How would I work out $d(\omega F)$ where $\omega$ is a 1-form and $F$ is a vector valued function?

I know that $d(f \omega) = df \wedge \omega + f d\omega$, for a smooth function $f$. I suppose this would also hold for a vector valued function, so $d(F \omega) = dF \wedge \omega + F d\omega$.

Is there a way of computing $d( \omega F)$ from this result? Would it be the same as $d( F \omega)$?

up vote 2 down vote accepted

Let $F$ be a vector-valued form then $F = (F_1,F_2, \dots , F_n)$ where $F_1,F_2, \dots , F_n$ are functions. The definition of $dF$ is that we exterior differentiate component-wise, $dF = (dF_1,dF_2, \dots , dF_n)$. By definition, $$ \omega F = (\omega F_1, \omega F_2, \dots , \omega F_n)$$ and so, $$ d(\omega F) = (d(\omega F_1), d(\omega F_2), \dots , d(\omega F_n)).$$ Next, simply apply the usual product rule for forms on each component, $$ d(\omega F) = (d\omega \wedge F_1-\omega dF_1, d\omega \wedge F_2-\omega dF_2, \dots , d\omega \wedge F_n-\omega dF_n).$$ Finally, factor out the one-form $\omega$ etc.. to find: $$ d(\omega F) = d\omega \wedge F - \omega \wedge dF.$$ Some of these $\wedge$'s are at times omitted in texts. I think you can see that the same rule holds for $d(F\omega)$ assuming you define right-multiplication on the vectors of forms.

  • actually, for $d(F\omega)$ we find $d(F\omega) = dF \wedge \omega + F \wedge d\omega$ since there is no minus in the form-product rule when we hit the function first with $d$. The order of $F$ and $\omega$ matters in this regard. – James S. Cook Apr 26 '14 at 3:27

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