TITLE: $(M,\omega)$ not symplectomorphic to $(M,-\omega)$
QUESTION [22 upvotes]: Looking for an example of a symplectic manifold $(M,\omega)$ that is not symplectomorphic to $(M,-\omega)$.
In particular this means that $M$ must be chiral (i.e. doesn't admit an orientation-reversing diffeomorphism).
For a topological obstruction, I think it would be enough to find $(M,\omega)$ such that $\mathrm{Diff}(M)$ acts trivially on $H^2(M)\neq0$.
A complex projective variety defined by real equations won't work, because the complex conjugation map is antisymplectic.

REPLY [5 votes]: The following answer was a suggestion of Ivan Smith's. It seems like a very nice argument, although the proof is quite high-tech.
Suppose $X$ is a compact symplectic manifold.  By adding a small generic 2-form, we can ensure that the coefficients of $\omega$ with respect to a basis of $H^2(X;\mathbb{Z})$ are linearly independent over $\mathbb{Q}$.  Any diffeomorphism that reverses $\omega$ then has to act as $-1$ on $H^2$.
The question now is to find an $X$ such that no element of $\mathrm{Diff}$ acts as $-1$ on $H^2$, and Ivan said that a K3 surface would work.  $H^2$ has rank 22, and the signature of the intersection form is $(3, 19)$.  Apparently Donaldson theory means that $\mathrm{Diff}$ preserves the orientation of a positive-definite 3-dimensional subspace of $H^2$, so in particular cannot act as $-1$.  This is explained in Donaldson-Kronheimer (Corollary 9.1.4), but the basic idea is that a choice of such an orientation allows one to orient some gauge-theoretic moduli space.  Reversing the orientation reverses the sign of a corresponding Donaldson invariant, but this invariant is non-zero so it's not equal to itself with sign reversed.