0
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $F$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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