# Symplectic Forms

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?

## 1 Answer

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.

• Sorry I'm a bit confused. We have that $\omega \wedge \ldots \wedge \omega$ at $x$ is an alternating multilinear map $T_x M \times \ldots T_x M \longrightarrow \mathbb R$. In what sense is this map just $\operatorname{det}\omega_x$ and how do I see it? – Paul Slevin Apr 11 '13 at 22:52
• I can't make sense of this answer. What is the determinant $\det \omega_x$ of the $2$-form $\omega_x$? There is no way to define the determinant of a multilinear form on a vector space . – Georges Elencwajg Aug 18 '18 at 16:56