Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $N_d$ denote the number of rational curves in $\mathbf P^2$ passing through $3d-1$ points in general position. Maxim Kontsevich discovered a famous recursion for these numbers: $$ N_d = \sum_{k+l = d} N_k N_l k^2 l \left( l \binom{3d-4}{3k-2} - k \binom{3d-4}{3k-1}\right).$$ The proof of this recursion goes by interpreting $N_d$ as a Gromov-Witten invariant: one looks at the moduli space $\overline{M}_{0,3d-1}(\mathbf P^2, d)$ and pulls back the class of a point along each evaluation map. Taking the product of all these classes in the Chow ring produces a number. Using that $\mathbf P^2$ is homogeneous one can show that this number actually counts the number of stable maps where the markings are sent to the given points, and it is not hard to see that counting stable maps is the same thing as counting rational curves. Finally, the associativity of the quantum product can be translated to the WDVV differential equations for the Gromov-Witten potential, i.e. the generating function of all Gromov-Witten invariants. These differential equations translate into the above recursion.

At least, this is how the proof is stated in Kontsevich's "Enumeration of rational curves via torus actions" and Konstevich-Manin "Gromov-Witten classes, quantum cohomology and enumerative geometry", which seem to be the earliest published sources.

However, there is also a beautiful streamlined proof which avoids the use of the quantum product. Here one instead works with $\overline M_{0,3d}(\mathbf{P}^2,d)$ (one more marking) and takes the pullback of the classes of two lines in $\mathbf P^2$ along the first two markings and $3d-2$ classes of points for the remaining. The intersection of these are a curve in the moduli space and one then computes the intersection of this curve with two different linearly equivalent boundary divisors. These two intersection numbers can easily be computed by hand, producing the recursion. The proof that these two boundary divisors are linearly equivalent uses the forgetful map to $\overline{M}_{0,4}$ which is also used in the proof of the WDVV equations so in some sense it seems like this proof inlines the particular case of WDVV that is needed in a very clever way.

This latter version of the proof appears for instance in the book of Kock and Vainsencher, in Abramovich's "Lectures on Gromov-Witten invariants on orbifolds", and in the lecture notes I found here. But where is it from originally? All these sources just refer to it as "Kontsevich's proof" without attribution. Did Kontsevich also come up with this streamlined version but did not see it as worth publishing?

share|improve this question
    
the link is broken... –  Dmitri Mar 8 '11 at 16:29
    
Should $N_d$ appear on both sides of the given recursion? –  Daniel Litt Mar 8 '11 at 17:18
    
That's harmless, since all terms where $k$ or $l$ are equal to $d$ will vanish anyway. Perhaps it would have been clearer to write the sum over all $k+l = d$, $k \geq 1$, $l \geq 1$. –  Dan Petersen Mar 8 '11 at 17:33
    
@Dan: I think Daniel's referring to the typo $N_d$ on the RHS. –  ndkrempel Mar 8 '11 at 19:16
    
Oh. Thanks. Yes, that was a typo. –  Dan Petersen Mar 8 '11 at 19:49

1 Answer 1

up vote 8 down vote accepted

In Manin's book "Frobenius manifolds, quantum cohomology, and moduli spaces", section 0.6.2, the argument ("an old trick of enumerative geometry") is attributed to Kontsevich. I would guess that Kontsevich probably came up with the "streamlined proof" but preferred to talk about it within the framework of WDVV because it more readily generalizes this way, and because it leads to nice-sounding statements like "the enumerative formula is equivalent to associativity of quantum multiplication". That's what I imagine at least. Also, I would be inclined to trust Manin's attribution, as he collaborated with Kontsevich on these early works.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.