Let $V$ be a three dimensional euclidian vector space which is oriented. Because of the orientation, we can define the cross product $\times: V^2 \rightarrow V$ uniquely by: $<v\times w,u> = \det(v,w,u)$. Now suppose we have another real vector space $W$ and let $c:V^2\rightarrow W$ be an alternating bilinear form. Intuitively, $\times$ should have some kind of universal property, namely that we get a unique $\phi_c:V\rightarrow W$, such that $\phi_c(\times) = c$.
Intuitively this makes sense, but how would one go of proving this? I'm struggling to see which notable parts of $c$ to use to construct $\phi$. Note that this is part of an assignment, so hints rather than solutions would be appreciated.
Thank you!
There is another universal property that is related and much more well known, which is the universal property of the exterior algebra. In particular, there is a universal property for each $\Lambda^k(V)$, and the one we need is the following:
Universal Property of the Exterior Product: Let $V$ be a vector space. Given any alternating bilinear map $\mu: V\times V\rightarrow W$, there exists a uniquely induced linear map $\tilde{\mu}:\Lambda^2(V)\rightarrow W$ such that $\tilde{\mu}\circ\wedge = \mu$, where $\wedge:V\times V \rightarrow \Lambda^2(V)$ is the exterior product map.
The exterior product is intimately related to the cross product. You should be able to translate this universal property into the one that you need.