Proving Exactness of a Differential Form with Poincaré's Lemma

[jeanf]

let f: R^n \rightarrow R^n be differentiable, with \sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0. show that \omega = \sum_{i=1}^n f_i dx_1 \Lambda ... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n is exact


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here's what i got so far:

\omega is a n-1 form, since the "^" on the dx_i indicates that this term is omitted. f is defined on R^n which is an open, star-shaped set. because of this, we can use poincare's lemma, which states that on an open starshaped set, every closed form is exact. therefore we only need to show that \omega is closed. that is, dw = 0.

when i write out dw, i get

dw = \sum_{i=1}^n \sum_{\alpha = 1}^n \frac{ \partial{f_i}}{\partial{x_\alpha}} dx_\alpha \Lambda dx_1 \Lambda... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n

how do i simplify this? how do i make this equal zero? if \sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0, does this mean that the expression i wrote for dw equals zero? i don't really see the connection here, since \sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} has an extra factor of (-1)^i...i'm not sure about the subscripts either. i hope someone can help me with this.

 

[HallsofIvy]

let f: R^n \rightarrow R^n be differentiable, with \sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0. show that \omega = \sum_{i=1}^n f_i dx_1 \Lambda ... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n

What is it you want to show? You can't show that \omega = \sum_{i=1}^n f_i dx_1 \Lambda ... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n without knowing how \omega is defined! You haven't said what \omega is. If, on the other hand, \omega = \sum_{i=1}^n f_i dx_1 \Lambda ... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n is the definition of \omega then I don't know what it is you want to show. You seem to be saying that you want to prove that \omega is exact.

 

[jeanf]

yes, the question was to show that \omega is exact - sorry, i forgot to type that in my original post. i have edited it above. thanks.