Consider a two-form $\gamma \in \Lambda^2(V)$ where $V$ is a real vector space. Now I would like to know the necessary and sufficient conditions for $\gamma$ to be expressible as an exterior product of two one-forms, $\gamma=\alpha \wedge \beta,\, \alpha,\beta \in \Lambda(V)$.


Obviously, a necessary condition for the decomposition to hold is that any exterior power of $\gamma$ vanishes, $\gamma\wedge...\wedge\gamma = 0$. But it is not clear to me whether this condition is sufficient.



What are the necessary and sufficient conditions for a two-form to be an exterior product of two one-forms?
While Plücker relations give the general theory, a direct answer to your question is the following: a 2-form $\gamma$ is decomposable (is a product of two 1-forms) iff $(\iota\_v \gamma) \wedge \gamma = 0$ for any vector $v$, where $\iota\_v \gamma$ is the usual contraction of a vector with a 2-form, that is, $(\iota\_v \gamma)(u) = \gamma(v,u)$ for any other vector $u$.


The proof is straightforward. Pick some vector $v$ such that $\alpha = \iota\_v \gamma \ne 0$. If it doesn't exist, then $\gamma = 0$, which is the trivial case. Then $\alpha \wedge \gamma = 0$ implies\* that there exists $\beta$ such that $\gamma = \alpha \wedge \beta$ and you are done.


\* If the implication is not obvious, extend $e\_1 = \alpha$ to a basis $e\_i$ in $\Lambda(V)$ and expand $\gamma$ in the induced basis $e\_i\wedge e\_j$. Clearly, $e\_1\wedge \gamma = 0$ iff the only non-zero coefficients in the expansion are in front of the terms $e\_1 \wedge e\_j$.


