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?](https://mathoverflow.net/questions/207662/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? 



Are genus zero Gromov Witten Invariants on Del-Pezzo surfaces enumerative?
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.


