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I was under the impression that all 2-forms are the wedge $(\wedge)$ of two 1-forms. Is it possible to have a 2-form that you can't write as $A\wedge B$ with $A,B$ 1-forms?

Hamou
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JimJones
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  • The question in your title is the opposite of the question in the body of your thread. Mariano is responding to the latter. – Ryan Budney Oct 06 '10 at 02:25

1 Answers1

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Yes, it is possible. (And you should find an example yourself: I will not deprive you of the joy of finding it :) )

Mariano Suárez-Álvarez
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    Would dA^dB+dC^dD (in R^4) be an example? It's a two form but it can't be written as A^B – JimJones Oct 06 '10 at 02:18
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    @JimJones: your example works, but you should not be satisfied until you have a proof of this. Just because you try to write it as the wedge of two one-forms and fail is of course not enough... – Pete L. Clark Oct 06 '10 at 02:44
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    @Mariano, awesome answer – BBischof Oct 06 '10 at 02:44
  • I still just don't understand – JimJones Oct 06 '10 at 12:51
  • @JimJones. I assume that $A,B,C,D$ in your first formula are functions... (Maybe I could add to Mariano's answer: simply look at the definition of 2-forms and remember something about linear independence.) – Agustí Roig Oct 06 '10 at 12:52
  • A,B,C,D are suppose to be 1-forms... is that wrong? – JimJones Oct 06 '10 at 13:01
  • @JimJones: you don't have to understand: you have to compute. Find the general form of a $2$-form which is a wedge product of two $1$-forms, and next show that not all $2$-forms are of that form. – Mariano Suárez-Álvarez Oct 06 '10 at 13:02
  • The definition I was given in class is "a k-form is a function taking p to a k-linear alternating map on TR^n. These look like linear combinations of dx^1^dx^2^..." I was under the impression that the definition of a 2-form is the wedge of two one forms... that's why I'm lost – JimJones Oct 06 '10 at 13:03
  • If you think a 2-form is that then what you need is to read up a bit on what forms are :) – Mariano Suárez-Álvarez Oct 06 '10 at 13:35
  • @JimJones. If $A,B,C,D$ are 1-forms, then $dA, dB, dC, dD$ are 2-forms. Then $dA\wedge dB$ and $dC\wedge dD$ are 4-forms. – Agustí Roig Oct 06 '10 at 14:47
  • @JimJones. As for your "look like linear combinations": that's ok. And you are making linear combinations of which kind of vectors? Can you find a linear combination of $dx\wedge dy, dy\wedge dz, dz \wedge dx$ equal to zero -different from the trivial one? – Agustí Roig Oct 06 '10 at 14:50