# Negative Gromov-Witten invariants

I understand the heuristic reason why Gromov-Witten invariants can be rational; roughly it's because we're doing curve counts in some stacky sense, so each curve $C$ contributes $1/|\text{Aut}(C)|$ to the count rather than $1$.

However, I don't understand why or how Gromov-Witten invariants can be negative. What is the meaning of a negative GW invariant? What are some simple examples?

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I wanna say that one should not be surprised to get a negative number when counting using intersection theory. What does it mean for a curve on a surface to have self intersection number -1? –  Yang Zhou Apr 10 at 7:04

Gromov--Witten invariants are designed to count the "number" of curves in a space in a deformation invariant way. Since the number of curves can change under deformations, the Gromov--Witten invariants won't have a direct interpretation in terms of actual numbers of curves, even taking automorphisms into account.

Here is an example of how a negative number might come up, though strictly speaking it isn't a Gromov--Witten invariant. Let M be the moduli space of maps from P^1 to a the total space of O(-4) on P^1. Call this space X. Note that I said maps from P^1, not a genus zero curve, so the source curve is rigid. That's why this isn't Gromov--Witten theory. Any such map factors through the zero section (since O(-4) has no nonzero sections), so this space is the same as the space of maps from P^1 to itself. I just want to look at degree one maps, so the moduli space is 3 dimensional.

We could also compute the dimension using deformation theory: the deformations of a map f are classified by $H^0(f^\ast T)$ where T is the tangent bundle of the target. The target in this case is O(-4), not just P^1, and the tangent bundle restricts to O(2) + O(-4) on the zero section. Thus $H^0(f^\ast T)$ is indeed 3-dimensional, as we expected. However, the Euler characteristic of $f^\ast T$ is not 3 but 0, which means that the "expected dimension" is zero.

The meaning of expected dimension is rather vague. Roughly speaking, it is the dimension of the moduli space for a "generic" choice of deformation. The trouble is that such a deformation might not actually exist. Nevertheless, we can still pretend that a generic deformation does exist and, if the expected dimension is zero, compute the number of curves that it "should" have.

What makes this possible is the obstruction bundle E on M. Any deformation of X gives rise to a section of E and the vanishing locus of this section is the collection of curves that can be deformed to first order along with X. Even though a generic deformation might not exist, the obstruction bundle does still exist, and we can make sense of the vanishing locus of a generic section by taking the top Chern class.

In our situation, the (fiber of the) obstruction bundle is $H^1(f^\ast T)$. Since O(2) does not contribute to H^1, the obstruction bundle is $R^1 p_\ast f^\ast O_{P^1}(-4) = R^1 p_\ast O_{P^3 \times P^1}(-4, -4)$ where $p : P^3 \times P^1 \rightarrow P^3$ is the projection. By the projection formula, this is $O(-4)^{\oplus 3}$ and the top Chern class is -64. This is the "Gromov--Witten invariant" of maps from P^1 to $O_{P^1}(-4)$.

Unfortunately, I don't have anything to say about what this -64 means...

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Gromov Witten invariant is supposed to "count" the number of curves but it can happen that the dimension of the space of curves that you want to count is larger than you expect. For example on a 3-dimensional Calabi-Yau variety the expected dimension of any curve is 0, so in an ideal situation curves should be isolated and you should just count the number of curves of a given degree and genus. In practice this does not happen all the time and often you can deform curves (though Clemence conjectured that RATIONAL curves on generic CY 3-fold are rigid) For example if you take a quintic in $CP^4$, you can intersect it with a 2-plane and you obtain a curve that moves in the quintic. While you excpect it to be rigid. What you need to do in this case to count the GW invariant is to conisder on the moduli space of curves the obstruction bundle, which will be of the same dimesnion as the moduli space and then you need to calculate the Euler class of the bundle (this is in the best case, when the moduli space is smooth).

Now, the Euler class that you calculate has no particular reason to be positive. Of course if the curve were isolated it would count with the positive weight, but it is not. I can not provide an explicite example where you get a negative answer, but there are a lot of examples when you get zero, while there are curves. This happens on algebraic surfaces. You can take a minimal surface of general type and consider a curve C of zero expected dimention, i.e. $C^2-KC=0$, where K is the canonical divisor. Then the GW invariant of this curve will be non zero only if C is a canonical divisor. This is a corrolary of Seiberg Witten theory. But there are examples when you do have curves C that satisfy $C^2-CK-0$ and $C$ is different from $K$ these curves contribute 0 to GW invariant.

So, what is the meaning of negative GW invariant? If you are doing symplectic geometry, everything can be perturbed so that the final number of curves that you get is finite. It is exactly the same as to say that if you have a map of two manifolds $M^n\to N^n$, for a generic point the number of perimages is finite. Some of them count with + and some with -, add together and you get degree. Negative GW= negative degree.

But if you are doing Algebraic geometry negative GW invariant has an additional consequence -- namely that the moduli space of curves (this time algebraic and not almost complex) that you consider MUST has exessive dimention. Just as in the example with CY quintic that is desribed above. Indeed, if all curves were isolated they would contribute positively and GW would be positive.

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Be specific, this happens on the Fermat quintic, as I recall. However, for a generic one, the curves should have no deformations. –  Charles Siegel Dec 5 '09 at 1:53
No Charles, it is clear that the example that I decirbed works for every quntic. You can always move a plane in CP^4 and its intersection with the quintic will move. I guess, you were thinking about RATIONAL curves, they indeed should be rigid on a generic quintic according to Clemmence conjecture. But the example that I give is not genus zero. This is a generic phenomena for ALL Calabi-Yaus. Take two divisors of high degree that move on CY and intersect them –  Dmitri Dec 5 '09 at 2:00
Ahh, ok. Yeah, I only think about GW invariants in genus zero here, where they're generically rigid. (and it's Clemens, just for anyone who might google names, and the paper arxiv.org/abs/alg-geom/9510015 has some good info on it, if I remember right) –  Charles Siegel Dec 5 '09 at 2:44
While what you say makes sense, Dmitri, it doesn't seem to answer the question as to the meaning of negative invariants. –  Simon Rose Dec 5 '09 at 4:36
Simon, I added two pragraphs on the meaning of the negative tsign - as much as I can say :) Charles, really sorry for misspelling the name of Clemence. –  Dmitri Dec 5 '09 at 8:57

Let me use an example. If I recall correctly (I am too lazy to look), the GW invariant for "local P^2" (i.e. the canonical bundle O(-3) over P^2) in genus zero and degree 1 is 3, and in degree 2 is -45/8. Now -45/8 breaks down, after trying to account for multiple covers, to 3/2^3 - 6, giving an effective number of degree 2 curves as -6. What up? This number is how many degree 2 curves the P^2 should "account for," when it pops into existence within a family of Calabi-Yau's. (Here of course we have a family of maps into the P^2.) Then you should ask, "Does this mean that for a compact CY to have just rigid genus zero curves and an embedded P^2 (and no other surface) and lie in a family that includes a CY with only rigid genus zero curves and no other surface, that it must have at least 6 degree 2 rational curves somewhere else?" Presumably.

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The simplest example that I could find is: blow-up $P^3$ at one point and consider the following correlator $\langle pt, E^2, E^2\rangle_l=-1$ , where $l$ is the pull-back of a line from $P^3$.

Proof: use the splitting axiom for the following 4 classes $(pt, E^2|E,E)_l$. To unravel this notation take a look at page 5 in http://www.mathematik.uni-kl.de/~gathmann/pub/blowup-9804043.pdf

The splitting axiom gives:

$\langle pt, E^2, E^2\rangle_l=\langle pt, E, E\rangle_{l-E'}\cdot\langle E^2, E^2, E\rangle_{E'}=1\cdot(-1)=-1$

I hope this is corect.

Does anybody know an explicit example in dimension 2?

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