# Wedge Product of a Vector Space

I have the following problem: Let $V$ be a four-dimensional vectorial space over the field $\mathbb{R}$ or $\mathbb{C}$ with basis $\{e_1,e_2,e_3,e_4\}.$ Show that an element of $\bigwedge^2 V$ of the form $$\sum_{1 \leq i < j \leq 4} p_{ij}e_i \wedge e_j$$ can be written as $v \wedge w,$ with $v,w \in V,$ iff $$p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=0.$$ One implication is easy, but I can't prove the other one. Thank you to everybody!

• What implication have you done? Commented Mar 21, 2017 at 14:58
• That if an element of $\bigwedge^2 V$ of the form $\sum_{1 \leq i < j \leq 4} p_{ij}e_i \wedge e_j$ can be written as $v \wedge w,$ then $p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=0$ holds. Commented Mar 21, 2017 at 15:02

Writing $V^4=V^2\times V^2$ the canonical map $\nu: V^4\ \longrightarrow\ \bigwedge^4V$ restricts to alternating bilinear maps on $V^2\times\{0\}\cong V^2$ and $\{0\}\times V^2\cong V^2$, yielding a bilinear map $$\bigwedge^2V\times\bigwedge^2V\ \longrightarrow\ \bigwedge^4V:\ (v_1\wedge w_1,v_2\wedge w_2)\ \longmapsto\ v_1\wedge w_1\wedge v_2\wedge w_2.$$ It is alternating because $\nu$ is, which means that for every $x\in\bigwedge^2V$ you have $x\wedge x=0$.
What does this tell you when $x=\sum_{1\leq i<j\leq 4}p_{ij}e_i\wedge e_j$?