Wedge Product of Differential Forms on Euclidean Space

Let $n\in\mathbb{N}$ and let $k,\ell\in\mathbb{N}\cup\{0\}$. Let $U\subseteq \mathbb{R}^n$ be open. Let $\alpha$ be a differential $k$-form on $U$, and let $\beta$ be a differential $\ell$-form on $U$. The wedge product $\alpha\wedge\beta$ is the differential $(k+\ell)$-form on $U$ defined as follows: for each $x\in U$ and each collection of vectors $v_1,\dots,v_{k+\ell}\in\mathbb{R}^n$,

$$(\alpha\wedge\beta)_x(v_1,\dots,v_{k+\ell}) =\frac{1}{k!\,\ell!}\sum_{\sigma\in S_{k+\ell}} \operatorname{sgn}(\sigma)\, \alpha_x(v_{\sigma(1)},\dots,v_{\sigma(k)})\, \beta_x(v_{\sigma(k+1)},\dots,v_{\sigma(k+\ell)}),$$

where $S_{k+\ell}$ is the set from the definition of permutations, $\operatorname{sgn}(\sigma)$ is the sign from the sign definition, and one uses the convention $0!=1$.

If $r\in\mathbb{N}$ and $\omega_1,\dots,\omega_r$ are differential forms on $U$ such that each successive wedge product is defined, then

$$\omega_1\wedge\cdots\wedge\omega_r$$

means the left-associated iterated wedge product

$$(((\omega_1\wedge\omega_2)\wedge\omega_3)\wedge\cdots)\wedge\omega_r.$$