Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm learning about Gromov-Witten invariants for the first time, and I'm trying to do some recursive computations using Kontsevich's method of counting stable maps. (And I guess this is really a question about counting stable maps.) For $r\le3$ I have no problem, but there's a subtlety for $r>3$ that I'm confused about.

During the computation, I have to figure out the following. (This particular example would come up in the somewhat icky computation of $I_4(h^4,h^4,h^4,h^4,h^4,h^3,h^3,h^2,h^2)$. It's really the general case I'm interested in.)

How many stable maps in $\overline{M}_{0,10}(\mathbb{P}^4,4)$ are of the form: irreducible of degrees (2,2), meeting at a point, with one component containing the marks $(h^4,h^4,h^3,h^3)$ and the other containing the marks $(h^4,h^4,h^4,h^2,h,h)$. The issue here is that neither curve is determined by the marks, but rather they each lie in 1-parameter families, and the 2-dimensional subvarieties of $\mathbb{P}^4$ traced out by these families intersect in finitely many points corresponding to the image of the node.

Thanks in advance, and let me know if I'm being unclear. (This one is kind of messy to explain.)

share|cite|improve this question
You may want to try this question on – Samuel Reid Dec 3 '12 at 18:01
I was going to, but I figured maybe I should try here first in case they said it wasn't of research interest. (As indeed it isn't.) – Rob Silversmith Dec 3 '12 at 20:58

Figured it out - I was maybe being a little silly. If we know the degrees of those 2-dimensional subvarieties, we can figure out the number of intersections. The degree is simply the number of intersection points with a generic plane, i.e. (in the first case) the Gromov-Witten invariant $I_2(h^4,h^3,h^3,h^2)$ and (in the second case) $I_2(h^4,h^4,h^4,h^2,h,h,h^2)$. Thus the number of points in the intersection of the two surfaces is the product of these, and the recursive computation can be continued.

share|cite|improve this answer

Your Answer


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.