## Quantum cohomology rings as invariants

Let $X$ and $Y$ manifolds. What kind of relations between them (like homeomorphism, diffeomorphism, homotopy equivalence) gives an isomorphic quantum cohomology rings?

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 You need a symplectic structure to define quantum cohomology. – Bruno Martelli Nov 25 2010 at 14:44

I am not an expert in the field, but I guess that there are examples of diffeomorphic symplectic manifolds with non-isomorphic quantum cohomologies (hence homotopy equivalence, homeomorphism, or even diffeomorphism alone is not enough to give an isomorphism on quantum cohomology).

Two simply connected smooth 4-manifolds having the same intersection form are homotopy equivalent by a theorem of Milnor. More than that, they are homeomorphic by a famous theorem of Freedman. However, very often these manifolds are not diffeomorphic. An example of homeomorphic but non-diffeomorphic pair is the Barlow surface and the blow-up of the complex plane in 8 points. Note that both manifolds admit a Kahler structure.

In the book of McDuff - Salamon J-holomorphic curves and quantum cohomology, they describe in Example 7.3.6 a construction due to Ruan of two diffeomorphic non-deformation equivalent 6-manifolds. The manifolds are the two 4-manifolds just described, both multiplied by $\mathbb{CP}^1$ (by a result of Wall, these two homeomorphic 4-manifolds become diffeomorphic after such a stabilization).

Ruan uses quantum cohmology to prove that these two diffeomorphic symplectic 6-manifolds are not deformation-equivalent, so my guess is that he uses that their quantum cohomologies are not isomorphic as a tool.

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If $X$ and $Y$ are Calabi-Yau threefolds which are obtained from each other from a simple flop, then their quantum cohomologies are isomorphic (Li and Ruan proved this). Lee-Lin-Wang prove the analog of this in higher dimensions. One of the things that makes this interesting is that the classical cohomology rings are not (in general) isomorphic, but the quantum cohomology rings are.

More generally, if $X$ and $Y$ are birational and $K$-equivalent (i.e. there is a resolution of the birational map $X\leftarrow W \to Y$ such that $K_X$ and $K_Y$ are isomorphic when pulled back to $W$), one expects their Gromov-Witten theories to be "equivalent". In some situations, "equivalent" will imply isomorphic quantum cohomologies, but not in general. To formulate the general equivalence properly requires Coates and Givental's Lagrangian cone formalism. Crepant resolutions are a special case of this and so the "crepant resolution conjecture" is part of this set of equivalences. Here are some papers on the subject (please forgive my laziness at only putting arXiv numbers and not journal references):

Li-Ruan arXiv:math/9803036

Bryan-Graber arXiv:math/0610129

Iritani arXiv:0809.2749

Lee-Lin-Wang math.AG/0608370

Coates-Ruan arXiv:0710.5901

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Quantum cohomology, and more generally Gromov Witten invariants, are invariants under deformation. Given a family $X$ of compact symplectic manifolds over a base $B$, to a path in $B$ one can associate an isomorphism between the quantum cohomology of the fiber $X_0$ over the starting pooint of the path and the qc of the fiber $X_1$ over the end point of the path. Homotopic paths (homotopic wrt end points) induce the same isomorphism. In particular $\pi_1(B,0)$ acts on the quantum cohomology of $X_0$.

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 Presumably GW invariants of smooth projective varieties are invariant under deformation in the sense of algebraic geometry. But GW invariants of compact symplectic manifolds are invariant under deformation in the sense of ... ? – Kevin Lin Nov 25 2010 at 20:06 Also, I don't see how you're getting your action of $\pi_1$. Are you thinking of the Gauss-Manin connection? – Kevin Lin Nov 25 2010 at 21:13 I once wrote something on MO similar to what you are saying... mathoverflow.net/questions/2269/… – Kevin Lin Nov 26 2010 at 7:25 I tried to explain the $\pi_1$-action in my reply to the same question that you are referring to: mathoverflow.net/questions/2269/… It indeed comes from the Gauss-Manin connection (i.e. the Gauss-Manin connection gives an action on the cohomology, and GW-invariants are invariant under this action). – ABayer Nov 26 2010 at 17:00 Thanks, Arend. But I still wonder what notion of "deformation" or "family" for compact symplectic manifolds Barbara has in mind here... – Kevin Lin Nov 30 2010 at 6:03
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