What does an exponent of a differential form mean?

Question

I know that the wedge product is alternative, $\omega \wedge \omega =0$, but I still see people talking about $\omega^{\wedge n}$ like when the power of a symplectic form is considered as the volume form. I assume that this volume form is meant to be non zero for higher dimensions but I can’t see how. Is this just some weird short hand for something else?

Answer 1

If $\omega$ is a $p$-form and $\eta$ is a $q$-form, then $\omega\wedge\eta = (-1)^{pq}\eta\wedge\omega$. In particular, if $p$ is odd, then $\omega\wedge\omega = -\omega\wedge\omega$ and hence $\omega\wedge\omega = 0$. However, if $p$ is even, then $\omega\wedge\omega$ need not be zero. For example, if $\omega$ is a symplectic form on a $2n$-dimensional manifold with $n > 1$, then $\omega\wedge\omega \neq 0$; in fact, $\omega^n = \underbrace{\omega\wedge\dots\wedge\omega}_{n\ \text{times}}$ is non-zero.

For an explicit example, consider $\omega = dx^1\wedge dx^2 + dx^3\wedge dx^4$ on $\mathbb{R}^4$. Note that

$$\omega\wedge\omega = 2dx^1\wedge dx^2\wedge dx^3\wedge dx^4 \neq 0.$$