6

1

I wonder if there is any situation where one can talk about Gromov-Witten invariants or quantum multiplication for singular varieties. Ideally, I would like have a situation where for a singular variety $X$ one can define quantum multiplication operators by elements of ORDINARY cohomology of $X$ on the INTERSECTION cohomology of $X$ (I have some examples where I know what I want the answer to be, but I don't know how to ask the question).

In fact, I will be ready to start with the following simple example: assume that $X$ just has quotient singularities, i.e. locally it looks like $Y/G$ where $Y$ is smooth and $G$ is a finite group. In this case the intersection cohomology coincides with the ordinary cohomology, so my question is whether in this case one can define quantum multiplication. One warning: I am talking about quantum cohomology of $X$ itself, not about what is called "orbifold quantum cohomology" (which in many cases coincides with the quantum cohomology of a good resolution of $X$).

flag

1 Answer

1

Since no one else has said anything, let me make two naive comments.

  1. The only case I know of where there is a well developed notion of Gromov-Witten invariants for a singular variety is the very special case when the target admits a gluing $X \cup_D Y$ where X and Y are smooth and projective and D is a smooth divisor.

  2. You write that you are not interested in the orbifold Gromov-Witten invariants, but perhaps it is possible to define something similar to what you want in terms of the orbifold invariants. Let $X = Y/G$ and set $\mathscr X = [Y/G]$. The Gromov-Witten invariants of $\mathscr X$ are given by maps $$ I_{g,n,\beta} : H^\ast(\overline I \mathscr X)^{\otimes n} \to H^\ast(\overline M_{g,n}),$$ where $\overline I \mathscr X$ is the rigidified inertia stack of $\mathscr X$. However there is also a natural map $H^\ast(X) \to H^\ast(\overline I \mathscr X)$ induced by: (i) the rigidification morphism $I \mathscr X \to \overline I \mathscr X$, in particular the isomorphism it induces on cohomology; (ii) the forgetful map $I \mathscr X \to \mathscr X$; and (iii) the coarse moduli space map $\mathscr X \to X$. So by precomposition we get a collection of maps $H^\ast(X)^{\otimes n} \to H^\ast(\overline M_{g,n})$, and even though it seems they will not satisfy the axioms required of GW invariants maybe they are what is needed in your situation.

link|flag
Thanks. In fact I only care about genus 0 case. I have a question: suppose that $X$ has a crepant resolution $\widetilde{X}$. Then $H^*(X)$ maps to $H^*(\widetilde{X})$ in the obvious way. Question: if one assumes the crepant resolution conjecture, will one get the same construction from this embedding as from yours? – Alexander Braverman Apr 6 2011 at 0:13

Your Answer

Get an OpenID
or

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