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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. –  Ted Oct 3 '12 at 7:40
    
@Ted, thanks, should be $j$. –  AsinglePANCAKE Oct 3 '12 at 7:43
    
@AsinglePANCAKE I've had a quick look at it and I think this is true only if $f_{i}(\vec{x})$ is changed to $f_{i}(x_{i})$. –  in_wolfram_we_trust Oct 3 '12 at 8:43
    
@in_wolfram_we_trust: Mind elaborating? I'm not sure I really see what you're saying.. –  AsinglePANCAKE Oct 3 '12 at 9:10
    
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. –  in_wolfram_we_trust Oct 3 '12 at 9:18
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1 Answer

up vote 3 down vote accepted

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.

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