Under what conditions can a general 2-form be written as a wedge product of two 1-form

Question

Assume we have a 2-form $\omega \in \Lambda^2\mathbb{R}^n$. It is usually stated one can write $$\omega = \alpha \wedge \beta,$$ with $\alpha, \beta \in \Lambda^1\mathbb{R}^n$ only for $n < 4$. How to prove this statement?

Answer 1

For $\omega\in\wedge^k V$, the minimal number $s$ such that $\omega$ has an expression $$ \omega=\omega^{(1)}+\dotsb+\omega^{(s)} $$ where each $\omega^{(i)}$ is of the form $$ \omega^{(i)}=\omega_1^{(i)}\wedge\dotsb\wedge\omega_k^{(s)} $$ is called the rank of $\omega$.

Now, if $\{e_1,\dotsc,e_n\}$ is a basis for $V$, then every $\omega\in\wedge^2 V$ has a unique expression $$ \omega=\sum_{i,j}a_{ij} \,e_i\wedge e_j $$ where $a_{ij}=-a_{ji}$. A general fact from linear algebra states that the rank of the skew-symmetric matrix $[a_{ij}]$ is twice the rank of $\omega$.

Thus you wish to prove that every $\omega\in\wedge^2\Bbb R^n$ has rank one if and only if $n<4$. To do so, note that the matrix $[a_{ij}]$ is an $n\times n$ skew-symmetric matrix. Since every $n\times n$ skew-symmetric matrix with $n<4$ is rank zero or rank two it is clear that the rank of $\omega$ is one if $n<4$. Then note that for $n\geq 4$ any skew-symmetric matrix with rank greater than two provides a two-form with rank greater than one.

See more about the rank of a $k$-vector:

https://en.wikipedia.org/wiki/Exterior_algebra#Rank_of_a_k-vector

Answer 2

If $\omega=\alpha\wedge\beta$, that is $\omega$ is decomposable, then $\omega\wedge\omega=0$.

For $n\leq 3$ the wedge product of any two form with itself vanishes, so we have chances to prove that any 2-form is decomposable. For $n\geq 4$ you can check that if $\omega=e_1\wedge e_2+ e_3\wedge e_n$ you have $\omega\wedge \omega\neq 0$, hence there are forms which are non decomposable.

Let us now prove decomposability. For $n=0,1$ there is nothing to prove, and for $n=2$ you have dim($\Lambda^2(\mathbb{R}^2))=2$ hence any 2-form is a multiple of $e_1\wedge e_2$.

For $n=3$ if $\omega\wedge\omega=0$ we have contracting with some vector $v$ $0=\iota_v\ (\omega\wedge\omega)= 2\iota_v\ \omega\wedge\omega$ which implies $\omega=\alpha\wedge\iota_v\ \omega$ for some 1-form $\alpha$.