Why is the wedge product of a 1-form and itself $0$? Why doesn't this apply to 2-forms?
closed as off-topic by Aaron Maroja, Semiclassical, Leucippus, Shailesh, choco_addicted May 10 '16 at 1:36
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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 does not apply. Indeed, for any finite-dimensional vector space $\Bbb V$ of dimension $\geq 4$, 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 $\alpha, \beta$.
On a vector space of dimension $<4$ the wedge product of any form with itself is zero.