symplectic automorphism groups can’t really be the answer. This clarification may make the whole endeavour seem quixotic. Still, if one looks at it from the point of view of quantum cohomology $\mathit{QH}^*(M)$, the situation is less clear-cut. Passage to quantum cohomology generally reduces the grading to ${\mathbb{Z}}/2$, putting all odd degree cohomology formally on equal footing (but degree one classes retain a more direct connection to geometry, because their quantum product remains equal to the classical cup product). In that vein, it turns out that one can give a partially positive answer to the question above, at least if one is willing to settle for an invariant which is somewhat more obscure, lacking the simplicity and geometric elegance of the flux group.

[**The examples.**]{} As an application, we consider a particular pair of $28$-dimensional simply-connected symplectic manifolds (the following is only an outline of the construction, omitting many details and assumptions). Let $K$ be a $K3$ surface, and $T \subset K$ a symplectically embedded two-torus. Take $K^7$, the product of seven copies of $K$, and blow up the $12$-dimensional submanifold $T^2 \times K^2$. Denote the outcome by $B^{\mathit{triv}}$. This has a more interesting cousin $B$, defined in the same way 