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I'm attempting to show that $dg(\vec{x})=\alpha$, where $\alpha=\Sigma^n_{i=1}f_idx_i$ and $g(\vec x)={1\over{p+1}}\Sigma_{i=1}^nx_if_i(\vec x)$...and $d$ is the exterior derivative. The $f_i$ are all smooth and homogeneous of the same ($p\neq-1)$) degree, i.e. $f_i(t\vec x)=t^pf_i(\vec x)$.

This latter fact gives us that $\large\Sigma_{j=1}^nx_j{\partial f_i\over \partial x_j}(\vec x)=pf_i(\vec x)$.

I don't know whether it's all the summations involved or what, but I can't seem to get this to work out. It so happens that I always have the j's and i's mixed in the wrong way...This might indicate that the question isn't true as stated...or more likely that I'm failing at some fairly basic bookkeeping.

In the end I keep ending up with $\large dg={1\over{p+1}}\Sigma^n_{j=1}\Sigma^n_{i=1}x_i{\partial f_i\over \partial x_j}dx_j+f_i$, which won't really allow me to use any of the nice identities I've earned.

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    Something is wrong with your equation after "This latter fact gives us..." because $i$ is the index of summation on the left side, yet appears as a free variable on the right side.2012-10-03
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    @Ted, thanks, should be $j$.2012-10-03
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    @AsinglePANCAKE I've had a quick look at it and I think this is true only if $f_{i}(\vec{x})$ is changed to $f_{i}(x_{i})$.2012-10-03
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    @in_wolfram_we_trust: Mind elaborating? I'm not sure I really see what you're saying..2012-10-03
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    Consider $f_{1} = x_{1}$ and $f_{2} = x_{1} + x_{2}$. If you calculate $\textrm{d}g$ you will find $\textrm{d}g \neq \alpha$. If you impose the condition that $f_{i}$ is only a function of $x_{i}$ you should find that your problem with "mixed up" $i$'s and $j$'s disappears.2012-10-03
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    @in_wolfram_we_trust Certainly, since we're claiming $\alpha$ is the exterior derivative of some other form, $\alpha$ must be exact, and so must be closed. Hence $d\alpha=0$, which doesn't hold for your counterexample...2012-10-03
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    We're not claiming $\alpha$ is exact, we're trying to show that it is, i.e. given some conditions on $f_{i}$ (smooth, homogeneous), will $\alpha$ be exact (and hence closed)? No, we have a counterexample. Choose $f_{i}$ as suggested, then $\alpha$ is neither exact, nor closed.2012-10-03
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    @in_wolfram_we_trust, I should probably have stated by hypothesis that $\alpha$ is closed, but if $\alpha$ is not closed, then no $g$ will work, let alone the one specified. So choosing an $\alpha$ that's not closed doesn't really tell us much about this particular g.2012-10-03
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    You're right of course, we have to take $\alpha$ closed. I guess this is a cautionary tale about taking a 'quick look' at things.2012-10-03

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Let $$\alpha = \sum_{i = 1}^{n} f_{i}\ \textrm{d}x_{i}$$ be a closed one-form with functions $f_{i}$ smooth and homogeneous of degree $p$.

Since $\alpha$ is closed, $\textrm{d}\alpha = 0$ and so $$\sum_{j = 1}^{n}\sum_{i = 1}^{n}\frac{\partial f_{i}}{\partial x_{j}} \textrm{d}x_{j}\wedge \textrm{d}x_{i} = 0, $$ $$\Rightarrow \frac{\partial f_{i}}{\partial x_{j}} = \frac{\partial f_{j}}{\partial x_{i}}.$$

Let $g$ be defined by $$g = \frac{1}{p+1} \sum_{i = 1}^{n}x_{i}f_{i}.$$

Then $$\textrm{d}g = \frac{1}{p+1}\left(\sum_{j=1}^{n}f_{j}\ \textrm{d}x_{j} + \left(\sum_{i=1}^{n} x_{i}\frac{\partial f_{i}}{\partial x_{j}} \right)\textrm{d}x_{j}\right).$$ Use the relationship we derived from the condition that $\alpha$ is closed to swap the indices inside the innermost sum: $$\textrm{d}g = \frac{1}{p+1}\left(\sum_{j=1}^{n}f_{j}\ \textrm{d}x_{j} + \left(\sum_{i=1}^{n} x_{i}\frac{\partial f_{j}}{\partial x_{i}} \right)\textrm{d}x_{j}\right).$$ Now use the identity you derived in your question (after "The latter fact...") $$\textrm{d}g = \frac{1}{p+1}\left(\sum_{j=1}^{n}f_{j}\ \textrm{d}x_{j} + p f_{j}\ \textrm{d}x_{j}\right),$$ $$\textrm{d}g = \sum_{j = 1}^{n} f_{j}\ \textrm{d}x_{j}$$ as required.