Differential K-Forms on Rn: Meaningless w/ k>n?

[TranscendArcu]

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 dx_is 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?

 

[HallsofIvy]

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 \begin{pmatrix}n \\ k\end{pmatrix} 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.

 

[TranscendArcu]

Okay. Could you please show that there exist \begin{pmatrix}n \\ k\end{pmatrix} k-forms, or otherwise direct me to the proof of this?

 

[TranscendArcu]

So I found this quote that seems to explain what I'm asking about:

There are no nonzero differential forms of degree > n on an open subset of Rn. This is because is deg x_I > n, then in the expression dx_I at least two of the 1-forms dx_i must be the same, force dx_I = 0.

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

Thanks!