Prove the exterior derivative is a linear transformation of vector spaces.

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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: $$ d(\omega_1+\omega_2)=d(\omega_1)\wedge\omega_2+(-1)^k\omega_1\wedge d(\omega_2) $$