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There is a famous recursion formula by Kontsevich to find the number of genus zero degree $d$ curves in $\mathbb{CP}^2$ through $3d-1$ points. My question is the following: Let $S$ be a complex surface and $A$ a fixed homology class in $H^2(S,\mathbb{Z})$. Is there any hope of answering this question:

``How many genus zero curves are there (through the right number of points) that represent the homology class $A$ in $S$?''

Does Kontsevich's argument rely heavily on the fact that $S$ is $\mathbb{P}^2$? Are there some large classes of $S$ for which his argument might go through?

Note that to find ``the right number of points'' you have to calculate the dimension of the moduli space (for which there is a formula).

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The whole Gromov-Witten formalism works in quite great generality. The more delicate issue is whether the Gromov-Witten invariants have any enumerative significance. For $\mathbf P^2$ they certainly count curves, but it's not obvious: one uses that since the target space is convex, there is an honest-to-god fundamental class, and moreover one can use the action of $\mathrm{PGL}_3$ on $\mathbf P^2$ and Kleiman's transversality theorem to guarantee that intersections are sufficiently generic. I don't know off hand under how general conditions it is known that GW invariants actually count curves. – Dan Petersen Mar 27 '12 at 8:57
To answer Dan Petersen's question (not Ritwik's), there has been much work proving genus 0 GW invariants are enumerative, e.g., for a Fano hypersurface or Fano complete intersection which is sufficiently general in moduli, in about 2/3rds of the cases. This is work of Beheshti-Kumar (building on earlier work of Harris-Roth-Starr and Coskun-Starr). – Jason Starr Mar 27 '12 at 12:18
See also Aleksey Zinger's early work for a symplectic approach for the relationship between GW and enumerative invariants. I think Eleny Ionel also has an early paper on this in genus 1 for $CP^2$. – Jonny Evans Mar 27 '12 at 13:05
up vote 11 down vote accepted

The Kontsevich recursion formula is not special to $\mathbf{CP}^2$. It is a particular application of the more general associativity formula for quantum cohomology, which is something true for symplectic manifolds in general. Going from associativity to actual numbers counting curves is quite combinatorially involved, I guess, and gets worse as the classical cohomology ring of your space gets more complicated. A nice worked example is the quadric 3-fold which can be found in Fulton-Pandharipande "Notes on stable maps and quantum cohomology" and more examples are in this paper of Di Francesco-Itzykson.

The key property that $\mathbf{CP}^2$ has which won't necessarily be true more generally is that the recursion formulae only require a finite amount of input to give everything: in the case of $\mathbf{CP}^2$ it suffices to know that there's a unique line through two points to determine all the other numbers. This is because $\mathbf{CP}^2$ is Fano and so as you increase the degree of a curve it moves in a higher and higher dimensional moduli space: the dimension of the moduli space of genus 0 curves in a class $A\in H_2(X;\mathbf{Z})$ with 3 marked points in a $2n$-manifold with first Chern class $c_1$ is $$2n+2c_1(A)$$ and Fano means $c_1(A)$ increases with the symplectic area of $A$. The Gromov-Witten invariants are obtained by pushing forward the fundamental class of this 3-point moduli space along an evaluation map to $X^3$. Eventually (i.e. for curves of large enough area) the 3-point moduli space becomes so high-dimensional that its fundamental class lives in homological degree larger than the fundamental class of the product $X\times X\times X$ (which is the target of the 3-point evaluation map), therefore the 3-point Gromov-Witten invariants vanish for degree reasons if the area is big enough. By Gromov compactness there are only finitely many moduli spaces of curves with bounded area, so in the Fano case you only ever get finitely many quantum contributions for a given product.

By contrast, for Calabi-Yau 3-folds the expected dimension of a holomorphic sphere is zero no matter what the degree, so you end up having to count infinitely many curves to compute a single quantum product (you get around this by weighting them with Novikov coefficients to encode their areas).

Of course, to answer your specific question, for a completely general algebraic surface you often expect to have no genus 0 curves. For example, a generic K3 surface has no genus 0 curves because you can perturb the complex structure to ensure that there are no $(1,1)$-classes in the rational cohomology. Fano is a good condition for ensuring the existence of genus 0 curves because Mori theory tells us they always exist - a smooth Fano variety is rationally connected (indeed the proof, which I'm told proceeds by passing to a field of characteristic $p$ and then using the Frobenius automorphism of the field to increase the degree of the curve, uses morally the same property that I mentioned above: high degree curves move in high-dimensional moduli spaces).

Note: Kontsevich actually has several proofs of this recursion formula. One of the most beautiful is the one in his paper on toric localisation. That method certainly works more generally than $\mathbf{CP}^2$, e.g. for complete intersections in weighted projective spaces, as he explains there. This method doesn't really have anything to do with associativity, as far as I can see.

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