TITLE: Symplectic Forms
QUESTION [5 upvotes]: Let $(M, \omega)$ be a symplectic manifold, so that $\omega$ is a non-degenerate 2-form. If $\dim M = 2n$ why does $\omega$ being non-degenerate imply that $\underbrace{\omega \wedge \ldots \wedge \omega}_{n \text{ times}}$ is non-vanishing?

REPLY [5 votes]: Let $x$ be a point in $\rm M$. Then because $\omega$ is non degenerate at $x$, the antisymmetric matrix $\omega_x$ has full rank on the tangent space at $x$. Now $\omega \, \wedge\,  ... \wedge \, \omega$ at $x$ is just $\det \omega_x$ which is not zero because $\omega_x$ has full rank.