Q1) Let $n=3$ and let $$\omega =F_1dx_1+F_2dx_2+f_3dx_3.$$ Then $$d \omega =K_1 dx_2\wedge dx_3+K_2 dx_3\wedge dx_1+K_3 dx_1\wedge dx_2,$$ where $(K_1,K_2,K_3)=\operatorname{Curl}(F)$. Therefore $d\omega $ gives again the operation of the curling.

What does it mean ? I don't really understand. Because the curling is indeed a differential operator but is not a differential form.

Q2) Same if $\omega $ is a $p-$form we have for $f\in \mathcal C^1$ that $$d(f\omega )=df\wedge \omega +f(d\omega ),$$ which is equivalent to $$\nabla (fg)=f\nabla g+g\nabla f$$ if $g\in \mathcal C^1$. Again, le gradient is a differential operator, what is the connection with differential form ?

  • $\begingroup$ If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $d\omega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$\star(d\omega) = K_1\,dx_1 + K_2\,dx_2 + K_3\,dx_3.$$ $\endgroup$ – Ted Shifrin Aug 28 '18 at 21:31
  • $\begingroup$ No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $f\vec F$. $\endgroup$ – Ted Shifrin Aug 29 '18 at 6:21

Well, you've already written the answer yourself: if you take the coefficients of the differential forms $\omega$ and $d\omega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.

  • $\begingroup$ ok, but $d\omega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ? $\endgroup$ – MathBeginner Aug 26 '18 at 15:39
  • $\begingroup$ I don't think I really understand what you mean, but $d\omega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ... $\endgroup$ – Hans Lundmark Aug 26 '18 at 15:53
  • $\begingroup$ @HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :) $\endgroup$ – Ted Shifrin Aug 28 '18 at 21:30
  • $\begingroup$ @TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl” was supposed to mean. $\endgroup$ – Hans Lundmark Aug 29 '18 at 6:03

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