Antisymmetric Bilinear Forms and Wedge Products

Question

I want to show that every antisymmetric bilinear form on $\mathbb{R^3}$ is a wedge product of two vectors. In other words, suppose we have a basis $\vec{u}$, $\vec{v}$, and $\vec{w}$ for $\mathbb{R}^{3}$. Let $F:\mathbb{R^3}\times\mathbb{R^3}\rightarrow\mathbb{R}$ be an antisymmetric bilinear form on $\mathbb{R^3}$. How would I show that there exists covectors $\alpha$ and $\beta$ such that (for scalars $a$, $b$, $c$, $d$, $e$, $f$), $$\alpha\wedge\beta(a\vec{u}+b\vec{v}+c\vec{w}, d\vec{u}+e\vec{v}+f\vec{w})=F(a\vec{u}+b\vec{v}+c\vec{w},d\vec{u}+e\vec{v}+f\vec{w})$$

Answer 1

I'm going to give you a clue firstly. The point is that $F$ has rank two, i.e., $\operatorname{dim}\operatorname{im} F =2 $. Pick a basis for $\operatorname{im}F$, say $\{a_1,a_2\}$. Now choose another vector $v$ such that $\{a_1,a_2,v\}$ is a basis for $\mathbb R^3$ (this always can be done) and expand $F$ in terms of this basis and see what happen. As a second hint, use the fact that the vector $v$ ins in the kernel of $A$: $A(v)=0$.

Remark. Your statement isn't true in general: a $p$- form can't be written as a product of $p$ $1$-forms. However, as Peter Szekeres (Mathematical Physics 2004) shows in an exercise, if $V$ is an $n$-dimensional vector spaces, every $n-1$-form can be expanded in such a way. And this is your case precisely for $n=3$.