Rays on the null space of the quadratic form given by wedge product on $\wedge^2(\mathbb{R}^4)$ are oriented 2-planes

Question

Let $U$ be a real 4-dimensional vector space. On the second exterior power, $\wedge^2(U)$ consider the quadratic form given by the wedge product $\wedge^2(U) \times \wedge^2(U) \rightarrow \wedge^4(U)$ composed with the isomorphism $\wedge^4(U)\cong \mathbb{R}$ given by fixing a basis (one non-zero vector) of $\wedge^4(U)$. In the book `The Geometry of Four-Manifolds', on page 8, it is stated that the rays in the null cone of this quadratic form have the geometric interpretation of being oriented 2-planes in $U$. I don't understand this statement.

Answer 1

Let $V$ be a vector space of dimension 4 over the field $F$ with characteristic not 2. Let $0\neq x \in \wedge^2 V$ be such that $q(x)=x \wedge x=0$. Then \begin{equation} x= v_1 \wedge v_2 \end{equation} with $v_1$, $v_2$ in $V$. Therefore $x$ is in correspondence with the plane determined by $(v_1,v_2)$.

We explain now why $x$ is a decomposable element.

Take $y=v_1\wedge v_2 + v_3 \wedge v_4$ in $\wedge^2 V$. Where the $v_i$ are chosen from a basis of $V$. If all the $v_i$ are different then $y \wedge y \neq 0$. If $v_i = v_j$ for any $i \neq j$, then we can write $y$ as $v_i \wedge a$ ($y$ is not $0$ if we are not in characteristic $2$) and therefore $y \wedge y = 0$. When three of the $v_i$ are equal $y=0$ (by assumption $x\neq 0$). So when $y$ is a sum of 2 decomposable elements then the above implication holds.

If $y = v_1\wedge v_2 + v_3 \wedge v_4 + \lambda v_5 \wedge v_6$, with $\lambda \in F$, then at least 2 of the $v_i$ are equal and we can reduce it to a sum of two decomposable elements. So by a prove on induction we can assume that $y$ is a sum of two decomposable elements and the implication follows.