antisymmetric bilinear map and cross product Given an antisymmetric bilinear map, $g:\mathbb{R}^3\times\mathbb{R}^3 \rightarrow V$, where $V$ is a vector space, is it true that there is a function $f:\mathbb{R}^3\rightarrow V$ such that $g(u,v) = f(u\times v)\ \forall\ u,v\in\mathbb{R}^3$, $\times$ being the cross product?
There is a result that resembles this in Harley Flanders' Differential Forms book on page 7 for $p$-forms, but without proof.
 A: The universal property guarantees that any bilinear map $$h : \Bbb R^3 \times \Bbb R^3 \to V$$ factors through a linear map $$h' : \Bbb R^3 \otimes \Bbb R^3 \to V ,$$ so that $h({\bf x}, {\bf y}) = h'({\bf x} \otimes {\bf y})$.
If $h$ is alternating, then $h'({\bf x} \otimes {\bf x}) = h({\bf x}, {\bf x}) = {\bf 0}$ for all ${\bf x} \in \Bbb R^3$, so $h'$ descends to a map $\tilde h : (\Bbb R^3 \otimes \Bbb R^3) / I \to V$, where $I$ is the subspace spanned by the tensors ${\bf x} \otimes {\bf x}$, ${\bf x} \in \Bbb R^3$, but by definition we can identify $(\Bbb R^3 \otimes \Bbb R^3) / I$ with $\bigwedge^2 \Bbb R^3$ and so regard $\tilde h$ as a map $$\tilde h : {\bigwedge}^2 \, \Bbb R^3 \to V .$$
Now, consider the special case where $V = \Bbb R^3$ and $h$ is the cross product map $C: \Bbb R^3 \otimes \Bbb R^3 \to \Bbb R^3$. Here, the map $\tilde C$ is an isomorphism: For an oriented, orthonormal basis $(E_a)$ of $\Bbb R^3$ we have (up to an overall positive constant that depends on convention) that $\tilde C(E_2 \wedge E_3) = E_1$ and its cyclic permutations. Unwinding definitions, we have $$g = \tilde g \circ \tilde C^{-1} \circ C ,$$ or equivalently that $$g(u, v) = (\tilde g \circ \tilde C^{-1}) (u \times v) .$$


We can make this more concrete: If we write $u = \sum_{i = 1}^3 u_i E_i$ and $v = \sum_{i = 1}^3 v_i E_i$ we have (using antisymmetry of $h$) that
$$h(u, v) = (u_2 v_3 - u_3 v_2) h(E_2, E_3) + (\textrm{cyclic permutations}) .$$ But we recognize the coefficients from the usual formula
$$u \times v = (u_2 v_3 - u_3 v_2) E_1 + (\textrm{cyclic permutations}),$$
so the desired map $f : \Bbb R^3 \to V$ satisfying $g(u, v) = f(u \times v)$ is characterized by
$$f(E_1) = h(E_2, E_3)$$
and its cyclic permutations.
