Let $\omega_1$ be a $k$-form and $\omega_2$ be an $l$-form, both defined in an open subset $U\subset \mathbf R^3$. Let $d:\wedge^k(U)\rightarrow\wedge^{k+1}(U)$ be the exterior derivative of differential forms. Show that $d$ is a linear transformation of vector spaces.

I have the formula: $$ d\omega=\sum df_{i_1...i_n}\wedge dx_{i_1}\wedge...\wedge dx_{i_n} $$

It's my understanding that to do this I need to show $$ d(\omega_1+\omega_2)=d(\omega_1)+d(\omega_2) $$ and $$ d(k\omega_1)=k*d(\omega_1) $$ But the first property is false right below is a given property of the exterior derivative: $$ d(\omega_1+\omega_2)=d(\omega_1)\wedge\omega_2+(-1)^k\omega_1\wedge d(\omega_2) $$

  • $\begingroup$ The left hand side of the last equation should be $d(\omega_1\land\omega_2)$, so it doesn't contradict linearity. $\endgroup$ – Berci Mar 31 at 22:41
  • $\begingroup$ ohhh okay yeah I had that wrong and it was causing so much confusion $\endgroup$ – joseph Mar 31 at 22:50
  • $\begingroup$ Are those the only two properties I need to show it is a linear transformation $\endgroup$ – joseph Mar 31 at 22:51
  • $\begingroup$ Yes. But differentiation is a linear operator. $\endgroup$ – Berci Mar 31 at 22:55
  • $\begingroup$ I still don't quite understand how to represent $\omega_1+\omega_2$ $\endgroup$ – joseph Apr 1 at 20:00

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