2
$\begingroup$

I'm working on a project and I've used the Picard-Fuchs equation at a maximally unipotent monodromy point for a certain 1-dimensional family of Calabi-Yau 3-folds to calculate the A-model Yukawa coupling, and thus the generating function of the genus 0 GW invariants on the suspected mirror. I'm wondering if it's possible to have negative GW invariants even in genus 0. Also is it possible to have $n_d=0$ for all odd $d$, where the $n_d$ are defined to be the unique rational numbers fitting into the formula $N_l=\sum_{d|l} n_{l/d}d^{-3}$, where $N_d=I_{0,0,d}$ is the $d$-th genus 0, unpointed, GW invariant in degree d on a Calabi-Yau 3-fold with Picard rank 1. These numbers are also known as the instanton numbers.

To be more precise about my exact question: Suppose you have a CY 3-fold with Picard rank 1, so you can write any cohomology class in codim 2 as $dH$. Then we can write the GW invariants $$N_d=I_{0,0,dH}=\sum_{k|d} n_{d/k} k^{-3}.$$ Is it possible for these $n_{d/k}$'s to be negative for a few low degree terms, and zero for all odd degrees.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

It is certainly possible to have $n_d$ be negative or zero. For example if $X=K_{\mathbb{P}^2}$, the total space of the canonical bundle over $\mathbb{P}^2$, then $n_2=-6$. Your $n_d$ are genus 0 Gopakumar-Vafa invariants (or BPS numbers) and they have a description in terms of sheaves on $X$ which is (conjecturally) equivalent to your formula in terms of GW invariants. $n_d$ is given as the topological Euler characteristic (weighted by the Behrend function) of the moduli space of semi-stable sheaves $F$ on $X$ whose support is in the curve class dual to "$d$" and such that $\chi(F)=1$. The Behrend function is a integer valued constructible function on any scheme over $\mathbb{C}$ which takes the value $(-1)^n$ at non-singular points of dimension $n$. My point is that if the curves of degree $d$ move in a family, then this weighted Euler characteristic might easily be negative or zero. In the case of $X=K_{\mathbb{P}^2}$, $d=2$, the sheaves are given by the structure sheaves of conics in the plane and so the moduli space is a $\mathbb{P}^5$ (whose Behrend function is the constant -1) and so the weighted Euler characteristic is minus the topological euler characteristic of $\mathbb{P}^5$, so $n_2=-6$.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ the reference for this description of genus zero BPS invariants is the paper by Sheldon Katz: arxiv.org/pdf/math/0601193.pdf $\endgroup$
    – Jim Bryan
    Commented Jan 30, 2013 at 0:27
  • $\begingroup$ Two questions: 1) is your $K_{\mathbb P^2}$ Calabi-Yau? If you mean by the total space of the canonical bundle the scheme $Spec_{\mathbb P^2} Sym(\omgea_{\mathbb P^2})$ then I believe this has canonical bundle $\pi^*\mathcal O_{\mathbb P^2}(-6)$. So did you mean the total space of the line bundle which has sheaf of sections $\omega_{\mathbb P^2}$, which I agree would be Calabi-Yau. 2)Also, aren't the GV invariants conjectured to be nonnegative (for example see p. 19 of arxiv.org/abs/hep-th/9111025), so is that simply not believed anymore, or are they talking about different things? $\endgroup$
    – HNuer
    Commented Jan 30, 2013 at 14:27
  • $\begingroup$ The total space of a locally free sheaf $F$ on $X$ is given by $Spec(Sym(F^\vee))$ --- you are missing the dual in your comment above. The invariants $n^0_d$ are only expected to be positive integers when the number of genus 0 curves in the class $d$ is finite. $\endgroup$
    – Jim Bryan
    Commented Jan 30, 2013 at 17:51
  • 1
    $\begingroup$ This example, often called "local $\mathbf{P}^2$, is extensively studied in both the math and the physics literature. Here is one paper which analyses this example from the mirror symmetry / Picard-Fuchs point of view: arxiv.org/pdf/0710.0049v2.pdf Note that even though local $\mathbb{P}^2$ is non-compact, any compact CY3 $X$ which contains a $\mathbb{P}^2$ will have part of its GW theory identical to that of local $\mathbb{P}^2$. $\endgroup$
    – Jim Bryan
    Commented Jan 30, 2013 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .