Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
    
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
1  
You can answer your question. –  Davide Giraudo Nov 27 '12 at 21:49

1 Answer 1

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.

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.