Symplectic manifolds with dense group of periods

Question

Let $ (M, \omega) $ be a symplectic manifold. The de Rham class of $\omega$ induces a homomorphism $[\omega]: H_2(M) \to \mathbb{R}$, whose image $\Gamma_{\omega} \subseteq \mathbb{R}$ is called the group of periods of $\omega$. In the classical theory of prequantization one assumes that $\omega$ is integral, i.e. $\Gamma_{\omega} \subseteq \mathbb{Z}$. More generally, one can construct a prequantization principal $\mathbb{R} / \Gamma_\omega$ bundle if $\Gamma_\omega$ is a discrete subgroup of $\mathbb{R}$. The only other case is that the group of periods is dense.

Question: What are known examples of closed symplectic manifolds with a dense group of periods? Can one characterize such symplectic manifolds in a different way?

Answer 1

The discrete subgroups of $\mathbb{R}$ are the groups $a\mathbb{Z}$ for $a\in\mathbb{R}$; the group of periods of $\omega$ is of this form if and only if some (real) multiple of $\omega $ is integral. Any projective manifold with $h^{1,1}>1$ gives a counter-example. Indeed the Kähler classes form an open convex cone in $H^{1,1}_{\mathbb{R}}$, and a general line $\mathbb{R}\omega $ of this cone does not meet the integral lattice, so the group of periods of $\omega$ is dense.

Answer 2

Let $(M,\omega)$ be any compact symplectic manifold. The space of symplectic forms is open in the space of closed $2$-forms $\Omega(M)$, let $Symp(M)$ be the set of symplectic forms on $M$, the map $Symp(M)\rightarrow H^2(M,R)$ which sends $\omega$ to its $[\omega]$ is open. So in any neighborhood of a non trivial integral form there exists a symplectic form whose group of period is dense.

Answer 3

One can also build many simple examples as follows: Take $(M,\omega)$ be a symplectic manifold whose group of periods is non-trivial, and then $M\times M$ equipped with $\omega\oplus\alpha \omega$ for some irrational number $\alpha$, has a dense group of periods.