TITLE: Quantum cohomology rings as invariants
QUESTION [7 upvotes]: Let $X$ and $Y$ manifolds. What kind of relations between them (like homeomorphism, diffeomorphism, homotopy equivalence) gives an isomorphic quantum cohomology rings?

REPLY [2 votes]: 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.