$\omega$ a closed 2-form and $\bigwedge_{i=1}^n \omega \ne 0$ on a compact orientable smooth $2n$-manifold w/o boundary, $M$, then $H^2(M) \ne 0$.

Question

Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $\omega$ be a closed $2$-form such that $\bigwedge_{i=1}^n \omega_p \ne 0$ at every point $p$. Show that $H^2_{dR}(M) \ne 0$.

That is, $(M, \omega)$ is a symplectic manifold.

I'm not sure what to try with this. Any ideas on how to show this?

Answer 1

If the second de Rham cohomology were trivial, $\omega$ would be $d\alpha$ for some 1-form $\alpha$.

Then use the non-boundary condition and Stokes via

$$\int_M\omega^n=\int_Md(\alpha\wedge\omega^{n-1})=\int_{\partial M}\cdots=\cdots$$

to derive a contradiction with $\omega^n\neq0$.

Answer 2

This is a long comment. To emphasize that you must interpret the hypothesis as being that $\omega^n (p) \ne 0$ for all $p\in M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4\times S^4$, with obvious projection maps $\pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $\eta$ on $S^4$ with the property that $\eta^2 = \eta\wedge\eta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $\Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $\rho_1$ to be a smooth function that is $1$ on the unit ball in $\Bbb R^3$ and $0$ outside the ball of radius $2$, and let $\rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $\eta = d(\rho_1\,dx-\rho_2\,dy)$. Note that $\eta^2 = d\rho_1\wedge d\rho_2\wedge dx\wedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $\omega = \pi_1^*\eta + \pi_2^*\eta$. Then $\omega^4 = 6\pi_1^*(\eta^2)\wedge\pi_2^*(\eta^2)$ will be not identically zero on $S^4\times S^4$, and yet $H^2(S^4\times S^4) = 0$.