Find a closed simply-connected $4$-manifold with closed non-exact $2$-form

Question

Let $M$ be a closed simply-connected smooth $4$-manifold with a closed non-exact 2-form on it. $H^2_{dR}(M) \ne 0$.

By de Rham theorem, $H_{dR}^2(M) \cong H^2(M;\mathbb R) \cong \text{Hom}(H_2(M;\Bbb Z),\Bbb R)$.

Let $M = S^2 \times S^2$ with coordinates $((\theta_1,\varphi_1), (\theta_2,\varphi_2))$, how can we find such closed but non-exact form on it?

Edit:

How can we find such $2$-form explicity in coordinates $((\theta_1,\varphi_1), (\theta_2,\varphi_2))$?

Answer 1

Let $\omega$ be a closed non exact 2-form on $S^2$. Let $\pi:S^2\times S^2\to S^2$ be the usual projection on the first factor. Define $\hat \omega$ as the pullback of $\omega$ through $\pi$. $\hat\omega$ is closed, since $d(\pi^*\omega)=\pi^*(d\omega)=0$. It is not exact, since its restriction to the embedded submanifold $S^2\times \{(0,0,1)\}\simeq S^2$ is not exact.

To find an explicit example, take $\omega=\iota^*(xdy\wedge dz-ydx\wedge dz+z dx\wedge dy)$, where $\iota$ is the inclusion of $S^2$ in $\mathbb{R}^3$