Why is the wedge product of a 1-form and itself $0$?

Question

Why is the wedge product of a 1-form and itself $0$? Why doesn't this apply to 2-forms?

Answer 1

By definition, the wedge product of $1$-forms $\alpha, \beta$ is $$\alpha \wedge \beta = \alpha \otimes \beta - \beta \otimes \alpha .$$ When $\beta = \alpha$, this is zero.

The wedge product of $2$-forms has a different formula, so this conclusion does not apply in that case. Indeed, for any finite-dimensional vector space $\Bbb V$ of dimension $\geq 4$ (and whose underlying field has characteristic not $2$), fix a cobasis $(e^a)$ of $\Bbb V$; the $2$-form $$\omega := e^1 \wedge e^2 + e^3 \wedge e^4$$ satisfies $\omega \wedge \omega = 2 e^1 \wedge e^2 \wedge e^3 \wedge e^4 \neq 0$. A useful fact is that a $2$-form $\zeta$ satisfies $\zeta \wedge \zeta = 0$ iff $\zeta$ is decomposable, that is, if it can be written as $\zeta = \alpha \wedge \beta$ for some $1$-forms $\alpha, \beta$ (this assertion again requires that $\operatorname{char} \Bbb F \neq 2$).

On a vector space of dimension $<4$ the wedge product of any form with itself is zero.