# Differential forms in Physics

In the context of physics, I just read about the the symplectic 2-form $\omega$ on a symplectic manifold $M$ of dimension $2n$. Unfortunately, I could not follow a few arguments. I.e. it was said that since $\omega$ is non-degenerate, the $n$-fold wedge-sum $\omega \wedge \cdots \wedge \omega$ is a volume form and if $M$ would be a closed manifold $M$, then $\int_M \omega \wedge \cdots \wedge \omega \neq 0.$

Could anybody explain to me where these two conclusion actually come from, i.e. I see that the wedge sum is a top-degree volume form, but I don't see why this means that it is nowhere vanishing and especially why the integral is non-zero.

Nondegeneracy means precisely that $\underbrace{\omega\wedge\dots\wedge\omega}_{n\text{ times}}$ is nowhere $0$, so this is, by definition, a volume form. It follows in general that the integral is nonzero: The manifold must be orientable (use the nowhere-vanishing $2n$-form to decide whether a basis is "positive" or "negative") and then, assuming $M$ is connected, you can deduce from the intermediate value theorem that the integral cannot be zero.
• so nowhere 0 means that if I plug in for all $p \in M$ the basis of the respective tangent $\omega_p \wedge \cdot \wedge \omega_p( \partial_1|_p,...,\partial_{2n}|_p) \neq 0$ or how is this defined? – RealAnalysis May 11 '15 at 19:28
• Yes, for any basis of $T_pM$ we get a nonzero answer. – Ted Shifrin May 11 '15 at 19:31
If $\omega$ is nowhere degenerate then at every point $p$ in $M$ the volume form must be non-zero; otherwise $\omega$ is degenerate at $p$.
The second statement (the integral) relies on two unstated assumptions, namely continuity and the statement that $M$ is not only closed but also connected. With that, you show that for any two points $p_1$ and $p_2$ the vloume element has the same sign (by the intermediate value theorem), thus the integral cannot be zero.