Let $\alpha$ be a 1 -form and $\beta$ a 2 -form on $\mathbb{R}^{3}$. Then $$ \begin{array}{l} \alpha=a_{1} d x^{1}+a_{2} d x^{2}+a_{3} d x^{3} \\ \beta=b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2} \end{array} $$ Compute $\alpha \wedge \beta$.
solution : $\alpha \wedge \beta=a_{1} d x^{1}\wedge(b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2})+a_{2} d x^{2}\wedge(b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2})+a_{3} d x^{3}\wedge(b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2})=a_{1}b_{1} d x^{1}\wedge d x^{2} \wedge d x^{3}- a_{2}b_{2} d x^{1}\wedge d x^{2} \wedge d x^{3}+a_{3}b_{3} d x^{1}\wedge d x^{2} \wedge d x^{3}=(a_{1}b_{1}-a_{2}b_{2}+a_{3}b_{3})d x^{1}\wedge d x^{2} \wedge d x^{3}$
is this true ?
i think we have :
$a_{2}b_{2} d x^{2}\wedge d x^{3} \wedge d x^{1}= a_{2}b_{2} (d x^{2}\wedge d x^{3} )\wedge d x^{1}=-a_{2}b_{2} d x^{1}\wedge (d x^{2} \wedge d x^{3})$
is this true ?
Note that $dx^2\wedge dx^3\wedge dx^1 = - dx^2\wedge dx^1\wedge dx^3 = dx^1\wedge dx^2\wedge dx^3$. In general, if $\eta$ is a $k$-form and $\zeta$ is an $l$-form, then $\eta\wedge\zeta = (-1)^{kl}\zeta\wedge\eta$.
So the correct answer is $\alpha\wedge\beta = (a_1b_1 + a_2b_2 + a_3b_3)dx^1\wedge dx^2\wedge dx^3$.