May 28, 2015 at 12:22

1 Answer 1

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I believe that the paper http://arxiv.org/abs/alg-geom/9611012 by Pandharipande and Gottsche addresses exactly this question. The short answer is yes, the genus 0 invariants are enumerative up to k=8.

For k>8, one can still conclude that the genus 0 GW invariants are weakly enumerative, meaning that for $\delta_\beta$ generic points, there are a finite number of curves that the genus 0 GW invariants count, but possibly with positive integer multiplicities. See section 4.4 of http://www.math.ubc.ca/~jbryan/papers/survey.pdf

You might look at Gottsche and Pandharipande's algorithm to figure how to get the 12. They have an explicit algorithm and the 12 appears several times on their table of numbers. Conjecturally, all occurrences of the number 12 in that table are equivalent to N_3 = the number of rational cubics passing through 8 points.

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    $\begingroup$ Thank you for pointing out this paper; indeed this paper answers precisely the question I asked. The algorithm given in theorem 3.6 also produces all the numbers. $\endgroup$
    – Ritwik
    May 29, 2015 at 14:22

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