I am trying to understand how to create a 3-form from 1 triple wedge product of 1-forms.
I understand that $\alpha \wedge \beta = \alpha \otimes \beta - \beta\otimes\alpha$ where the outer product, $\otimes$, is dropped for brevity later.
In creating the triple wedge product we get:
$\alpha\wedge\beta\wedge\gamma = \alpha(\beta\gamma-\gamma\beta)+\beta(\gamma\alpha-\alpha\gamma)+\gamma(\alpha\beta-\beta\alpha)$
How is this triple wedge product made? I'm trying to prove it but quickly run into the following:
$\alpha\wedge\beta\wedge\gamma = (\alpha\wedge\beta)\gamma-\gamma(\alpha\wedge\beta)= (\alpha\beta-\beta\alpha)\gamma-\gamma(\alpha\beta-\beta\alpha)$
Am I missing an important step? Where can I go from here?
The formula $\alpha\wedge\beta=\alpha\otimes\beta-\beta\otimes\alpha$ is only true when $\alpha$ and $\beta$ are $1$-forms. The more general formula for a $k$-form $\mu$ and a $l$-form $\nu$ is $$ \mu\wedge\nu=\frac{(k+l)!}{k!l!}\operatorname{Alt}(\mu\otimes\nu)\ \ \ \ \ \mu\in\Lambda^kV^*,\nu\in\Lambda^lV^* $$ Here $\operatorname{Alt}$ denotes the antisymmetrization operator. $$ \operatorname{Alt}(\omega)(v_1,\cdots,v_m)=\frac{1}{m!}\sum_{\sigma\in S_m}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)},\cdots,v_{\sigma(m)})\ \ \ \ \ \omega\in T_mV $$ Where $S_m$ is the set of permutations of $(1,\cdots,m)$ and $\operatorname{sgn}$ is the sign of a permutation.
It might be worth noting that some authors choose to define the wedge product of alternating tensors without the prefactor $\frac{(k+l)!}{k!l!}$. This is just a matter of convention, but one to be mindful of. In this alternate convenntion, the product of $1$-forms would be $\alpha\wedge\beta=(\alpha\otimes\beta-\beta\otimes\alpha)/2$.