Do the GromovWitten invariants count the morphisms from a curve to a variety over $\mathbb{C}$?

4$\begingroup$ fpqc, don't be a jerk. I'm sure you're aware that many mathematicians and grad students aren't native English speakers, cut them some slack. $\endgroup$ – Charles Siegel Dec 8 '09 at 5:40

3$\begingroup$ @fpqc: there are many nonnative speakers of English in the site, myself included. Instead of making fun of people, you could suggest a way of rewriting it in a manner that is more correct. $\endgroup$ – Alberto GarcíaRaboso Dec 8 '09 at 5:41

$\begingroup$ You're right. I was trying to point out that the way the question was worded didn't make any sense while trying to also be funny. I'm sorry, HYYY and Alberto. I didn't mean anything by it. $\endgroup$ – Harry Gindi Dec 8 '09 at 6:15
To further qualify Charles's yes: that these moduli spaces are orbifolds instead of manifolds does result in rational numbers, but this is quite natural and not much of a problem. The orbifolds here are just resulting because we're counting things that have automorphisms, here, for instance, the map from P^1 to P^1 given by the polynomial z^d has Z_d as its automorphisms: we can multiply a point in P^1 by a dth root of unity and not change where it maps to). Whenever you count things with automorphisms it's quite natural to count each thing weighted by 1/(the size of its automorphism groups), or to rigidify the things we're counting by adding some kind of extra structure so they no longer have automorphisms.
As an example: Cayley's formula that there are n^(n2) trees on n labeled vertices  the labeling of the vertices guarantees that the objects we're counting do not have automorphisms, and we get an integer  we've rigidified the problem. If we wanted to count the number of trees on n unlabeled vertices, the problem is much more difficult. However, if we weight each such tree by the inverse of its automorphism group, then the problem has a nice answer again: it's simply n^(n2)/n!. My point is: the rationality is not the ugly part of what's going on.
The ugly part is that these moduli spaces of maps are not even orbifolds: they have much worse singularities, and can have different components of different dimension. From deformation theory, we expect these moduli spaces to have a certain dimension. To get a finite number, we put conditions on the map that cut this dimension down until its zero. Geometrically, you should think of each of these conditions as a cycle on the moduli space, and we want to intersect them. Doing this intersection naively doesn't work when the space is singular, and furthermore the moduli space might be smooth but have a dimension different than what we were expecting. But a lot of hard work shows that these spaces have a "virtual fundamental class" of the dimension that we expect, and using this we can proceed as above to get a number. But in doing this, we've lost the sense in that we're counting something.
But it strikes me that perhaps that's not necessarily what the questioner was after; most typically this is done for smooth, projective varieties of C, but somehow the part that really matters is the symplectic structure: GromovWitten invariants can be defined for any symplectic manifold  they will all have almost complex structures J that "play nicely" with the symplectic form omega, and we're "counting" these maps. Or: all this works for orbifolds (which are really smooth objects), but not singular spaces.
The over $\mathbb{C}$ bit is pretty necessary, I think  people have looked a little at doing in positive characteristic, but one big problem is that the orbifold stuff, which I was just telling you isn't really a problem, can be a big problem in positive characteristic if the order of your automorphisms aren't coprime with the characteristic.

$\begingroup$ Paul, my understanding is that it's still conjectural that algebraic GW and symplectic GW give equivalent info, along with pretty much every other curvecounting scheme. If they've been shown to be the same (even, say, for genus 0, where things don't get SO bad) could you provide a reference? $\endgroup$ – Charles Siegel Dec 8 '09 at 12:50

$\begingroup$ Charles: See this paper by Siebert arxiv.org/abs/math/9804108 and this paper by LiTian arxiv.org/abs/alggeom/9712035 $\endgroup$ – Kevin H. Lin Dec 8 '09 at 12:55

$\begingroup$ Charles: I think there has been a lot of recent progress in showing equivalence of GromovWitten, DonaldsonThomas, and PandharipandeThomas theories, but I don't know much about this stuff and I don't know any references off the top of my head. $\endgroup$ – Kevin H. Lin Dec 8 '09 at 13:00

$\begingroup$ The equivariant GW/DT correspondence has been proven for toric 3folds in MOOP: front.math.ucdavis.edu/0809.3976. There's been a huge explosion in the study of the DTtype sheaf theories recently  the names here are Joyce and KontsevichSoibelman  but I know less about this than I should, and any link I'd give you would be almost random. $\endgroup$ – Paul Johnson Dec 8 '09 at 13:19
HYYY, the answer is a qualified "yes." I'm not an expert (read a few papers at the beginning of the year before deciding that enumerative geometry wasn't going to be my area) and I know that the answer is a definitive yes for rational curves in homogeneous spaces. However, more generally than that, the Kontsevich moduli space of stable maps isn't a manifold, but merely an orbifold, and so has rational cohomology but not integral. So you get rational numbers. Worse, if your curves aren't rigid, they might count negatively, so there will be rational and negative GromovWitten invariants, in general.
This paper is a great introduction, and works out the first big theorem that GW invariants proved, in a case where they do, in fact, count curves.

$\begingroup$ For more on negative GW invariants, see this question mathoverflow.net/questions/7823/… $\endgroup$ – Kevin H. Lin Dec 8 '09 at 12:57