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

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

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