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From Munkres "Analysis on Manifolds" Consider the form $ \omega = xydx + 3dy -yzdz $. Check by direct computation that $ d(d\omega) = 0 $. Can someone show me how to do it, because I don't seem to be getting how to compute these differentials...

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If $\omega= f(x,y)dx+g(x,y)dy$, can you compute $d\omega$ in function of the partial derivatives of $f$ and $g$? The idea is the same for three variables. – Davide Giraudo Nov 27 '12 at 20:59
If $\omega = \sum f_idx_i$, then $d\omega = \sum (df_i)\wedge dx_i$. – Neal Nov 27 '12 at 21:25
Ok, I got it. I was a little confused with variables, but I think I finally understood it. – Ormi Nov 27 '12 at 21:32
You can answer your question. – Davide Giraudo Nov 27 '12 at 21:49

Differentiating means exactly what you do with usual functions. The only additional rule is: $d^2x=d^2y=d^2z=0$. Therefore, $dx\wedge dy=-dy\wedge dx$, etc.

$$ d\omega = d(xy\,dx) + 3\,d^2y - d(yz\,dz) \\ = dx\wedge y\,dx +x\,dy\wedge dx + 0 - dy\wedge z\,dz - y\,dz^2 =\\ = 0 -x\,dx\wedge dy - z\,dy\wedge dz - 0=\\ = -x\,dx\wedge dy - z\,dy\wedge dz.$$

$$ d^2\omega = -dx\wedge dx\wedge dy - dz\wedge dy\wedge dz = 0+0=0.$$

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