If $\omega$ is a non-vanishing 1-form on $\mathbb R^2$, then for any a point $p\in \mathbb R^2$, can we find an open neighborhood $U$ of $p$ and two functions $f,g$ on $U$ such that $\omega=fdg$ on $U$?
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Note this is essentially the same question (if you bring $f$ to the other side) as asking whether you can find an "integrating factor" for a differential equation. – Oct 18 '13 at 03:50
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Consider the equation $d(h\omega) = 0$. It means that $dh \wedge \omega =0$. If we write $\omega = adx + bdy$, this means that $b\partial h/\partial x - a\partial h /\partial y = 0$. Can you find a nonvanishing function $h$ satisfying this equation?
Provided you can, this means that the form $h\omega$ is exact, so by the Poincaré lemma, it is equal to $dg$ for some $g$. Then you'll have shown that $\omega = h^{-1} dg$.
Bruno Joyal
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Shouldn't $d(h\omega)=0$ be equivalent to $dh\wedge \omega +h d\omega =0$? My plan is to set $f=\text {log}h$, then the equation is $df \wedge \omega=-d\omega$. WLOG we can assume $\omega=dx+ady$, then the equation becomes $a \partial f/\partial x- \partial f/\partial y=- \partial a/\partial x$. But I cannot find a solution to this. – Summer Oct 18 '13 at 18:09