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

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_addictedMay 10 '16 at 1:36

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aaron Maroja, Semiclassical, Shailesh, choco_addicted
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• In its current form, this question lacks context and therefore may be closed. Please consider improving your question. – Michael Albanese May 9 '16 at 21:01
• The question seems kind of self-explanatory -- I don't see what more context would be necessary. – Chill2Macht May 10 '16 at 1:33
• @William Well, consider the tags of the question. The first tag is "calculus". This would suggest that one is talking about de Rham forms, I guess? But the second tag is "exterior-algebra". So are we talking about exterior algebras (the assumption that the answer used), maybe? Which is it? This is the kind of vagueness that would be eliminated by a tiny bit of context. It's not asking much! Even a link to Wikipedia would be a good start! – Najib Idrissi May 10 '16 at 18:28
• Oh to be fair I didn't look at the tags, but I can see now how they would be misleading. I had just assumed it was talking about differential forms but hadn't really thought through the terminological ambiguity. – Chill2Macht May 10 '16 at 18:46
• @NajibIdrissi: Would the difference in context (purely differential vs. purely algebraic) change anything? No, the proof would be the same: $(\alpha \wedge \alpha) (u, v) = \frac 1 2 (\alpha (u) \alpha (v) - \alpha (v) \alpha (u)) = 0$. After all, the exterior differential algebra is an exterior algebra. Let's not be picky just because we can. – Alex M. May 10 '16 at 19:16

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.