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.



$(M,\omega)$ not symplectomorphic to $(M,-\omega)$
Let $n \geq 2$ be a natural number and $M$ a torus of dimension $2n$. Then a generic element of $H^2(M, \mathbb R)$ comes from a symplectic form, because we can take a $2$-form invariant under the torus action representing it, and it is a symplectic, and a generic such form is nondegenerate.


Thus it is sufficient to show that for a generic $\omega \in H^2(M,\mathbb R)$, the diffeomorphism group does not send $\omega$ to $-\omega$.


The action of the diffeomorphism group factors through the representation $\wedge^2$ of $GL\_{2n}(\mathbb Z)$. Because $2n>2$, no element acts as $-1$ on this representation: It would have to have the product of any two different eigenvalues be $-1$, but there cannot be $3$ or more numbers with this property.


For each $\sigma \in GL\_{2n}(\mathbb Z)$, the $\omega$ that are sent to $-\omega$ by it form a proper subspace in $H^2(M, \mathbb R)$, and there are countably many such. Choosing $\omega$ outside all these subspaces gets you an example. For instance, a random $\omega$ succeeds with probability $1$.


