We start out with $(\mathbb{R}^{2n},\omega_0)$, the "standard" symplectic manifold. We define $\Omega=\omega_0\wedge\dotso\wedge\omega_0$, i.e. $\Omega$ is the product of $\omega_0$ with itself $n$ times. Then we have $H:\mathbb{R}^{2n}\to\mathbb{R}$ and $S$ a compact regular energy surface for $H$. Hofer-Zehnder then says, on page 19, that:
Since $dH\neq0$ on a neighborhood $U$ of $S$ we find a $(2n-1)$-form $\alpha$ on $U$ satisfying $$\Omega=dH\wedge\alpha.$$
How do we find this form? I'd expect it to end with $n-1$ $\omega_0$'s, so all we need is a 1-form $\gamma$ satisfying $\omega_0=dH\wedge\gamma$, but how do we find one? And is this guess correct?
In general, consider a one form $\beta = \sum_i \beta_i dx^i $ on $U\subset\mathbb R^m$. If
$$\| \beta\|^2 = \beta_1^2 + \cdots + \beta_m^2 \neq 0$$
on $U$, then the $(n-1)$-form
$$\alpha = \frac{1}{\|\beta\|^2} \sum_i (-1)^{i-1} \beta_i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots \wedge dx^m$$
on $U$ safisfies
$$\beta \wedge \alpha = dx^1 \wedge \cdots \wedge dx^m.$$