# Showing a form is exact

1. Aug 8, 2005

### 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.

Last edited: Aug 9, 2005
2. Aug 9, 2005

### HallsofIvy

Staff Emeritus
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.

3. Aug 9, 2005

### 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.