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

John B
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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 09 '16 at 21:01
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    The question seems kind of self-explanatory -- I don't see what more context would be necessary. – Chill2Macht May 10 '16 at 01:33
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    @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
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    @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
  • @AlexM. Did you read Travis's answer? Try to make sense of it assuming that $\alpha$ and $\beta$ are differential forms; it doesn't. What you've written isn't completely rigorous either, since differential forms are sections of the exterior algebra on the cotangent bundle (not space)... The thing you've written only holds locally. So you see, knowing whether we're talking about a plain exterior algebra or differential forms matters. And in any case, there are standards for a minimum amount of context, regardless of whether the question is comprehensible. – Najib Idrissi May 10 '16 at 19:21
  • @AlexM. And hey, maybe it's a graded manifold, in which case the 1-form could have even degree and not square to zero! Or maybe we're working in characteristic 2, and then it depends on the precise definition of the exterior algebra. Who knows, the question doesn't say. – Najib Idrissi May 10 '16 at 19:24
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    @NajibIdrissi: I totally understand your point about the context, I just tend to be on the permissive side. Anyway, what I've written holds both on a vector space, and for global sections on a manifold. I don't use trivializations anywhere, think of $\alpha, u, v$ as being a global $1$-form and two vector fields. That definition of the wedge product is global, too. I agree about supermanifolds and characteristic $2$, but I feel that your argument is a bit stretched. – Alex M. May 10 '16 at 19:47

1 Answers1

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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 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.

Travis Willse
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