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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  • $\begingroup$ In its current form, this question lacks context and therefore may be closed. Please consider improving your question. $\endgroup$ – Michael Albanese May 9 '16 at 21:01
  • $\begingroup$ The question seems kind of self-explanatory -- I don't see what more context would be necessary. $\endgroup$ – Chill2Macht May 10 '16 at 1:33
  • 1
    $\begingroup$ @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! $\endgroup$ – Najib Idrissi May 10 '16 at 18:28
  • $\begingroup$ 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. $\endgroup$ – Chill2Macht May 10 '16 at 18:46
  • $\begingroup$ @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. $\endgroup$ – 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.


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