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Differential K-Forms

  1. Dec 6, 2011 #1
    This is more a concept question than an actual homework problem but why is it more or less meaningless to define differential k-forms on Rn in this case of k > n?

    I'm not really sure why this is so, but I might guess that since two dxis are the same in the product of such terms, then that whole term equals 0. If that's the case, is a k-form with k > n just the same as a k-form with k=n?
  2. jcsd
  3. Dec 6, 2011 #2


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    Yes, that is the case but no, a k-form with k> n is not "just the same as a k-form with k= n". What is true is that there are NO k-forms with k> n. In fact, one can show that in n-dimensional space, there exist [itex]\begin{pmatrix}n \\ k\end{pmatrix}[/itex] k-forms.

    For example, if n= 1, there exist one 0-form, the empty form, and 1 1-form, dx. If n= 2, there exist one 0-form, two 1 forms, dx and dy, and one 2 form, dxdy. If n= 3, there exist one 0-form, three 1 forms, dx, dy, and dz, three 2-forms dxdy, dydz, and dzdx, and one 3-form, dxdydz.
  4. Dec 6, 2011 #3
    Okay. Could you please show that there exist [itex]\begin{pmatrix}n \\ k\end{pmatrix}[/itex] k-forms, or otherwise direct me to the proof of this?
  5. Dec 6, 2011 #4
    So I found this quote that seems to explain what I'm asking about:

    The only problem is that I don't know what "deg xI > n" means. What, specifically, is "deg"?

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