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?

  • $\begingroup$ The question in your title is the opposite of the question in the body of your thread. Mariano is responding to the latter. $\endgroup$ – Ryan Budney Oct 6 '10 at 2:25

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

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    $\begingroup$ 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 $\endgroup$ – JimJones Oct 6 '10 at 2:18
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    $\begingroup$ @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... $\endgroup$ – Pete L. Clark Oct 6 '10 at 2:44
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    $\begingroup$ @Mariano, awesome answer $\endgroup$ – BBischof Oct 6 '10 at 2:44
  • $\begingroup$ I still just don't understand $\endgroup$ – JimJones Oct 6 '10 at 12:51
  • $\begingroup$ @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.) $\endgroup$ – Agustí Roig Oct 6 '10 at 12:52

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